text:
\begin{document}

\maketitle

\begin{abstract}
Agents acting in the natural world aim at selecting appropriate actions based on noisy and partial sensory observations. Many behaviors leading to decision making and action selection in a closed loop setting are naturally phrased within a control theoretic framework. Within the framework of optimal Control Theory, one is usually given a cost function which is minimized by selecting a control law based on the observations. While in standard control settings the sensors are assumed fixed, biological systems often gain from the extra flexibility of optimizing the sensors themselves.  However, this sensory adaptation is geared towards control rather than perception, as is often assumed. In this work we show that sensory adaptation for control differs from sensory adaptation for perception, even for simple control setups. This implies, consistently with recent experimental results, that when studying sensory adaptation, it is essential to account for the task being performed.

\end{abstract}

\section{Introduction}

Biological systems face the difficult task of devