text:
\begin{document}

\title{A Generalized Framework for Simultaneous Long-Short Feedback Trading}

\author{Joseph~D.~O'Brien}
\author{Mark~E.~Burke}
\author{Kevin~Burke}
\affil{MACSI, Department of Mathematics and Statistics, University of Limerick, Ireland}
\date{}

\maketitle

\begin{abstract}
	We present a generalization of the Simultaneous Long-Short (SLS) trading strategy described in recent control literature wherein we allow for different parameters across the short and long sides of the controller; we refer to this new strategy as Generalized SLS (GSLS). Furthermore, we investigate the conditions under which positive gain can be assured within the GSLS setup for both deterministic stock price evolution and geometric Brownian motion. In contrast to existing literature in this area (which places little emphasis on the practical application of SLS strategies), we suggest optimization procedures for selecting the control parameters based on historical data, and we extensively test these procedures across a large number of real stock price trajectories (495 in total). We find that the implementation of such optimization procedures greatly improves the performance compared with fixing control parameters, and, indeed, the GSLS strategy outperforms the simpler SLS strategy in general.
\end{abstract}

\section{Introduction\label{sec:intro}}

The use of feedback in a control-theoretic scenario has been well studied within a variety of different fields -- its use dates back at least two millennia where the flow of water was regulated to improve the accuracy of water clocks. The application of feedback models became widespread during the Industrial Revolution and since then their use has become ubiquitous in engineered systems \cite{kumar2014control}. In recent years, the methodology has been applied in the setting of equity trading wherein a closed-loop feedback system is used to modulate the investment level, $I(t)$, in response to changes in the equity price, $p(t)$. As this basic system reacts to changes in price, rather than making predictions about future price movements, the resulting strategy is often described as ``model free''.

The initial framework was originally developed from a purely long investment perspective \cite{barmish2008trading}, but was later extended so that one takes both long and short positions in the equity -- the so-called ``simultaneous long-short'' (SLS) strategy \cite{barmish2011performance}.  The key feature of the SLS strategy is that the resulting gain function, $g(t)$, is guaranteed to be positive under (potentially restrictive) assumptions of deterministic $p(t)$ (albeit this is not assumed to be known), continuous trading, perfect liquidity, and the absence of transaction fees (for more details see \cite{barmish2011performance}). Further developments in this area included the consideration of interest rates and collateral requirements \cite{barmish2011arbitrage, barmish2012market, malekpour2013stock, barmish2015stock}. This work culminated in \cite{barmish2016new} which laid the foundations for many future research directions such as using a controller with delay \cite{malekpour2016stock}, different price process models \cite{baumann2017simultaneously, baumann2017stock}, time varying price evolution parameters \cite{primbs2017robustness} and also the use of a proportional-integral (PI) controller rather than the proportional controller which was used in the original SLS strategy \cite{malekpour2018generalization}.

While various extensions to the SLS framework have been proposed to date, the basic underlying model structure has remained essentially unchanged in the sense that the initial investment and feedback parameters are the same for both the long and short investments. However, a real trader may wish to tune these components of the controller in different ways. We also note that in the current literature, although much theoretical work has been done (mainly on investigating the positive gain property under different scenarios), testing such SLS strategies in practical situations has been much more limited. Specifically, testing has been typically carried out on a very small number of stock price series (one or two), and, hence, there is very little sense of the general performance of these control-based strategies. Furthermore, there has also been a lack of guidance on how one should select the feedback parameter in practice, where, apparently, this choice has been quite arbitrary in applications shown to date. However, with the level of gain being quite modest in some applications (as was mentioned in \cite{primbs2017robustness, malekpour2018generalization}), one wonders whether or not greater gains can be made through a more objective selection process, e.g., by optimizing some criterion.

The aim of the current article is to tackle the issues raised in the previous paragraph by:
\begin{enumerate}[a)]
\item generalizing SLS to allow parameters which differ across the short and long side -- the resulting strategy we call ''Generalized SLS'' (GSLS), thereby permitting greater flexibility beyond the classical SLS approach,
\item proposing an optimization procedure for selecting control parameters based on historical data, providing an objective process for their selection, and
\item extensively analyzing the performance of GSLS (and SLS) with our proposed optimizer on a large number (495) of real stock price trajectories to determine the practical usefulness of the control-based trading concept.
\end{enumerate}
The rest of this article is set out as follows. In Section \ref{sec:SLS} we introduce the classical SLS strategy of \cite{barmish2016new}, after which, in Section \ref{GSLS}, we extend to our proposed GSLS strategy, deriving a number of new analytical results. Section \ref{Optim} provides a suggestion on how one might objectively choose control parameters within GSLS (and, hence, SLS and other varieties thereof) which is then tested on historical data in Section \ref{sec:testing}. Finally, we conclude with some discussion in Section \ref{sec:discussion}.

\section{Simultaneous Long-Short (SLS) Strategy\label{sec:SLS}}

As is customary, we assume that the strategy is applied in the setting of an ``idealized market'' where the primary assumptions are: (i) continuous-time trading is possible, (ii) the equity is perfectly liquid so that shares can be purchased with no gap between the bid and ask prices, and (iii) there are no transaction fees or interest rates. See \cite{barmish2011performance} for further details.

We now consider the key components of the SLS strategy. Let $p(t)$ represent the price of the equity at time $t \geq 0$. Furthermore, let $I(t)$ be the level of investment at time $t$, and $g(t)$ be the gain function, i.e., the cumulative trading gain over the period $[0,t]$ where $g(t) < 0$ represents an overall trading loss; also note that, by definition, $g(0) = 0$. In the simpler long strategy of \cite{barmish2008trading} (i.e., no short component), the amount invested is given by

\begin{equation*}
I(t) = I_0 + Kg(t),
\end{equation*}
where $K \ge 0$ is the feedback parameter and $I_0>0$ is the initial investment. The gain made on a given stock is the amount invested in this stock multiplied by the percentage change in stock price, giving the incremental gain equation

\begin{equation*}
dg = \frac{dp}{p}I = \frac{dp}{p}(I_0 + Kg)
\end{equation*}
whose solution yields

\begin{equation*}
g(t) = \frac{I_0}{K}\left\{q(t)^K - 1\right\},
\end{equation*}
where $q(t) = p(t)/p(0) > 0$ is the ratio of the current and initial stock price, and, hence, we see that the investment level is $I(t) = I_0 q(t)^K$.

The SLS strategy extends the above (but follows along the same lines) by introducing simultaneous long and short investments, $I_L(t)$ and $I_S(t)$ respectively, and their associated cumulative gain functions, $g_L(t)$ and $g_S(t)$. The strategy is defined by

\begin{align*}
I_L(t) &= I_0 + Kg_L(t), \\
I_S(t) &= -I_0 - Kg_S(t),
\end{align*}
where

\begin{align*}
g_L(t) &= \frac{I_0}{K}\left\{q(t)^K - 1\right\},\\
g_S(t) &= \frac{I_0}{K}\left\{q(t)^{-K} - 1\right\}.
\end{align*}

Note that the overall investment is $I(t) = I_L(t) + I_S(t) = K\{g_L(t) - g_S(t)\}$ and, in particular, that $I(0) = 0$,  i.e., the simultaneous long and short positions (which are opposite and equal when $t=0$ with magnitude $I_0$) lead to an initial net zero investment position. As shown in \cite{barmish2011performance}, under the idealized market, the cumulative gain at time $t$ for this SLS strategy is

\begin{equation}
\label{totalgain}
g(t) = g_L(t) + g_S(t) = \frac{I_0}{K}\left\{q(t)^{K} + q(t)^{-K} -2\right\},
\end{equation}
and, hence, $g(t) > 0$ provided that  $q(t) \ne 1$, i.e., this strategy always makes a profit except when $p(t) = p(0)$.

\section{Generalized SLS Strategy\label{GSLS}}

As seen in Section \ref{sec:SLS}, the basic SLS strategy is composed of the simpler long strategy with the addition of a short component. In particular, the short component mirrors the long component in the sense that it shares the same initial investment, $I_0$, and feedback parameter, $K$. However, we now view this as an interesting special case of a more general framework in which the long and short components have distinct parameters as follows:
\begin{align*}
I_L(t) &= I_0 + K g_L(t), \\
I_S(t) &= -\alpha I_0 -\beta K g_S(t),
\end{align*}
where $\alpha,\beta>0$, and
\begin{align*}
g_L(t) &= \frac{I_0}{K}\left\{q(t)^K - 1\right\},\\
g_S(t) &= \frac{\alpha I_0}{\beta K}\left\{q(t)^{-\beta K} - 1\right\}.
\end{align*}
We refer to this as the Generalized SLS (GSLS) strategy, within which $\alpha = \beta = 1$ corresponds to SLS, and whose overall gain function is
\begin{equation}
g(t) = \frac{I_0}{K} \left[q(t)^K - 1 + \frac{\alpha}{\beta}\left\{q(t)^{-\beta K} - 1\right\}\right]
\label{ComboEq}
\end{equation}
Notwithstanding the guaranteed positive gain property of the important SLS special case, in practice, this specific strategy cannot uniformly yield the optimal gain. We will discuss this in the following sections.

\subsection{GSLS with $\beta=1$\label{sec:gslsb1}}

Before investigating the GSLS strategy, it is instructive to first consider the special case where $\beta=1$, i.e., the case where the long and short initial investments differ, but the feedback parameter is the same in both components. In this specific case note that the gain is given by
\begin{equation}
g(t) = \frac{I_0}{K} \left[q(t)^K - 1 + \alpha\left\{q(t)^{-K} - 1\right\}\right],
\label{beta1}
\end{equation}
where we see that $g(t) = 0$ when $q(t) = 1$ and when
$q(t) = \alpha^{1/K}$. Furthermore, viewed as a function of $q(t)$, the gain function has a single turning point (a global minimum of $-I_0(\alpha^{1/2}-1)^2/K \le 0$) at $q(t) = \alpha^{1/(2K)}$ which lies between the two roots. Note that in the classic SLS case, where $\alpha=1$, these three points coincide to become a single root and global minimum of zero at $q(t) = 1$, yielding the associated positive gain property.

Assume that $\alpha < 1$, so that we invest in the short component to a lesser extent than the long component. In this case, $g(t) > 0$ provided that $q(t) \not\in [\alpha^{1/K}, 1]$. This can be seen as a trade-off between risk and reward. First consider the classic SLS strategy where $\alpha = 1$. This is, theoretically, risk free in the sense that $g(t) > 0$ provided that $q(t) \ne 1$. On the other hand, in selecting $\alpha < 1$ and, hence, investing more on the long side, we are essentially admitting a belief that the stock price will increase in the future, i.e., $q(t) > 1$. Indeed, it is easy to show that $g_{\alpha<1}(t) > g_{\alpha=1}(t)$ when the price rises ($q(t) > 1$) so that the reward goes up by choosing $\alpha < 1$ -- but, of course, the risk also goes up since there is a chance that $g(t) < 0$.

Now consider a more standard trading strategy (i.e., not control-based) where we simply go long, i.e., buy at time zero and sell at time $t$. In this case, the investment is fixed at $I(t) = I_0$ and $g(t) = I_0\left\{q(t) - 1\right\}$. In a similar manner to GSLS with $\alpha < 1$, we are anticipating a price rise. However, we are guaranteed to make a loss if $q(t) < 1$, whereas, with GSLS, we can still profit if $q(t) < \alpha^{1/K}$. In other words, we are reducing the risk compared with simply going long. Furthermore, although with GSLS the risk is reduced relative to going long, the gain is not uniformly lower (for all control parameter values) even when $q(t) > 1$ as demonstrated by Figure \ref{fig:gslsvlong}.

\begin{figure}[h]
\centering
\begin{tikzpicture}[scale = 1]
\begin{axis}[
axis lines = center,
xlabel=$q$,ylabel=$g$,

y label style={at={(axis description cs:-0.1,1)},anchor=north},
x label style={at={(axis description cs:1,0.05)},anchor=east},
legend style={at={(0.1,1.0)},
	anchor=north west,legend columns=1},
domain=0.5:2,

legend style = {nodes={scale=0.5, transform shape}}]
\addplot[samples = 100, blue, smooth, thick] {(1/1)*(x^1 - 1 + (0.5/1)*((x)^(-1*1) - 1))};
\addplot[samples = 100, blue, dashed, thick] {(1/3)*(x^3 - 1 + (0.5/1)*((x)^(-1*3) - 1))};
\addplot[samples = 100, red, smooth, thick] {(1/1)*(x^1 - 1 + (1/1)*((x)^(-1*1) - 1))};
\addplot[samples = 100, red, dashed, thick] {(1/3)*(x^3 - 1 + (1/1)*((x)^(-1*3) - 1))};
\addplot[samples = 100, green, smooth, thick] {x-1};
\legend{$\alpha=0.5,\ \beta=1,\ K= 1$\\$\alpha=0.5,\ \beta=1\, K= 3$\\$\alpha=\beta=1,\ K= 1$\\$\alpha=\beta=1,\ K= 3$\\ ``going long''\\};
\end{axis}
\end{tikzpicture}
	\caption{Plots of the gain for various values of $q(t)$ when the trading strategy is GSLS ($\alpha=0.5$, $\beta=1$) (blue), SLS (red), and simply going long (green). Furthermore, for GSLS and SLS both $K=1$ (solid) and $K=3$ (dashed) cases are shown.\label{fig:gslsvlong}}
\end{figure}

The above discussion relates to the case where $\alpha < 1$. When $\alpha > 1$, $g(t) > 0$ provided that $q(t) \not\in [1, \alpha^{1/K}]$, and, moreover, we can show that $g_{\alpha>1}(t) > g_{\alpha=1}(t)$ when the price falls ($q(t) < 1$). Therefore, in general, the choice of $\alpha=1$ is sub-optimal in the sense that, if one possesses knowledge on the likely direction of the stock price, a greater return can be made by choosing $\alpha \ne 1$. In practical situations, a trader is likely to require some input into his/her trading strategy such that GSLS may be more attractive than SLS. Furthermore, GSLS may be seen as lying somewhere between SLS and standard trading in that trader knowledge can be incorporated as in standard trading, but where the risk is reduced as in SLS. Interestingly, since $\alpha^{1/K} \rightarrow 1$ as $K\rightarrow \infty$, the condition for positive gain when $\alpha \ne 1$ is simply that $q(t) \ne 1$ when $K$ is large. This suggests that a large enough $K$ value can overcome a poor $\alpha$ choice; indeed, GSLS with large $K$ essentially behaves as SLS.

\subsection{Positive Gain in GSLS}

In the previous section we discussed GSLS under the restriction that $\beta = 1$. We now consider the more general scenario where $\beta$ is not necessarily equal to $1$. In particular, we investigate the conditions for positive gain within GSLS.

\begin{theorem}
	In the idealized market, the gain made by the GSLS strategy is positive $\forall K,\beta$ provided that \mbox{$(1-\alpha)\ln\{q(t)\} \ge 0$}.
\end{theorem}

{\bf Proof}:
First we write the gain function, (\ref{ComboEq}), as
\begin{equation}
g(t) = I_0 \frac{q(t)^K - 1}{K} + \alpha I_0  \frac{q(t)^{-\beta K}-1}{\beta K}. \label{eq:gainfn2}
\end{equation}
Now, consider the inequality
\begin{equation}
q^x \geq 1 + x \ln q \label{eq:ineq}
\end{equation}
for $q > 0$ and $x \in \mathbb{R}$, which is due to the fact that the function $m(x) = q^x-1-x \ln q$ is convex with global minimum of zero at $x = 0$.

Then, from (\ref{eq:ineq}), setting $x=K$ yields
\begin{align}
\frac{q^K - 1}{K} &~\ge~ \ln q \label{eq:ineqgL},
\end{align}
while setting $x=-\beta K$ yields
\begin{align}
\frac{q^{-\beta K}-1}{\beta K} &~\ge~ - \ln q.\label{eq:ineqgS}
\end{align}
Hence, combining (\ref{eq:gainfn2}), (\ref{eq:ineqgL}), and (\ref{eq:ineqgS}), we get
\begin{equation*}
g(t) \ge I_0 (1 - \alpha) \ln\{q(t)\}
\end{equation*}
and, since $I_0 > 0$, we require that $(1 - \alpha) \ln\{q(t)\} \ge 0$ to ensure that $g(t) \ge 0$. $\square$

The above result generalizes the positive gain result of SLS which we recover by setting $\alpha=\beta=1$. Furthermore, note, in fact, that the result does not depend on the value of $\beta$ (provided that it is finite) so that positive gain is assured once $\alpha=1$ even if the feedback parameters on the short and long side differ. The requirement that $(1-\alpha)\ln\{q(t)\} \ge 0$ can be interpreted as follows: if we expect $q(t) > 1$, we should set $\alpha < 1$, and if we expect $q(t) < 1$, we should set $\alpha > 1$. It is also worth noting that, for large $K$, the behaviour of $g(t)$ is as follows:
\begin{align*}
g(t) \sim \left\{
\begin{array}{lr}
I_0 q(t)^K \ln\{q(t)\}, & q(t) > 1 \\[0.2cm]
-\alpha I_0 q(t)^{-\beta K} \ln\{q(t)\}, & q(t) < 1 \\
\end{array}
\right.
\end{align*}
which is positive in both cases. This suggests that a sufficiently large value of $K$ can yield positive gain irrespective of $\alpha$ and $\beta$. These findings mirror those of Section \ref{sec:gslsb1} which were for GSLS with $\beta=1$. Indeed GSLS with $\beta = 1$ is qualitatively similar to the general case.

\begin{theorem}
The GSLS gain function is increasing in $K$ and $\beta$.
\end{theorem}
{\bf Proof}:
In this case we write the gain function, (\ref{ComboEq}), as
\begin{equation}
g(t) = I_0 \frac{q(t)^{K_L} - 1}{K_L} + \alpha I_0  \frac{q(t)^{-K_S}-1}{K_S}. \label{eq:gainfn3}
\end{equation}
where $K_L$ and $K_S$ are the feedback parameters on the long and short sides respectively, and the incremental gain equation is given by
\begin{equation}
dg = \frac{\partial g}{\partial K_L} dK_L + \frac{\partial g}{\partial K_S} dK_S, \label{eq:ig}
\end{equation}
where
\begin{align*}
\frac{\partial g}{\partial K_L} &= I_0 \frac{q^{K_L}(K_L \ln q -1)+1}{K_L^2}, \\[0.2cm]
\frac{\partial g}{\partial K_S} &= \alpha I_0 \frac{q^{-K_S}(-K_S \ln q -1)+1}{K_S^2}. \\
\end{align*}
By replacing $x$ with $-x$ in (\ref{eq:ineq}) and multiplying by $q^x$, we can establish another inequality,
\begin{equation}
q^x(x \ln q - 1) + 1 \geq 0  \label{eq:ineq2}.
\end{equation}
Using (\ref{eq:ineq2}) with $x=K_L$ and $x=-K_S$, respectively, gives $\partial g / \partial K_L \ge 0$ and  $\partial g / \partial K_S \ge 0$. Therefore, the incremental gain, (\ref{eq:ig}), is positive whenever $dK_L$ and $dK_S$ are positive, so that $g(t)$ is increasing in $K_L$ and $K_S$, and, hence, in $K$ and $\beta$ (since $K_L = K$ and $K_S = \beta K$).$\square$

While the above result suggests that a trader can simply increase the feedback parameters to increase profits, this may not be feasible in practice. Firstly, large feedback parameters will cause the controllers (and associated investments) to vary wildly in response to (potentially small) changes in gains which introduces a large degree of variability into the system. Moreover, one of either the long or short investments will become large, and, of course, all traders will have limited resources; for the same reason, one cannot simply increase $I_0$ arbitrarily.

The basic results of this section can be seen in Figure \ref{fig:gainK} for some specific parameter values.

\begin{figure}[h]
\centering
\begin{tikzpicture}[scale = 1]
\begin{axis}[
axis lines = center,
xlabel=$K$,ylabel=$g$,

y label style={at={(axis description cs:0.55,1)},anchor=north},
x label style={at={(axis description cs:1,0.35)},anchor=east},
legend style={at={(0.65,1.0)},
	anchor=north west,legend columns=1},
domain=-3:3,
legend pos  = north west,
legend style = {nodes={scale=0.5, transform shape}}]
\addplot[samples = 100, blue, smooth, thick] {(1/x)*((0.5)^x + (0.25/1)*(0.5)^(-1*x) - (0.25/1) - 1)};
\addplot[samples = 100, yellow, smooth, thick] {(1/x)*((2)^x + (0.25/1)*(2)^(-1*x) - (0.25/1) - 1)};
\addplot[samples = 100, red, smooth, thick] {(1/x)*((0.5)^x + (2/1)*(0.5)^(-1*x) - (2/1) - 1)};
\addplot[samples = 100, green, smooth, thick] {(1/x)*((2)^x + (2/1)*(2)^(-1*x) - (2/1) - 1)};
\legend{$\alpha=0.25,\ q= 0.5$\\$\alpha=0.25,\ q= 2$\\$\alpha=2,\ q= 0.5$\\$\alpha=2,\ q= 2$\\};
\node[label={},circle,fill,inner sep=2pt, blue] at (axis cs:2,0) {};
\node[label={},circle,fill,inner sep=2pt, green] at (axis cs:1,0) {};
\node[label={},circle,fill,inner sep=2pt, red] at (axis cs:-1,0) {};
\node[label={},circle,fill,inner sep=2pt, yellow] at (axis cs:-2,0) {};
\end{axis}
\end{tikzpicture}
	\caption{GSLS gain viewed as a function of $K$ with $\beta = 1$, $I_0 = 1$. The dots are given at $K = \log_{q(t)} \alpha$ which is a root of (\ref{beta1}). The cases where this root lies in the negative $K$ region satisfy the requirement that $(1-\alpha)\ln\{q(t)\} \ge 0$. In these cases, positive gain is assured for all $K$ values since negative $K$ values are infeasible.\label{fig:gainK}}
\end{figure}

\subsection{GSLS under Brownian Motion Price Evolution\label{sec:GBM}}

We now show that the GSLS strategy is robust in the case where the price evolution of the equity is determined by Geometric Brownian Motion (GBM) such that
\begin{equation*}
\frac{dp}{p} = \mu \ dt + \sigma \ dW
\end{equation*}
where $W$ represents a Weiner process, $\mu$ is the drift and $\sigma$ is the volatility \cite{hull2016options},\cite{oksendal2003stochastic}. Our proof follows along similar lines to that of \cite{barmish2016new}.

As in \cite{barmish2016new}, the gain made by the long controller is
\begin{equation*}
g_L(t) = \frac{I_0}{K}\left\{q(t)^{K}e^{\frac{1}{2}\sigma^2(K-K^2)t} - 1\right\}
\end{equation*}
and, similarly, the gain made by the short controller is
\begin{equation*}
g_S(t) = \frac{\alpha I_0}{\beta K}\left\{q(t)^{-\beta K}e^{\frac{1}{2}\sigma^2(\beta K-(\beta K)^2)t} - 1\right\}
\end{equation*}
and, hence, the total gain is
\setlength{\arraycolsep}{0.0em}
\begin{eqnarray*}
g(t)&{}={}&\frac{I_0}{K}\left[ q(t)^{K}e^{\frac{1}{2}\sigma^2(K-K^2)t} - 1 \right.\nonumber\\
&&\left. + \frac{\alpha}{\beta}\left\{q(t)^{-\beta K}e^{\frac{1}{2}\sigma^2(\beta K-(\beta K)^2)t} - 1\right\}\right].
\end{eqnarray*}
\setlength{\arraycolsep}{5pt}
The expected gain can then be derived by noting that the $k$th moment of a log-normally distributed random variable $X$ with $\log(X) \sim \mathcal{N}\Big(\big(\mu - \frac{\sigma ^2}{2}\big)t, \ \ \sigma^2 t\Big)$ is given by
\begin{equation*}
\mathbb{E}\left[X^k\right] \ = \ e^{\big(k\mu - \frac{\sigma ^2}{2}\big)t + \frac{1}{2}k^2 \sigma^2 t}.
\end{equation*}
Using this result, it can be shown that
\begin{equation}
\mathbb{E}\left\{g(t)\right\} \ = \frac{I_0}{K}\left\{e^{K\mu t} - 1 +\frac{\alpha}{\beta}\left(e^{-\beta K \mu t} - 1\right) \right\}.
\label{Eg}
\end{equation}
As the expected gain function under GBM, (\ref{Eg}), is of the same form as the gain function under deterministic price evolution, (\ref{ComboEq}), with $q(t) = e^{\mu t}$. Theorem 1 follows immediately and, hence, positive expected gain follows when $(1 - \alpha) \mu \ge 0$. Similarly, Theorem 2 applies so that the expected gain increases in $K$ and $\beta$.

Note that the variance of the gains may also be calculated and it is given by
\begin{align}
\mathrm{Var}\left\{g(t)\right\} &=
\mathbb{E}\left\{g(t)^2\right\} - \left[\mathbb{E}\left\{g(t)\right\}\right]^2 \notag \\
&= \frac{I_0^2}{K^2}\left\{e^{\ 2K\mu t}\left(e^{\ \sigma^2 K^2 t} - 1\right) +\frac{\alpha}{\beta}e^{\ -\beta K \mu t}\Big(e^{\ \beta K^2 \sigma^2 t} \right. \notag \\
& \left. \qquad \qquad -1\Big)\left(\frac{\alpha}{\beta}e^{\ -\beta K \mu t} + 2 e^{\ K\mu t - \beta K^2 \sigma^2 t}\right)\right\}.
\label{VarG}
\end{align}
We also note that these results extend to the time-varying price dynamics case described in \cite{primbs2017robustness}.

\section{Selection of Parameters\label{Optim}}

While results such as those in the previous section provide insight into the behaviour of feedback-based strategies (especially the conditions leading to positive gain), they fall short of yielding a practical implementation in the sense of suggesting values of the control parameters in real applications. Therefore, in this section, we focus on possible criteria that one may optimize in order to select the control parameters within GSLS (and, hence, also SLS).

As a first step we will make the assumption that the price evolution process can be modelled by GBM since this assumption yields explicit solutions for both $\mathbb{E}\left\{g(t)\right\}$ and $\mathrm{Var}\left\{g(t)\right\}$, and our suggested criteria will be based on these quantities; of course other price evolution models could be used. Now, let $g^*_t$ be some prespecified target gain for time-point $t$, and define the trading ``bias'' as
\begin{equation}
\text{bias}\left\{g(t)\right\} = \mathbb{E}\left\{g(t)\right\} - g^*_{t}\label{eq:bias}
\end{equation}
which is the expected difference between the realized gain, $g(t)$, and the target gain, $g^*_t$. Under the GBM assumption, this quantity will depend on the parameter $\mu$ as well as the control parameters, $\alpha$, $\beta$, and $K$. Therefore, as a first step, the GBM parameters can be estimated using standard inference procedures \cite{ait2002maximum}, and, following this, the control parameters can be selected by minimizing, $\left[\text{bias}\left\{g(t)\right\}\right]^2$, i.e., these are the control parameters which minimize the difference between $\mathbb{E}\left\{g(t)\right\} \text{and} \ g^*_{t}$.

The suggestion above does not take account of the volatility of the stock price, as evidenced by the fact that the objective function does not depend on $\sigma$. Thus, an alternative criterion would involve both the so-called bias and the variability in gains. We therefore propose the trading ``mean squared error'' (MSE) as
\begin{equation}
\text{MSE}\left\{g(t)\right\} = \mathbb{E}\left\{g(t) - g^*_t\right\}^2 = \left[\text{bias}\left\{g(t)\right\}\right]^2 + \text{Var}\left\{g(t)\right\}\label{eq:mse}
\end{equation}
which depends on both $\sigma$ (via the variance term) in addition to $\mu$, and can be minimized with respect to the control parameters. Such control parameters might result in $\mathbb{E}\left\{g(t)\right\} < g^*_t$, but where the lower variation in gains justifies the choice, i.e., control parameters selected via $\text{MSE}\left\{g(t)\right\}$ will, in any given run, tend be closer to the target gain due to the lower variability in gains.

\section{Testing GSLS with Optimized Control Parameters\label{sec:testing}}

\subsection{Simulation Study\label{sec:sim}}

Before testing our proposed methods on real data, we first carry out a simulation study. We simulated stock prices according to GBM with drift $\mu = 0.1$, and a range of different volatility values, $\sigma = \{0.05, 0.1, 0.2\}$. Since $I_0$ simply plays the role of scaling up and down the gain, we fix $I_0 = 1$, and optimize both (\ref{eq:bias}) and (\ref{eq:mse}) with respect to the control parameters wherein the known values of the GBM parameters are inserted prior to this optimization. Furthermore, we assume that, after one year of trading (over 252 days), the trader is aiming for a 15\% return, i.e, $g^*_1 = 0.15$, so that he/she beats the market drift ($\mu=0.1$) by 5 percentage points.

Although the two objective functions can be optimized using standard optimization algorithms, since there are only three parameters $(K, \ \alpha, \ \beta)$, we performed a discrete grid search over the parameter space. In particular, we used 10 equally spaced values for each parameter with $K \in [0, 5]$, $\alpha \in [0, 5]$, and $\beta \in [0, 5]$ yielding 1000 parameter combinations. For each of the 3 simulation scenarios, we found the optimal control parameters according to both bias and MSE minimization, and then applied the resulting strategy over 252 trading days for each of 1000 simulated GBM trajectories.

\begin{figure*}
\subfloat[Bias Optimizer\label{sfig:BiasDensity}]{
	\includegraphics[width=.49\linewidth]{plots/Bias_density}
}\hfill
\subfloat[{MSE Optimizer}\label{sfig:MSEDensity}]{
	\includegraphics[width=.49\linewidth]{plots/MSE_density}
}
\caption{Density of gains over 1000 simulated GBM stock prices over one year where control parameters were selected on the basis of (a) bias and (b) MSE. The target gain here is $g^*_t = 0.15$ and is shown by the dashed black line. \label{fig:biasmse}}
\label{}
\end{figure*}

Figure \ref{fig:biasmse} shows the results based on optimizing both bias and MSE in terms of a density plot of the realized gains over each of the 1000 simulation replicates. We can see from Figure \ref{BiasDensity} that optimizing the bias does not appear to perform well as the volatility increases, with most of the mass being on negative gain for the larger volatility value, $\sigma = 0.2$. However, this is perhaps to be expected since this approach does not take account of the volatility. In contrast, minimizing MSE (see Figure \ref{MSEDensity}) yields more consistent results in terms of the target gain even as the volatility parameter increases (albeit the variability of the gain naturally increases with the volatility). This suggests that parameter selection using MSE as an objective criterion may be preferable than using bias.

\subsection{S\&P 500 Stock Prices}

As mentioned, in the existing literature, the degree to which feedback-based trading strategies have been tested on real data has been somewhat limited, i.e., the number of stock price series has been very small, and the choice of control parameters has been apparently quite arbitrary. Thus, more extensive testing is required if real-world traders are to be convinced that the adoption of such strategies can be fruitful in general.

With the above goal in mind, we consider the daily closing prices for 495 members of the Standard and Poor's 500 (S\&P 500) index over the course of two-year period, January 2016 - December 2017 (of the current 500 members, it was not possible to obtain complete data for the time period in question in 5 cases which were, therefore, omitted).
The justification for choosing this data is based on the fact that the members of the S\&P 500 are the largest companies traded in the United States and eligibility is based on a number of factors including market capitalization\footnote{Market capitalization of \$5.3bn or more.}, liquidity\footnote{Annual dollar value traded to float-adjusted market capitalization should be greater than 1, and the stock should trade a minimum of 250,000 shares in each of the six months leading up to the evaluation date.}, and the company being domiciled\footnote{The company must be a U.S company as defined by the US Index Committee.} \cite{SP500}. Thus, these members have highly liquid stocks so that feedback-based trading strategies could plausibly be implemented in a real-world scenario, and, moreover, the members of this index are intended to provide a reasonably good representation of the stock market as a whole.
Figure \ref{SP500} shows the median adjusted closing price calculated over all 495 members on each day over the two-year period, along with 2.5\%, and 97.5\% quantiles; it is clear that the majority of the stock prices have increased over time.

\begin{figure}
	\centering
	\includegraphics[width = 0.5\textwidth]{plots/SP_quantiles}
	\caption{The 2.5\% (bottom, dashed), 50\% (middle, solid), and 97.5\% (top, dashed) quantiles of closing prices for 495 members of the S\&P 500 over the period January 2016 - December 2017.}
	\label{SP500}
\end{figure}

For each of the stock price series, we take the year 2016 to be the training period for which the GBM parameters are estimated using maximum likelihood \cite{ait2002maximum}, and then, based on the estimated GBM parameters, we select the GSLS control parameters via the bias and MSE optimization approaches described in Section \ref{Optim} with a target gain of $g^* = 0.15$ for each stock. Following Section \ref{sec:sim}, when optimizing GSLS control parameters, we use a grid search over 1000 parameter combinations resulting from 10 equally spaced values for each parameter, $K \in [0, 5]$, $\alpha \in [0, 5]$, and $\beta \in [0, 5]$. The optimized trading strategy (either bias and MSE) for that stock price series is then applied over the course of the year 2017, i.e., the testing period, where a discrete-time version of the strategy is implemented (cf.~\cite{barmish2016new}). While the procedure as described is carried out separately on each stock price series, we will summarize results by way of aggregating over all series. Note: in all of our testing, we fix $I_0=1$ since this simply scales the gains.

Before applying any optimized strategies, we first test the classical SLS strategy in a similar way to previous literature, i.e., a value for the control parameter is simply selected and tested; we use $K = \{1,2,3,4,5\}$. The advantage here, however, is that we are testing over a much larger sample of price series than that of previous literature. So that the results are comparable with the case where we carry out optimization, we test these five strategies in the year 2017 only (2016 is not used to provide insight since we are simply fixing $K$ from the offset). Figure \ref{SLSgains} displays the average gain on each trading day (average was computed over the 495 stock series). We can see that the average gain increases with the value of $K$, however the trajectory of the gain is quite erratic over time; most of the growth appears as a sharp jump occurring in the last c.50 days.

\begin{figure}[h!]
	\centering
	\includegraphics[width = 0.5\textwidth]{plots/SLS_gain}
	\caption{Average gain (computed over all stock series) on each trading day in 2017 from the SLS strategy with various $K$ values.}
	\label{SLSgains}
\end{figure}

We then applied each of the 1000 GSLS parameter combinations to all stocks in the year 2017 (again without using 2016 to optimize), and computed the average gain over all stocks at the end of the year; the results are visualized in Figure \ref{ABK} along with the five values from the five SLS strategies mentioned in the previous paragraph. Interestingly, there are a large number of GSLS strategies which lead to a loss. These mainly correspond to cases where $\alpha > 1$ which is not a surprise since most of these stock prices are increasing on average over time as evidenced by Figure \ref{SP500}. Note also that, while the classic SLS case does outperform these particular combinations, there are clearly many other GSLS cases which outperform SLS.

\begin{figure}[h!]
	\centering
	\includegraphics[width = 0.5\textwidth]{plots/ABKg_mean}
	\caption{Average gain (computed over all stocks) after 252 days of trading in 2017 for each of 1000 GSLS strategies. The colour of the points represent the $\beta$ value, while their size represents the $\alpha$ value. Also shown are five SLS strategies (the dark coloured points).}
	\label{ABK}
\end{figure}

The tests described above (both SLS and GSLS) mimic those of previous literature, i.e., we have simply set the control parameters at the start of 2017 without drawing insight from historical data. In other words, the parameters could be thought of as essentially randomly selected, and are not tuned on a per-stock basis. It is unlikely that a real-world trader would adopt such a strategy from which he/she is quite detached. We, therefore, implement our optimization approach for each stock based on the 2016 data (as described above), and apply the optimized strategies to the 2017 data. Figure \ref{total_gains} shows the average gain using optimized strategies (both bias and MSE) over each trading day, where we find that the MSE approach performs better than the bias approach, and gets quite close to the target, $g^* = 0.15$. It is interesting to note that, in both cases, the gain increases reasonably steadily over time (in contrast to those of Figure \ref{SLSgains}).
\begin{figure}[h!]
	\centering
\includegraphics[width = 0.5\textwidth]{plots/mean_gain}
	\caption{Average gain (computed over all stocks) on each trading day in 2016 for optimized strategies. Both the bias-optimized (red) and MSE-optimized (black) strategies were based on the data from 2016. A horizontal green reference line highlights the target gain, $g^* = 0.15$.}
	\label{total_gains}
\end{figure}

To get a sense of the variability in gains, Figure \ref{gain_quantiles} displays the 2.5\% and 97.5\% quantiles for the gains over all the stocks. Unsurprisingly, the level of variability associated with the bias-optimized strategies is higher than that of the MSE-optimized strategies. Furthermore, note that the median gain is essentially zero for the bias-optimized strategies, whereas, for the MSE-optimized strategies, it is very close to the average gain observed in Figure \ref{total_gains}. Clearly, from the quantiles shown, the distribution of gains is somewhat skewed for the bias-optimized strategies with some ''lucky'' cases having done very well over the period in question. However, the performance of the MSE approach is more consistent which is in line with the findings of Section \ref{sec:sim}.

\begin{figure}[h!]
	\centering
\includegraphics[width = 0.5\textwidth]{plots/optim_results}
	\caption{Median gain (computed over all stocks) for both the bias-optimized (red, solid) and MSE-optimized (black, solid) strategies, along with 2.5\% and 97.5\% quantiles (dashed). A horizontal green reference line highlights the target gain, $g^* = 0.15$.}
	\label{gain_quantiles}
\end{figure}

\begin{figure}[h!]
	\centering
\includegraphics[width = 0.5\textwidth]{plots/mean_DiffTargetG}
	\caption{Average gain (computed over all stocks) on each trading day in 2016 for optimized strategies with $g^* = |\mu| + 0.05$. Both the bias-optimized (red) and MSE-optimized (black) strategies were based on the data from 2016. Although the target gain changes here on a stock-by-stock basis, we show a horizontal green reference line at $g = 0.15$ for comparison with earlier results.}
	\label{mean_DiffTargetG}
\end{figure}

In the optimization procedures discussed above, the target gain was fixed at 15\% for all stocks. However, the GBM parameters could also be used in order to determine a reasonable target gain, for example, a target of 15\% might be an unobtainable level of gain for some of the stocks in question, or, indeed, an underestimate of the potential gain for other stocks. Thus, one might consider varying the target on a stock-by-stock basis. A simple suggestion in this direction would be to set the target gain, for stocks $i=1,\ldots,495$, at $g_{t,i}^* = |\hat\mu_i| + C$ where $\hat\mu_i$ is the estimated GBM drift parameter and $C$ is some constant which describes the amount by which we wish to beat the inherent drift in the price evolution. The reason we take the absolute value of the drift is that profit can also be made in the case of negative drift via the short component. Figure \ref{mean_DiffTargetG} displays the average gain over time with $C = 0.05$. While the average gain in the MSE-optimized case does not change much from that of Figure \ref{total_gains}, the average gain for the bias-optimized case is almost doubled -- and is much closer to that of the MSE case. This finding is perhaps to be expected since the bias optimization is solely chasing a target, whereas the MSE approach also takes account of variability.

\begin{table*}[htb]
	\centering
	\caption{Summary of gains over all stock at the end of the year 2017 using different strategies}
	\label{Gtable}
	\renewcommand{\arraystretch}{1.3}
	\resizebox{\textwidth}{!}{\begin{tabular}{|cr|rr|rr|rrr|rr|rr|}

		\hline
& Strategy		& \multicolumn{4}{|c|}{\textbf{GSLS}} & \multicolumn{7}{|c|}{\textbf{SLS}}  \\
\hline
                                 & Parameters   & MSE    & MSE     & Bias   & Bias   & $K=1$    & $K=2$    & $K=5$    & MSE      & MSE      & Bias    & Bias    \\
\hline
                                 & Target       & Fixed  & Varied  & Fixed  & Varied & ---      & ---      & ---      & Fixed    & Varied   & Fixed   & Varied  \\
\hline

\multirow{3}{*}{Summary} &		1st Quartile    &  0.004 &  0.006  & -0.158 & -0.154 & -0.029   & -0.057   & -0.131   & -0.008   & -0.016   & -0.083 & -0.114  \\
\multirow{3}{*}{of}      &            Median    &  0.155 &  0.160  & -0.006 &  0.026 &$<\!0.001$& -0.002   & -0.020   &$<\!0.001$&$<\!0.001$& -0.006 & -0.017  \\
\multirow{3}{*}{Gains}   &           Average    &  0.161 &  0.164  &  0.077 &  0.142 &  0.026   &  0.053   &  0.138   &  0.015   &  0.022   &  0.108 &  0.124  \\
		                 &      3rd Quartile    &  0.293 &  0.300  &  0.185 &  0.264 &  0.060   &  0.115   &  0.237   &  0.023   &  0.039   &  0.128 &  0.217  \\

                         & Inter-Quartile Range & 0.289  &  0.294  &  0.343 &  0.418 &  0.089   &  0.172   &  0.368   &  0.031   &  0.055   &  0.211 & 0.331 \\

		\hline
	\end{tabular}}
\vspace{0.2cm}

\footnotesize{``MSE'' and ``Bias'' indicate that parameters were selected by optimizing these criteria for each stock based on 2016 data where ``Fixed'' means that $g^* = 0.15$ and ``Varied'' means $g^* = |\mu| + 0.05$. Inter-Quartile Range $=$ 1st Quartile $-$ 3rd Quartile.}
\end{table*}

Table \ref{Gtable} contains a summary of the gains for the various strategies considered so far, along with optimized versions of SLS (i.e., $\alpha=\beta=1$ and optimization is done with respect to $K$ only). We can see that GSLS with MSE optimisation is the strategy with the highest mean and median gain. Furthermore, it is the only strategy where the mean and median gains are numerically very similar (e.g., SLS with $K=5$ has an average gain of 0.138 but a median gain of -0.02); in fact, the median gains for all other strategies are either negative or negligibly low. Also GSLS with MSE optimisation is the only strategy with a first quartile gain which is non-negative (although it is close to zero). In terms of variability of gains (via the inter-quartile range), the bias optimizer has higher variability than the MSE optimizer which we would expect. It is noteworthy that SLS with MSE optimization has the lowest level of variability, but also the lowest level of gain.

\section{Discussion\label{sec:discussion}}

The GSLS strategy expands classical SLS so that the parameters of the long and short controllers can differ (i.e., the initial investment and feedback parameters). By allowing different initial long and short investments such that the overall net initial investment is nonzero, $I(0) \ne 0$, the GSLS strategy can be thought of as a paradigm lying between SLS and more standard trading strategies (e.g., simply going long or short). This provides the opportunity to include some insight into the trading strategy while lowering risk relative to the standard trading strategies.

The robustness of GSLS with differing long and short initial investments ($\alpha\ne1$) is weakened relative to SLS in the sense that theoretical (expected) gain is no longer guaranteed to be positive; interestingly, GSLS with differing feedback parameters ($\beta\ne1$) maintains robustness provided that $\alpha=1$. However, should the trader possess some knowledge of the price evolution, and select $\alpha\ne1$, a greater level of gain can be achieved than that of SLS. Indeed, the primary driver of any realized gain is, as one would expect, the price evolution of the stock itself -- which is why one might wish to make estimates about its likely evolution at the expense of some robustness.

Throughout the current paper trader ``knowledge'' entered in the form of a GBM assumption of price evolution which is common in financial literature \cite{black1973pricing, merton1974pricing}. Of course, this is a simplifying assumption, and, in practice, more general models could be used which could include other forms of market knowledge exogenous to the historical price series itself. The secondary reason for using the GBM model is that it admitted a closed form solution for the expected gain and the variance of the gain. In more complex models, these quantities could be obtained by simulation. The model assumption could potentially be avoided altogether by using some model-free estimate of $\mathbb{E}\left\{g(t)\right\}$ such as the last realized gain value achieved in a historical sequence $\{g_1, g_2, \ldots, g_n\}$. Another alternative might be a weighted average over these realized gains $\sum_{j=1}^n w_j g_j$ where the weights, $w_j$, grow with $j$ so that more weight is placed on more recent observations. In any case, in our practical application, the methods appeared to perform well despite the GBM assumption.

Selection of control parameters which are ``optimal'' in some sense has not previously been considered in the literature to our knowledge. The lack of such selection procedures presents a major hurdle in the wider adoption of feedback-based trading strategies. To this end, we have proposed two possibilities which we call bias and MSE optimization. The bias optimization focusses on the expected gain, whereas the MSE approach additionally takes account of the variation of gains. Under the GBM assumption, the expected gain is connected to the drift of the process, while the variation incorporates the process volatility. While our suggestion involves specifying a target gain (for which a reasonable value can be informed by the estimated GBM model), we might, alternatively, have simply selected the best performing strategy over the testing period. This is equivalent to minimizing bias with some arbitrarily large target gain. Note also that our MSE objective function, whilst being a very natural quantity in itself, places equal weight on the bias and variance of the gain; more generally, a tuning parameter might control the trade-off between these two quantities. Another approach still (akin to the so-called ``efficient frontier'' in portfolio optimization \cite{markowitz1952portfolio}) would be based on maximizing the expected gain subject to a fixed level of variance, or minimizing the variance subject to a fixed expected gain. We could also imagine an optimization procedure which updates throughout time such that the control parameters are altered dynamically (albeit the initial investments are obviously fixed from the start). Thus, while we have provided the some initial suggestions for selecting control parameters, which perform well in our applications, there are a variety of avenues for extending these; this is a focus of our future work.

Overall, despite being a relatively new area of research, feedback-based trading strategies clearly have interesting theoretical properties as well as promising performance in practice. Through the extension of classical SLS to GSLS, the investigation of parameter selection procedures, and the extensive analysis of performance in a large sample of real stock price data, this paper contributes to a greater understanding of the potential of such strategies. Moreover, it moves the theory towards the goal of the wider adoption of feedback-based trading strategies by real-world traders.

\section*{Acknowledgment}
The authors would like to thank James P. Gleeson (MACSI, University of Limerick) for his helpful comments on an earlier draft of this paper. This work was funded by Science Foundation Ireland Grant 16/IA/4470 (J.D.O'B.).

\bibliographystyle{abbrv}
\bibliography{Bibliography}

\end{document}


TITLE: Galois group of a subextension is a subgroup of the Galois group of the extension?
QUESTION [1 upvotes]: Let $\overline{\mathbb{F}}_{5}$ be the separable closure of $\mathbb{F}_{5}$ and let $G=G(\overline{\mathbb{F}}_{5}/\mathbb{F}_{5})$ be its Galois group.
Say we pick a finite subextension of $\overline{\mathbb{F}}_{5}/\mathbb{F}_{5}$ such as $\mathbb{F}_{5}(\sqrt{2})/\mathbb{F}_{5}$, which has Galois group $\{\varphi_{1},\varphi_{2}\}$, where $\varphi_{1}$ is identity and $\varphi_{2}$ is conjugation.
How can I see this Galois group inside the big Galois group $G$?
I'm supposed to see it as a quotient of $G$ by some subgroup or simply as a subgroup of $G$? This is still very confusing. Sorry if the question is so easy.

REPLY [1 votes]: Do you know that the Galois group $G(\overline{\Bbb F}_5/\Bbb F_5)$ is topologically generated by the Frobenius automorphism of $\overline{\Bbb F}_5$, which sends any element $\alpha$ to $\alpha^5$? You can think of it as an infinite cyclic group, but completed with respect to a certain topology.
Calling the Frobenius automorphism $\mathbf f$, its square $\mathbf f^2$ (defined by $\alpha\mapsto\alpha^{25}$) generates a subgroup of index two. That’s the subgroup you’re looking for.


TITLE: What is the need for angular magnification
QUESTION [1 upvotes]: We've been told that angular magnification is defined as the ratio of the angle subtended by the image to the angle subtended by the object. But why would we need a quantity called angular magnification? Won't the simple magnification formula do?

REPLY [3 votes]: The angle subtended by an object is called the visual angle and it determines the size of the image formed on the retina.

When you look at the Moon with the naked eye it subtends an angle of about $0.5 ^\circ$.
When looking through a pair of binoculars the angle subtended by the image of the Moon might be $5^\circ$.
The angular magnification is $\dfrac{5}{0.5}=10$
What does that mean?
It means that the image of the Moon formed on the retina of the eye when looking through binoculars is ten times bigger than the image formed on the retina when the Moon is  looked at directly.
So the Moon appears to be bigger.
In such a context the idea of linear magnification would not be useful because the size of the image of the Moon formed by the binoculars is not ten times the actual size of the Moon.


TITLE: Real-valued measurability vs. Two-valued measurability in determining whether $CH$ holds or not
QUESTION [0 upvotes]: The following fact is known:

If there is a measurable cardinal, then there are only countably many constructible reals.

It is also known that if $ZFC$ + "There is a (two-valued) mesurable cardinal" is consistent, then $ZFC$ + "There exists a (two-valued) measurable cardinal" + $CH$ is also consistent.
If one replaces "There exists a (two-valued) measurable cardinal" with "The cardinality of the continuum is a real-valued measurable cardinal", is
i)$ZFC$ + "The cardinality of the continuum is a real-valued measurable cardinal" + $CH$ is consistent if $ZFC$ + "The cardinality of the continuum is a real-valued measurable cardinal" is consistent (it is not, by a result of Banach and Kuratowski)? Also,
ii) Does $ZFC$ + "The cardinality of the continuum is a real-valued measurable cardinal" imply that there are only countably many constructible reals?
Since i) is inconsistent, how is it possible that $ZFC$ + "The cardinality of the continuum is a real-valued measurable" implies $\lnot$$CH$ while $ZFC$+ "There exists a (two-valued) measurable cardinal" is consistent with $CH$ or $\lnot$$CH$?

REPLY [4 votes]: First of all, it is not the case that if ZFC is consistent, then ZFC + CH + "There is a measurable" is consistent; a measurable cardinal has much greater consistency strength than ZFC alone. What is true is that if ZFC + "There is a measurable" is consistent, then so is ZFC + CH + "There is a measurable;" I think this is what you mean.
A bit of clarification, just to pin things down: when we say "If $A$ is consistent, then $B$ is consistent," this is usually shorthand for "$T$ proves $Con(A)\implies Con(B)$" where $T$ is some reasonable weak theory (PRA is almost always enough); such an interpretation is needed to avoid stupidity (if ZFC is consistent, then the sentence "PA is consistent implies ZFC is consistent" is technically true, even though what we mean by it clearly isn't). So, elaborating a bit on the previous paragraph, here are a couple things PRA proves:

"If ZFC + "There is a measurable" is consistent, then ZFC + CH + "There is a measurable" is consistent."
"If PRA proves "If ZFC is consistent, then ZFC + "There is a measurable cardinal" is consistent," then ZFC is inconsistent."

No, that second one wasn't a typo.

Now re: question 1, you ask:

how is it possible that ZFC + "The cardinality of the continuum is a real-valued measurable" implies $\neg$CH while ZFC+ "There exists a (two-valued) measurable cardinal" is consistent with CH since the two theories are equiconsistent?

There's no tension here - equiconsistent theories can disagree. Equiconsistency doesn't mean that they're consistent together, but rather that each is consistent if and only if the other is consistent. For example, ZFC+CH and ZFC+$\neg$CH are equiconsistent, but clearly inconsistent with each other.
So there's no issue.

Re: question 2, off the top of my head I'm not sure, but if memory serves Solovay showed that if there is a real-valued measurable, then there is an inner model with a measurable (it is certainly true that he showed that "there is a real-valued measurable" and "there is a measurable" are equiconsistent over ZFC; but it's possible he build an inner model of a forcing extension, instead of an actual inner model, which would be a problem here). If my memory is correct, then the answer to 2 is "yes:" letting $M$ be that inner model, we have $M\models \vert\mathbb{R}^L\vert=\aleph_0$. But $L$ (hence $\mathbb{R}^L$) and $\aleph_0$ are absolute, so $V\models\vert\mathbb{R}^L\vert=\aleph_0$.
Note that this is stronger than what you ask for in (2): the conclusion is that if there is a real-valued measurable at all, regardless of its comparison with the continuum, then $\mathbb{R}^L$ is countable.


TITLE: How to solve the Schrödinger equation with magnetic field?
QUESTION [4 upvotes]: The Schrödinger equation of electron in a magnetic field is
$$
\frac{1}{2m} \left(-\mathrm{i}\hbar\nabla+\frac{e}{c}\mathbf{A}\right)^2 \psi + V\psi = E\psi
$$
where $V=-e\phi$ and $\phi$ is the scalar potential.
And the solution is
$$
\psi(\mathbf{r}) = \psi_0(\mathbf{r}) \exp\left[-\frac{\mathrm{i}e}{\hbar c} \int^{\mathbf{r}}_{\mathbf{r}_0} \mathbf{A}(\mathbf{r}') \cdot \mathrm{d}\mathbf{r}'\right].
$$
It's easy to check this solution by putting into the previous equation. But how can I achieve the solution directly from Schrödinger equation?

REPLY [3 votes]: If $A$ is not a pure gauge, or more weakly if $\nabla \times A \neq 0$, then it is false that $$\nabla_r \int_{r_0}^r A(r')\cdot dr' = A(r)\tag{0}$$ Without this result, by direct inspection you see that your statement is false. Otherwise it is  true. The proof is easy, it is the direct generalization of this elementary identity
$$\left(-i\frac{d}{dx} + f(x)\right)  \psi(x) =-ie^{+i \int^x_0 f(y) dy} \frac{d}{dx} e^{-i \int^x_0 f(y) dy}\psi(x) \tag{1}\:.$$
From (1), implementing once more the identity you have
$$\left(-i\frac{d}{dx} + f(x)\right) \left(-i\frac{d}{dx} + f(x)\right)  \psi(x) =(-i)^2e^{+i \int^x_0 f(y) dy} \frac{d}{dx} e^{-i \int^x_0 f(y) dy} e^{+i \int^x_0 f(y) dy} \frac{d}{dx} e^{-i \int^x_0 f(y) dy}\psi(x)\:,$$
that is
$$\left(-i\frac{d}{dx} + f(x)\right)^2 \psi(x)=  e^{+i \int^x_0 f(y) dy} \frac{d^2}{dx^2} e^{-i \int^x_0 f(y) dy}\psi(x)$$
and finally
$$ e^{-i \int^x_0 f(y) dy} \left[\left(-i\frac{d}{dx} + f(x)\right)^2 + U(x)\right]\psi= \left(-\frac{d^2}{dx^2} +U(x) \right)e^{-i \int^x_0 f(y) dy}\psi(x)\:.$$
The crucial point in the computations above is that
$$\frac{d}{dx}\int_0^x f(y) dy = f(x).\tag{2}$$
Unfortunately, the procedure does not work in dimension $>1$, since the generalization of $\int^x_0 f(y) dy$ depends on the integration path unless the vector field $A$ which replaces $f$ has zero curl. Also  fixing arbitrarily a path, the generalisation (0) of (2) generally does not hold in dimension greater than $1$.
Validity of (2) for dimension greater than $1$ is equivalent to saying that $A$ is a pure gauge at least locally, and however $B = \nabla \times A =0$.


TITLE: $f(X)\subset \{0\}\cup f(\text{supp}(f))$ vs $f(X)\subset f(\text{supp}(f))$
QUESTION [1 upvotes]: Suppose $f$ has compact support and is continuous (say we are in the simplest case of $f:\mathbb R\rightarrow \mathbb R$).
I was reading this answer, and can't understand why there needs to be a $\{0\}$ in the expression $$f(X)\subset\{0\}\cup f(\text{supp}(f))$$
Even if $\{x\ : f(x)=0\}\ne\emptyset$, won't we have that $0\in f(\text{supp}(f))?$
tryin to think of counterexamples with a function such that $\{x\ :\ f(x)\ne 0\}=(a,b)\cup(b,c)$ for some $a<b<c$, we get that $\text{supp}(f)=\overline{(a,b)\cup(b,c)}=[a,c]$, therefore $b\in \text{supp}(f)$ and $0=f(b)\in f(\text{supp}(f))$, so there would be no need for the $\{0\}$.
So either there is a counterexample, in which case I'd like to know it, or the $\{0\}$ is unnecessary.

REPLY [3 votes]: If you consider a function $f: X \to \Bbb R$ with $X \subseteq \Bbb R$ (like in the question you've linked) the $\{0\}$ is indeed necessary.
E.g. take $X: [0,1] \cup [2,3]$ and $f = 1_{[0,1]}$ the indicator function of $[0,1]$. Then $f: X \to \Bbb R$ is continuous and $$f(X) = \{0,1\} = \{0\} \cup f(\text{supp}(X))$$ but $$f(\text{supp}(X)) = f([0,1]) = \{1\}$$


TITLE: Determining the strength of information to vote for candidates.
QUESTION [0 upvotes]: Suppose you are trying to decide which of two candidates to vote for. You think there is a 60% chance that Candidate A is better than Candidate B. Then, you hear a news report that claims B is better than A.
a)Assuming you vote for whichever candidate you think is most likely to be better, how accurate does the information have to be for you to vote for Candidate B? [Hint: use ‘p’ to capture the accuracy of the news report]
b) Now, suppose that you receive an additional piece of information after hearing the news report: your friend tells you that she thinks A is better than B. How accurate does her information have to be for you to vote for Candidate A? Your answer will be in terms of ‘p’. [Hint: use ‘q’ to capture the accuracy of your friend’s information]

I'm not very good at probabilities, but for the first part, I was thinking that we have to determine a $p$ such that $.6*p=.5 \implies p=5/6 \approx 83.3 %$.  So the new information that we hear has to be $83.3%$ true.
For the second part, assuming the first part is correct, we do the same procedure to get $.6*p*q=.5 \implies q=\frac{5}{6p}$
Are these calculations correct?

REPLY [2 votes]: It rather depends on how $p$ and $q$ are stated.  What you would like is the likelihood of the reports saying A if A is better and of saying A is B is better, and of similarly of saying B if B is better and of saying B is A is better
Suppose for the newspaper, it says B is better when B is indeed better with conditional probability $p$ and it says B is better when A is better with conditional probability $1-p$.  These do not have to add up to one, though will do if the newspaper is unbiased
Then your revised view of the possibility of A being better, combining your previous view and the newspaper could be $\dfrac{0.6(1-p)}{0.6(1-p)+0.4p} =  \dfrac{3-3p}{3-p}$.  You might then vote for B when this expression is less than $\frac12$, which would require $p \gt \frac35$
This makes some kind of intuitive sense: you might change your vote if you think the newspaper is more accurate than you.
Now take into account your friend's view and similarly suppose she says A is better when A is indeed better with conditional probability $q$ and she says A is better when B is better with conditional probability $1-p$. Then you can update your opinion again, either with  $\dfrac{0.6(1-p)q}{0.6(1-p)q+0.4p(1-q)}$ or with    $\dfrac{\frac{3-3p}{3-p}q}{\frac{3-3p}{3-p}q +\left(1-\frac{3-3p}{3-p}\right)(1-q)}$, both of which simplify to $\dfrac{3pq-3q}{5pq-2p-3q}$ as your re-revised view of the possibility of A being better
That will greater than $\frac12$ if $q \gt \frac{2p}{3-p}$ at which point you switch back to supporting A. As a simple check, that cannot happen when $p=1$; meanwhile $p=q$ otherwise leads to you original value of $0.6$ since the newspaper and your friend balance each other out


TITLE: Direct proof that integral of a function does not depend on the $\sigma$-algebra used to define it?
QUESTION [2 upvotes]: If $\mathcal{G}\subset\mathcal{F}$ are two $\sigma$ algebras on a set $X$, $\mu$ is a nonnegative measure on $(X,\mathcal{F})$ and $f:X\to[0,+\infty]$ is $\mathcal{G}$-measurable, then there are two possible definitions of $\int f~d\mu$. The first is as an $\mathcal{F}$-measurable map:
$$\int_\mathcal{F} f~d\mu:=\sup\left\{\int\varphi~d\mu~\Big|\:\begin{array}{c}0\leq\varphi\leq f\text{ simple}\\\mathcal{F}\text{-measurable map}\end{array}\right\}$$
the second is as a $\mathcal{G}$-measurable map:
$$\int_\mathcal{G} f~d\mu:=\sup\left\{\int\varphi~d\mu~\Big|\:\begin{array}{c}0\leq\varphi\leq f\text{ simple}\\ \mathcal{G}\text{-measurable map}\end{array}\right\}$$
These two definitions coincide as follows for instance from an application of Fatou's lemma (for $\mathcal{F}$-measurable maps) and the dominated convergence theorem (for $\mathcal{G}$-measurable maps).

Question: is there a direct argument proving $\int_\mathcal{G} f~d\mu=\int_\mathcal{F} f~d\mu$ that I'm overlooking? Maybe on the basis of a monotone class argument?

Proof I had in mind: Note that by definition of both integrals the result is true for indicator functions $1_A$, $A\in\mathcal{G}$. Furthermore, by linearity of both $\int_\mathcal{G}\cdot~ d\mu$ and $\int_\mathcal{F}\cdot~ d\mu$, the result holds for all nonnegative $\mathcal{G}$-measurable simple functions ${\color{blue}{(1)}}$. Furthermore, since every $\mathcal{G}$-measurable simple function is a $\mathcal{F}$-measurable simple function, one always has
$$
\int_\mathcal{G} f~d\mu\leq\int_\mathcal{F} f~d\mu.
$$
Thus, if $\int_\mathcal{G} f~d\mu=+\infty$, the result is established. Now suppose $\int_\mathcal{G} f~d\mu<+\infty$. For every $n\in\Bbb{N}$ we set for instance
$$
f_n = \sum_{i=1}^{n2^n}2^{-n}1_{[i/2^n\leq f]}.
$$
Then $0\leq f_n\leq f$ is a nonnegative $\mathcal{G}$-measurable simple function which converges pointwise to $f$. We get that $f\in L^1(\Omega,\mathcal{F},\mu)$ and the reverse inequality since by Fatou's lemma $\color{red}{(2)}$ and dominated convergence $\color{green}{(3)}$
$$\begin{eqnarray}
    0
    \leq
    \int_\mathcal{F} f~d\mu
    =
    \int_\mathcal{F}\liminf_n f_n~d\mu
    & \overset{\color{red}{(2)}}\leq  &
    \liminf_n\int_\mathcal{F} f_n~d\mu\\
    & \overset{\color{blue}{(1)}}=  &
    \liminf_n\int_\mathcal{G} f_n~d\mu\\
    & \overset{\color{green}{(3)}}= &
    \int_\mathcal{G} f~d\mu.
\end{eqnarray}$$

Better proof. As noted in @KaviRamaMurthy's answer, the right tool is the monotone convergence theorem. We know by the first point ${\color{blue}{(1)}}$ that both definitions agree on nonnegative simple $\mathcal{G}$-measurable functions. Taking the $0\leq f_n\leq f_{n+1}\leq f$ defined above, one has by two applications of the monotone convergence theorem that
$$
\int_{\mathcal{F}} f~d\mu
\overset{\text{MCT}}=
\lim_n
\int_{\mathcal{F}} f_n~d\mu
\overset{{\color{blue}{(1)}}}=
\lim_n
\int_{\mathcal{G}} f_n~d\mu
\overset{\text{MCT}}=
\int_{\mathcal{G}} f~d\mu
$$

REPLY [1 votes]: In the books I have used $\int fd\mu$ is defined using a specific construction of simple functions approximating $f$: $f_n(x)=\frac {i-1} {2^{n}}$ if $\frac {i-1} {2^{n}} \leq f <\frac i {2^{n}}$, $i \leq N2^{n}$ and $f_n(x)=N$ for $i >N2^{n}$.  And $\int fd\mu=\lim \int f_n d\mu$. In the present case the same sequence is also a sequence of simple functions on $(X, \mathcal G)$ increasing to $f$ so the definition of $\int fd\mu$ is same whether you are using the sigma algebra $\mathcal G$ or $\mathcal F$.


TITLE: Upper bound of Wasserstein distance given by subvariables of codim 1
QUESTION [0 upvotes]: recently I am considering the upper-bound of Wasserstain distance. Say we have random vectors $X,Y$ of dimension $n$, and let $\tilde{X}_i (\tilde{Y}_i,$ resp.) be the $(n-1)$-dim random vector of $X (Y $, resp.) discarding the $i$-th component. For example, $n=3$ and $X=(X_1,X_2,X_3)$, then $\tilde{X}_2=(X_1,X_3)$.
My question is, can we formulate an inequality of the form $W_p(X,Y) \leq \sum \limits_{i=1}^na_i W_p(\tilde{X}_i,\tilde{Y}_i)$? I know that we can formulate similar inequality by using $1$-dim marginals refer here, hence I believe such inequality would hold for $(n-1)$-subvariables.

REPLY [1 votes]: $\newcommand\tX{\tilde X}\newcommand\tY{\tilde Y}\newcommand\D{\overset D=}$The answer is no. E.g., suppose that $n=2$, so that $\tX_1,\tX_2,\tY_1,\tY_2$  can be identified with $X_2,X_1,Y_2,Y_1$, respectively, and the inequality in question becomes
$$W_p(X,Y)\le a_1W_p(X_2,Y_2)+a_2W_p(X_1,Y_1).\tag{1}$$
Suppose also $X_1\D Y_1$, $X_2\D Y_2$, but $X\not\D Y$, where $\D$ denotes the equality in distribution. Then the right-hand side of (1) is $0$ but the left-hand side is $>0$. So, (1) fails to hold.


TITLE: Show that for any $n$, there are infinitely many cubes of the form $2^na - 9$.
QUESTION [6 upvotes]: Show that for any $n$, there are infinitely many cubes of the form $2^na - 9$.

Progress: We use induction on $n.$ For $n=1$ it works. Say it works for $n-1$. We will show for $n$. Note that $2^na-9$ is $$2^{n-1}2a-9.$$ If we have infinite $a$ such that $a$ is even for $n-1$, then we are done. Else, only finite amounts of $a$ are even.
So after a large constant $N$, we have $2^{n-1}a-9=x^3$ for only odd $a$.
So say $$2^{n-1}a_1-2^{n-1}a_2=u^3-w^3\implies 2^{n-1}|u^3-w^3$$
But I couldn't get anything further.
I thought about taking $v_2$, as in, if $2^na=x^3+9$ then we can take $x=y^2$. So we have $$2^na=y^2+3^2.$$

REPLY [4 votes]: Suppose we have some positive $b$ for which $2^nb-9=u^3$ for some positive integer $u$. Notice that $u$ must be odd. Note then that
\begin{equation}
2^n(b+3u^2+3u2^n+(2^{n})^2)-9=u^3+3u^22^n+3u(2^n)^2+(2^n)^3=(u+2^n)^3,
\end{equation}
and so $a=b+3u^2+3u2^n+(2^n)^2$ also satisfies the equation $2^na-9=v^3$ for some integer $v$.
However, $3u^2+3u2^n+(2^n)^2$ must be odd, as $u$ is odd. If $b$ was even, then $a$ is odd, and if $b$ was odd, then $a$ is even. Reiterating this argument, we find that if one solution $b$ exists, then we must in fact have an infinite amount of solutions, and that these solutions alternate signs, so that we also have an infinite amount of even solutions.
Your proof by induction then works.


TITLE: "Taylor" approximation of a function, based on 2 rather than 1 base points?
QUESTION [4 upvotes]: Normally, when we want to approximate a function $f(x)$ close to $x=a$, we do a Taylor approximation around $a$:
$$f(x)\approx f(a)+f'(a)(x-a)+...$$
However, what if we want to approximate a function $f(x)$ on the interval $[a,b]$, by an approximating function $g(x)$ where we require the following:

Contrary to Taylor approximation, where we only necessarily have that $g(a)=f(a)$ for some $a$. We now require two endpoints to coincide. That is, we need $g(a)=f(a)$ and $g(b)=f(b)$
Ideally: g(x) should be simple, easily integrable, analytically tractable, and (just like Taylor approximation) allow for arbitrary degrees of precision. Preferably polynomial.
We only care about the approximation on the interval $[a,b]$. That is, we need $g(x)\approx f(x)$ on $x\in[a,b]$ only.
We care whether the integrals of $f(x)$ and $g(x)$ on $[a,b]$ are approximately equal for arbitrary functions $f(x)$.
Moreover, the approximation should ideally be the "best" approximation in its class $\mathcal C$. By $\mathcal C$ I mean for example, the class of polynomial functions of degree $n$. If we want more precision, we can then increase $n$. I am open to different definitions of "best", but I am thinking of something like "the integral of $g(x)$ on $[a,b]$ should be the closest to that of $f(x)$ on that interval out of all possible g(x) in class $\mathcal C$".

Is there a canonical technique, similar to Taylor approximation, that satisfies these requirements? Or satisfy some of them?
The most simple approximation I can come up with is simply, $g(x)=A+Bx$ with $A,B$ such that $g(a)=f(a), g(b)=f(b)$. This determines a unique function.  However, when we add a quadratic term, there are now an infinite amount of functions satisfying the boundary conditions, so which one of them is the "best"?

REPLY [3 votes]: Here is another approach.  One way to think of approximating a function $f(x)$ with a Taylor polynomial of degree $n$ (at $x=a$) is that it is the unique $n$th degree polynomial $p_n(x)$ satisfying
$$p_n(a) = f(a)$$
$$p'_n(a) = f'(a)$$
$$p''_n(a) = f''(a)$$
$$\vdots$$
$$p^{(n)}_n(a) = f^{(n)}(a)$$
Think of those requirements as a specification of what "best" means in the context of a normal Taylor polynomial.  What are the equivalent specifications when you have two points?
Well, suppose we have two points $a$ and $b$, and we want to require that some polynomial $P(x)$ satisfies all of the following:
$$P(a) = f(a)$$
$$P'(a) = f'(a)$$
$$P''(a) = f''(a)$$
$$\vdots$$
$$P^{(n)}(a) = f^{(n)}(a)$$
$$P(b) = f(b)$$
$$P'(b) = f'(b)$$
$$P''(b) = f''(b)$$
$$\vdots$$
$$P^{(n)}(b) = f^{(n)}(b)$$
That's a set of $2n+2$ requirements, and in general we'd need a polynomial of $2n+1$ in order to satisfy all of them.  So let's propose a working definition of a two-point Taylor polynomial:
Definition: The two-point Taylor polynomial of order $n$ is unique polynomial of minimal degree satisfying the $2n+2$ requirements above.
The two-point Taylor polynomial of order $0$ is just the linear interpolation $L_{a,b}(x)$ between the two points $(a,f(a))$ and $(b,f(b))$.  This polynomial passes through the correct points, but does not match any of the derivatives.  Explicitly,
$$L_{a,b}(x)=f(a) + \frac{f(b)-f(a)}{b-a}(x-a)$$
If you want to not only match the values of $f(x)$ at $a$ and $b$, but also the derivative $f'(x)$ at both points, then you need the 1st-order two-point polynomial, which would be a 3rd degree polynomial.  Any 3rd-degree polynomial that passes through $(a,f(a))$ and $(b,f(b))$ can be written in the form
$$P(x) = L_{a,b}(x) + K(x-a)(x-b)(x-c)$$
where the parameters $K$ and $c$ will be chosen in order to meet the other requirements. Specifically, if we take the derivative of $P(x)$ we get
$$P'(x) = \frac{f(b)-f(a)}{b-a} + K\left((x-a)(x-b) + (x-a)(x-c) + (x-b)(x-c)  \right)$$
so
$$P'(a) = \frac{f(b)-f(a)}{b-a} + K(a-b)(a-c)$$
and
$$P'(b) = \frac{f(b)-f(a)}{b-a} + K(b-a)(b-c)$$
so we have to solve the pair of equations
$$\frac{f(b)-f(a)}{b-a} + K(a-b)(a-c) = f'(a)$$
$$\frac{f(b)-f(a)}{b-a} + K(b-a)(b-c) = f'(b)$$
for $K$ and $c$. The algebra is unpleasant but fairly straightforward.
If you want to match the 2nd derivatives also, you would need a 5th degree polynomial. I hesitate to work out the details as they are rather ugly, but the method generalizes more or less directly.


TITLE: Finding the roots of $x^2+x-6=0$ in $\mathbb{Z}_{14}$
QUESTION [0 upvotes]: I am currently studying ring theory and encountered the following problem $x^2+x-6=0$ In $\mathbb{Z}_{14}$. My instructor informed us to look out for zero divisors, but I cant see why
$x-2=7$
Or
$x+3=7$
Is two of the four options for the roots (the other being when those two equations $=0$)? I understand that if $xy=0$ in $\mathbb{Z}_{14}$ then x or y must be a multiple of 14, but wouldn’t that mean
$x-2=14$
Or
$x+3=14$
Instead of the former?

REPLY [1 votes]: You wrote

I understand that if xy=0 in Z14 then x or y must be a multiple of 14

That's not true.  The product $2\times7\equiv0\bmod14,$ but neither $2$ nor $7$ is a multiple of $14$.
Thus, for example, $4$ is a zero of $x^2+x-6=(x-2)(x+3)\bmod14,$
even though $4$ is not a zero of $x-2$ nor of $x+3$.
Euclid's lemma $p|xy\implies p|x$ or $p|y$ works for prime $p$.


TITLE: $G$ is an abelian group and $x=a_1a_2...a_n$
QUESTION [0 upvotes]: Assume that $G$ is an abelian group of order $n$ and $a_1,a_2,...,a_n$ are its elements and we have $x=a_1a_2...a_n$.
If $G$ has more than one element such as $b\neq e$ and $b^2=e$, prove that $x=e$.

Answer:
Assume that $a\neq b$, $a^2=e$, $b^2=e$, $a\neq e$ and $b\neq e$.
$$(ab)^2=(ab)(ab)=a^2b^2=e$$
If $$ab=e\implies a=b^{-1}\implies a=b$$
But we assume that $a\neq b$. contradiction!
So, $ab \neq e$ and $(ab)^2=e$.
So, what should I do now?
STUCK!

REPLY [3 votes]: $\newcommand{\Set}[1]{\left\{ #1 \right\}}$The main point is what happens in the subgroup
$$
H = \Set { x \in G : x^{2} = e },
$$
which by assumption is a vector space of dimension $k$ over the field $F$ with $2$ elements.
Write the elements of $H$ with respect to a (multiplicative) basis. Looking at the exponents with respect to any given basis elements $b$, half of the elements of $H$ will have exponent $0$, and half will have exponent $1$. But half of $2^{k}$ is $2^{k-1}$, which is even as by assumption $k > 1$, so the sum $s$ of the exponents will be even, so that $b^{s} = e$.


TITLE: Radius of convergence
QUESTION [2 upvotes]: What is radius of convergence of $\sum_{n=2}^{\infty} \left(1+\dfrac{1}{n}\right)^{n^2}z^n$?
I don't know anything

REPLY [4 votes]: Radius of convergence is given by $\dfrac{1}{R}= \lim_{n \to \infty}  |a_n|^{\dfrac{1}{n}}$
$\therefore$ $\dfrac{1}{R}=\lim_{n \to \infty} \left[\left(1+\dfrac{1}{n}\right)^{n^2}\right]^{\dfrac{1}{n}}$=$\dfrac{1}{R}=\lim_{n \to \infty} \left(1+\dfrac{1}{n}\right)^{n}=e$
$\therefore R=\dfrac{1}{e}$


TITLE: Calculation of $\int_0^\infty \frac{x^{1-a}}{x+c} dx $ using Gamma function
QUESTION [1 upvotes]: I am supposed to calculate some integrals using Gamma functions.
The simplest of the integrals is
$$ I(c)=\int_0^\infty \frac{x^{1-a}}{x+c} dx  $$
and the two relations I am supposed to use are
$$ \frac{\Gamma(b+1)}{a^{b+1}}=\int_0^\infty t^be^{-at}dt $$
and Euler's reflection formula,
$$ \Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)} $$
Try as I might, I have not managed to correlate the formulas to the integral in any significant way (in large part due to that pesky exponential I can't seem to shake). The supposed result of the above integral is $$I(c)=-\frac{\pi}{\sin(\pi a)}c^{1-a}$$
I definitely don't want to outsource my responsibilities (and once I understand how to go about this I'm confident I'll work out the more complicated integrals myself), but I've hit a wall here. I'd appreciate either a solution or a nudge in the right direction.

REPLY [2 votes]: This is an outline of a solution, and I hope it's detailed enough for you to fill in the gaps.
I assume you have attempted to use the two relations by substituting the first relation into the second one, which would get you something like
$$
\frac{\pi}{\sin(\pi b)} = \Gamma(b+1)\Gamma(-b) = a \int_0^\infty \int_0^\infty s^{-b-1} t^{b} e^{-a(s+t)} ds dt
$$
The hope is to somehow transform the right hand side into some form that is proportional to $I(c)$. This is in fact the correct direction to work towards.
You said that you can't get rid of the pesky exponential factor, but you can probably also guess that the key is to come up with some substitution where one of the new variables is $u = s+t$, so that the integral with respect to $u$ will get rid of the exponential factor. The substitution that will work is in fact $s = u v$, $t = u(1-v)$. Note that the $u$-dependent part of $s^{-b-1} t^{b}$ will simplify to $u^{-1}$. Combined with the Jacobian factor $u$, the only remaining $u$ dependence in the integrand is $e^{-au}$. Thus integrating with respect to $u$ will get rid of the exponential factor.
The remaining integral should be of the form
$$
\int_0^1 v^{-b-1} (1-v)^{b} dv\ .
$$
A substitution along the lines of $x = \frac{v^2}{1-v^2}$ will bring the integral closer to the form in which $I(c)$ is originally written down.
Note that the relation you are trying to prove can be thought of as a special case of the relation
$$
\mathrm{B}(x, y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}\ ,
$$
that is, the case when $x+y = 1$, and the integral $I(c)$ can be seen as a multiple of one of the integral representations of the beta function (see equation (22)),
$$
\mathrm{B}(2-a, a-1) = \int_0^\infty \frac{x^{1-a}}{1+x}dx\ .
$$
The proof outline I sketched above is a specialization of a standard proof the relation between the two functions. In fact the proof of the general case is more natural and suggests the first substitution used above.


TITLE: Proving that the shortest distance between two parabolas is along their common normal.
QUESTION [0 upvotes]: Can someone prove that the minimum distance between two non-intersecting parabolas is along their common normal (without calculus)? (If I understand how to prove it, I'll prove it myself for other curves too.)

REPLY [3 votes]: Assuming the existence of a shortest distance, let the shortest distance join points $A$ and $B$ and construct a circle with diameter $AB$. If the circle cuts either parabola at a point other than $A$ or $B$ there is a line joining the two curves which is shorter than the diameter. Else the circle must be tangent at points $A$ and $B$ and the diameter $AB$ is normal to both curves (because the diameter is perpendicular to the tangent).
There are some implicit assumptions here, but would this do? Note that if $A=B$ then the diameter doesn't have a fixed direction and the situation is different.


TITLE: For what kind of matrix $A$, there is a (symmetric) positive definite matrix $B$ such that $BA$ is symmetric
QUESTION [0 upvotes]: Let $A$ be a $n\times n$ real matrix. My ultimate goal is to find a sufficient condition on $A$ such that all the eigenvalues of $A$ are real.
Therefore, I want $A$ to be self-adjoint with respect to some inner-product. A general inner product associated with a positive definite matrix $B$ is given by $<x,y>=x^tBy$. Then $A$ is self-adjoint if $A^tB=BA$, i.e. $BA$ is symmetric.
So my problem reduces to "for what kind of matrix $A$, there is a (symmetric) positive definite matrix $B$ such that $BA$ is symmetric".
My idea: Given $A$, we want to find a positive definite $n\times n$ matrix $B$ by solving the equation $A^tB=BA$. There are $\frac{n(n+1)}{2}$ variables (not $n^2$) as $B$ is symmetric. There are $n^2$ equations but it seems that only $\frac{n(n-1)}{2}$ of them are independent. Therefore, the solutions should be abundant as there are more variables than equations. However, I don't know how to make sure the solution is positive definite. It is possible that the solution space does not intersect with the positive cone of positive definite matrices, except at $0$.

REPLY [1 votes]: Indeed, such a matrix $B$ exists if and only if $A$ has a real spectrum.
If $A$ has a real spectrum, then $A=P^{-1}DP$ for some real matrix $P$ and diagonal matrix $D$. Therefore, $B:=P^TP$ is positive definite and $BA=P^TDP$ is symmetric.
Conversely, suppose $S:=BA$ is symmetric for some positive definite $B$. Then $A=B^{-1}S$ is similar to $B^{-1/2}SB^{-1/2}$ and hence it has a real spectrum.


TITLE: Find the antiderivative F (x) for each function f(x)
QUESTION [1 upvotes]: $$f(x) = 8\sec x (\sec x - 4 \tan x)$$
Hi!
So I've already solved this but I just want to double check my answer. I got $8 (\tan(x) - 4\sec(x)) + C$ as my answer.

REPLY [0 votes]: Check by taking the derivative:
$$\dfrac{d}{dx}(8(\tan x - 4\sec x)+C) = 8\sec^2 x - 32\sec x \tan x = 8\sec x(\sec x -  4\tan x) = f(x)$$
So, yes, your answer is correct.


TITLE: Why is the magnitude of velocity negative in this example?
QUESTION [0 upvotes]: Magnitudes are positive values, but when I take, for example: the magnitude of a position vector: $r = 3 - 0.04t^2$ and try to take the derivate of it, the result will be $v = -2 * 0.04t$ which is a negative quantity. I don't really see the mistake here but that negative sign should not be there. Can a rate of change be negative in this case?

REPLY [1 votes]: Yes, a magnitude is positive, but the derivative of a magnitude may be negative. For example, speed is the magnitude of velocity, so speed is positive. But when I apply my car’s brakes the derivative of my speed is negative.


TITLE: How to prove $\frac{v}{\pi} \int_0^1 \frac{1}{x^2 \sqrt{1-x^2}} e^{\frac{-v^2}{2x^2}} dx =\frac{1}{\sqrt{2\pi}} e^{-\frac{v^2}{2}}$
QUESTION [2 upvotes]: If $v>0$ why
$$\frac{v}{\pi}  \int_0^1 \frac{1}{x^2 \sqrt{1-x^2}} e^{\frac{-v^2}{2x^2}} dx
=\frac{1}{\sqrt{2\pi}} e^{-\frac{v^2}{2}}$$.
I saw it in

I tried to solve it by $t=\frac{1}{x^2}-1$
$$\frac{v}{\pi}  \int_0^1 \frac{1}{x^2 \sqrt{1-x^2}} e^{\frac{-v^2}{2x^2}} dx
=\frac{v}{\pi}\int_0^{\infty} \frac{1}{\frac{1}{t+1} \sqrt{1-\frac{1}{t+1}}} e^{\frac{-v^2}{2}(t+1)} \, \frac{1}{2}(t+1)^{\frac{-3}{2}}dt $$
$$=\frac{v e^{\frac{-v^2}{2}} }{2\pi}\int_0^{\infty}\frac{1}{\sqrt{t}}  e^{\frac{-v^2}{2}t} \, dt $$
Thanks in advance for any help you are able to provide.

REPLY [2 votes]: To arrive to a gaussian integral
$$I= \int \frac{e^{\frac{-v^2}{2x^2}}}{x^2 \sqrt{1-x^2}}\,  dx$$
Let
$$x=\frac{1}{\sqrt{t^2+1}}\implies dx=-\frac{t}{\left(t^2+1\right)^{3/2}}\,dt$$ This makes
$$I=-\int e^{-\frac{1}{2} \left(t^2+1\right) v^2}\,dt=-\frac{\sqrt{\frac{\pi }{2}} e^{-\frac{v^2}{2}} \text{erf}\left(\frac{t
   v}{\sqrt{2}}\right)}{v}$$
$$J=-\int_0^\infty e^{-\frac{1}{2} \left(t^2+1\right) v^2}\,dt=-\frac{\sqrt{\frac{\pi }{2}} e^{-\frac{v^2}{2}}}{\sqrt{v^2}}$$


text:
\begin{document}
\publicationdetails{22}{2020}{4}{9}{6543}

\maketitle

\begin{abstract}
Given a graph $G$ and an integer $p$, a coloring $f : V(G) \to \mathbb{N}$ is \emph{$p$-centered}
if for every connected subgraph $H$ of $G$, either $f$ uses more than $p$ colors on $H$ or there is a color that appears exactly once in $H$.
The notion of $p$-centered colorings plays a central role in the theory of sparse graphs.
In this note we show two lower bounds on the number of colors required in a $p$-centered coloring.

First, we consider monotone classes of graphs whose shallow minors have average degree bounded polynomially in the radius, or equivalently (by a result of Dvo\v{r}\'ak and Norin), admitting strongly sublinear separators. We construct such a class such that $p$-centered colorings require a number of colors super-polynomial in $p$. This is in contrast with a recent result of Pilipczuk and Siebertz, who established a polynomial upper bound in the special case of graphs excluding a fixed minor.

Second, we consider graphs of maximum degree $\Delta$. D\k{e}bski, Felsner, Micek, and Schr\"{o}der recently proved that these graphs have $p$-centered colorings with $O(\Delta^{2-1/p} p)$ colors.
We show that there are graphs of maximum degree $\Delta$ that require $\Omega(\Delta^{2-1/p} p  \ln^{-1/p}\Delta)$ colors in any $p$-centered coloring, thus matching their upper bound up to a logarithmic factor.
\end{abstract}

\begin{textblock}{20}(0, 12.9)
\includegraphics[width=40px]{logo-erc}
\end{textblock}
\begin{textblock}{20}(-0.25, 13.3)
\includegraphics[width=60px]{logo-eu}
\end{textblock}

\section{Introduction}
Given a graph $G$ and an integer $p$, a coloring $f : V(G) \to \mathbb{N}$ is \emph{$p$-centered}
if for every connected subgraph $H$ of $G$, either $f$ uses more than $p$ colors on $H$ (i.e., $|f(V(H))| > p$) or
there is a color that appears exactly once in $H$ (i.e., there exists $i \in \mathbb{N}$ such that $|f^{-1}(i) \cap V(H)| = 1$).

The notion of $p$-centered colorings plays a crucial role in the theory of sparse graphs.
First, a $p$-centered coloring of a graph $G$ with small number of colors is very useful in algorithm design, for example in the task of finding or counting small subgraphs~\cite{experiments}.
Second, a classic result shows that a graph class has bounded expansion (in the sense of Ne{\v{s}}et{\v{r}}il and Ossona de Mendez~\cite{sparsity}) if and only if for every $p \in \mathbb{N}$ there exists $M(p)$ such that every graph $G$ admits a $p$-centered coloring with at most $M(p)$ colors~\cite{NesetrilM08}.

A recent experimental study~\cite{experiments} shows that the existing algorithms that find $p$-centered colorings find colorings with too many colors for the subsequent algorithms
to be efficient. This motivated a number of recent theoretical results trying to establish better upper and lower bounds on the number of colors needed for a $p$-centered coloring for various classes of sparse graphs~\cite{DFMS19,GroheKRSS18,PilipczukS19}.
In this note, we contribute two lower bounds to this direction.

\paragraph{Superpolynomial lower bound in graph classes of polynomial expansion.}
A \emph{minor model} of a graph $H$ in a graph $G$ is a collection $(I_h)_{h \in V(H)}$ of vertex-disjoint connected subgraphs of $G$ such that
$h_1h_2 \in E(H)$ implies that there is an edge of $G$ with one endpoint in $I_{h_1}$ and one endpoint in $I_{h_2}$. Fix an integer $r \geq 0$.
A minor model is \emph{$r$-shallow}
if every graph $I_h$ is of radius at most $r$.
For a graph $G$, by $\nabla_r(G)$ we denote the maximum density of $r$-shallow minors of $G$, that is, the maximum ratio $|E(H)|/|V(H)|$ over all graphs $H$
that admit an $r$-shallow minor model in $G$.
For a graph class $\mathcal{G}$, we denote $\nabla_r(\mathcal{G}) = \sup \{\nabla_r(G)~|~G \in \mathcal{G}\}$.
By definition, $\mathcal{G}$ is of \emph{bounded expansion} if for every $r \geq 0$ the value $\nabla_r(\mathcal{G})$ is finite.

The class $\mathcal{G}$ is of \emph{polynomial expansion} if there exists a polynomial $q$ such that $\nabla_r(\mathcal{G}) \leq q(r)$ for every $r \geq 0$.
The notion of polynomial expansion turned out to be pivotal for approximation algorithms:
a subgraph-closed graph class is of polynomial expansion if and only if it admits strongly sublinear separators~\cite{DvorakN16}
and such separators allow approximation schemes via local search for a number of optimization problems~\cite{Har-PeledQ17}.
It is natural to ask what other properties one can infer about graph classes of polynomial expansion, and in particular whether they admit $p$-centered colorings with a number of colors polynomial in $p$.
Pilipczuk and Siebertz~\cite{PilipczukS19} recently proved that such a polynomial upper bound holds in the special case of graphs excluding a fixed minor, which motivates our investigation.

In~\cite{GroheKRSS18} an intricate example of a graph class with polynomial expansion is shown to have super-polynomial (in the radius) weak coloring numbers.
(For precise definitions, we refer to the textbook~\cite{sparsity} or the recent lecture notes~\cite{sparsitynotes}.)
In Section~\ref{sec:polyexp} we show that the same graph class requires a super-polynomial (in $p$) number of colors for a $p$-centered coloring. More precisely, we show the following.
\begin{theorem}\label{thm:lb1}
There exists a graph class $\mathcal{G}$ and a constant $c>0$ such that
$\nabla_r(\mathcal{G}) \leq r+2$ for every integer $r \geq 0$, but
for every integer $p \geq 1$
there is $G \in \mathcal{G}$ such that every $p$-centered coloring of $G$
requires at least $2^{c p^{1/2}}$ colors.
\end{theorem}
The proof of Theorem~\ref{thm:lb1} builds substantially on a recent lower bound of~\cite{DFMS19}
for graphs of bounded treewidth.

We remark here that if $\mathcal{G}$ is of polynomial expansion,
then there exists a polynomial $q$ such that for every $G \in \mathcal{G}$
and $p \geq 1$ there exists a $p$-centered coloring of $G$ with at most $2^{2^{q(p)}}$ colors.
See Chapter~2 of the lecture notes~\cite{sparsitynotes} for an exposition.

\paragraph{Lower bound for graphs of bounded degree.}
A recent breakthrough result of D\k{e}bski, Felsner, Micek, and Schr\"{o}der~\cite{DFMS19} asserts
that graphs of bounded degree require much less colors for a $p$-centered coloring
than was anticipated.
\begin{theorem}[\cite{DFMS19}]\label{thm:bnddeg}
There exists a constant $C > 0$ such that for every integers $\Delta \geq 1$ and $p \geq 1$,
every graph $G$ of maximum degree at most $\Delta$ admits a $p$-centered coloring
with at most $C \cdot \Delta^{2-1/p} \cdot p$ colors.
\end{theorem}

In Section~\ref{sec:bnddeg} we show that the dependency on $p$ and $\Delta$ in Theorem~\ref{thm:bnddeg}
is optimal up to a logarithmic factor in $\Delta$.
\begin{theorem}\label{thm:lb2}
There exists a constant $c > 0$ such that for every integer $p \geq 1$ there exists $\Delta_p$ such that for all $\Delta \geq \Delta_p$ there exists a graph $G$ of maximum degree at most $\Delta$ such that every $p$-centered
coloring of $G$ requires at least $c \cdot \Delta^{2-1/p} p \ln^{-1/p}{\Delta}$ colors.
\end{theorem}

We remark that this lower bound is obvious for $p=1$, and was previously known for $p=2$~\cite{fertin2004star}.
Whether the $\ln^{-1/p}{\Delta}$ factor is necessary is an open problem.

\section{Graphs of polynomial expansion}\label{sec:polyexp}
\input{lb-poly-exp}

\section{Bounded degree graphs}\label{sec:bnddeg}
\input{bounded-degree}

\acknowledgements
Part of this research has been done at Structural Graph Theory Downunder workshop at MATRIX research center in November 2019.
We are very thankful to MATRIX for their hospitality.
We also thank Tereza Klimo\v{s}ov\'{a}, Jan Volec, David Wood, and Lena Yuditsky for insightful discussions.
Finally, we thank the two anonymous referees for their helpful comments on an earlier version of the paper.

\bibliographystyle{abbrv}
\bibliography{refs}

\end{document}

config: arxiv


text:
\section{Subset sum results}\label{app:subsetcor99432374}

\subsection{Product of uniform distributions contains a uniform distribution}

\begin{lemma} \label{lem:prodUnif}
Let $X\sim U[0,1]$ (or $X\sim U[-1,0]$) and $Y\in U[-1,1]$ be independent random variables. Then the PDF of the random variable $XY$ is
\begin{equation*}
    f_{XY}(z) = \begin{cases}
    \frac{1}{2}\log \frac{1}{|z|}& |z|\leq 1\\
    0 & \text{ otherwise}
    \end{cases}
\end{equation*}
\end{lemma}
\begin{proof}
It is easy to see why $f_{XY}(z)=0$ for $z>1$. We prove for $X\sim U[0,1]$. The proof for $X\sim U[-1,0]$ is similar.

Let us first try to find the CDF of $XY$.

Let $0\leq z\leq 1$ be a real number. Note that $XY\leq 1$. Now, if $XY\leq z$, and if $Y\geq z$, then $X\leq z/Y$. However, if $Y<z$, then $X$ can be anything in its support $[0,1]$. Thus,
\begin{align*}
    F_{XY}(z)
    &=\mathbb{P}(XY\leq z)&\\
    &=\int_{0}^z \frac{1}{2} \int_{0}^1 1 \text{d}x \text{d}y + \int_{z}^1 \frac{1}{2} \int_{0}^{z/y} 1 \text{d}x \text{d}y&\\
     &=\frac{z}{2} + \frac{1}{2}\int_{z}^1 \frac{z}{y} \text{d}y&\\
     &=\frac{z}{2} - \frac{z \log z}{2}.&
\end{align*}
Differentiating this, the pdf for $0\leq z\leq 1$ is
\begin{equation*}
    f_{XY}(z)=\frac{1}{2}\log \frac{1}{z}.
\end{equation*}

Now, because $XY$ is symmetric around $0$, we get that for $|z|\leq 1$
\begin{equation*}
    f_{XY}(z)=\frac{1}{2}\log \frac{1}{|z|}.
\end{equation*}

\end{proof}

\begin{corollary} \label{cor:prodUnif2}
Let $X\sim U[0,1]$ (or $X\sim U[-1,0]$) and $Y\in U[-1,1]$ be independent random variables.
Let $P$ be the distribution of $XY$.
Let $\delta_0$ be the Dirac-delta function.
Define a distribution $D = \frac{1}{2}\delta_{0} + \frac{1}{2}P$.

Then, there exists a distribution $Q$ such that
\begin{align*}
    P = \left( \frac{1}{2} \log 2\right) U\left[-\frac{1}{2},\frac{1}{2}\right]+\left( 1- \frac{1}{2} \log 2\right) Q
\end{align*}
\end{corollary}
\begin{proof}
The corollary follows from the observation that Lemma \ref{lem:prodUnif} shows that pdf of $P$ is lower bounded by $(\log 2)  U\left[-\frac{1}{2},\frac{1}{2}\right]$ on $\left[-\frac{1}{2},\frac{1}{2}\right]$.
\end{proof}

\subsection{Subset sum problem with product of uniform distributions}
\begin{corollary}[\cite{Lueker98}]\label{cor:LuekerCor}
Let $X_1, \dots ,X_n$ be i.i.d. from the distribution in the hypothesis of Corollary \ref{cor:prodUnif2}, where $n \geq C\log \frac{2}{\epsilon}$ (for some universal constant $C$). Then, with probability at least $1-\epsilon$, we have
\begin{align*}
    \forall z \in [-1, 1],  \qquad \exists S \subset [n] \text{ such that } \left|z - \sum_{i \in S}X_i \right| \leq \epsilon.
\end{align*}
\end{corollary}
\begin{proof}
This is a direct application of Markov's inequality on Corollary 3.3 from \cite{Lueker98} applied to the distribution in the hypothesis of Corollary \ref{cor:prodUnif2}.
\end{proof}


TITLE: Picking a special function from the set of random functions
QUESTION [3 upvotes]: Consider a fixed integer $q$. Consider the set of all functions from $\{0, 1\}^{n+1}$ to $\{0, 1\}^{m}$. Let us pick one function from this set uniformly at random.
Now, let's say we want functions $f$ with the following property.
There exists a $b_f \in \{0, 1\}^{m}$ (depending on the function $f$) such that for every $x \in \{0, 1\}^{n}$,
$$f(0, x) - f(1, x) = b_f~~~mod~q.$$
What is the probability we pick such a function?

REPLY [3 votes]: First note that $\ f(0,x)-f(1,x)\in\{-1,0,1\} \ $.   If $\ b\in\{0,1\}\ $ and  $\ q\ge3\ $, then
$$
f(0,x)-f(1,x)\equiv b\text{ mod}(q)\iff
f(0,x)-f(1,x)= b\ ,
$$
so the probability you're looking for will be the same for all $\ q\ge3\ $.  It's not the same for $\ q=2\ $, however, because $\ {-1}\equiv1\pmod{2}\ $.
Note also that if $\ X_{y,i}=f(y)_i\ $ for $\ (y,i)\in\{0,1\}^{n+1}\times$$\{1,2,\dots,m\}\ $, and $\ f\ $ is chosen uniformly at random, then $\ X_{y,i}\ $ will be independent Bernoulli variates with $\ P\big(X_{y,i}=0\big)=P\big(X_{y,i}=1\big)=\frac{1}{2}\ $.  Therefore, if $\ Y_{x,i}=X_{(0,x),i}-X_{(1,x),i}= f(0,x)_i-f(0,x)_i\ $ for $\ (x,i)\in\{0,1\}^n\times$$\{1,2,\dots,m\}\ $, then $\ Y_{x,i}\ $ will be independent random variables, with
$$
P\big(Y_{x,i}=-1\big)=P\big(Y_{x,i}=1\big)=\frac{1}{4}
$$
and
$$
P\big(Y_{x,i}=0\big)=\frac{1}{2}\ .
$$
If $\ q\ge3\ $, the probability you're looking for is
\begin{align}
P\left(\bigcup_{b\in\{0,1\}^m}\bigcap_{x\in\{0,1\}^n}\bigcap_{i=1}^m\big\{Y_{x,i}=b_i\big\}\right)&=\sum_{b\in\{0,1\}^m}P\left(\bigcap_{x\in\{0,1\}^n}\bigcap_{i=1}^m\big\{Y_{x,i}=b_i\big\}\right)\\
&=\sum_{b\in\{0,1\}^m}\left(\prod_{i=1}^m\left(\frac{1}{2}\right)^{1-b_i}\left(\frac{1}{4}\right)^{b_i}\right)^{2^n}\\
&=\sum_{b\in\{0,1\}^m}\left(\frac{1}{2^{2^n}}\right)^{m+|b|}\ \ \ \big(\text{where $\ |b|=\big|\{i\,|\,b_i=1\}\big|\ $}\big)\\
&=\sum_{k=0}^m\sum_{b\in\{0,1\}^m,\\|b|=k}\left(\frac{1}{2^{2^n}}\right)^{m+k}\\
&=\left(\frac{1}{2^{2^n}}\right)^m\sum_{k=0}^m{m\choose k}\left(\frac{1}{2^{2^n}}\right)^k\\
&=\left(\frac{1}{2^{2^n}}\right)^m\left(1+\frac{1}{2^{2^n}}\right)^m\\
&=\left(\frac{1}{2^{2^n}}+\frac{1}{2^{2^{n+1}}}\right)^m\ .
\end{align}
If $\ q=2\ $, then the probability you're looking for is instead
$$
P\left(\bigcup_{b\in\{0,1\}^m}\bigcap_{x\in\{0,1\}^n}\bigcap_{i=1}^m\big\{Y_{x,i}=b_i\text{ mod}(2)\big\}\right)\ ,
$$
and $\ P\big(Y_{x,i}=0\text{ mod}(2)\big)=P\big(Y_{x,i}=1\text{ mod}(2)\big)=\frac{1}{2}\ $.  So for $\ q=2\ $
\begin{align}
P\left(\bigcup_{b\in\{0,1\}^m}\bigcap_{x\in\{0,1\}^n}\bigcap_{i=1}^m\big\{Y_{x,i}=b_i\text{ mod}(2)\big\}\right)&=\sum_{b\in\{0,1\}^m}P\left(\bigcap_{x\in\{0,1\}^n}\bigcap_{i=1}^m\big\{Y_{x,i}=b_i\text{ mod}(2)\big\}\right)\\
&=\sum_{b\in\{0,1\}^m}\left(\frac{1}{2}\right)^{2^nm}\\
&=2^m\left(\frac{1}{2}\right)^{2^nm}\\
&=\left(\frac{1}{2}\right)^{(2^n-1)m}\
\end{align}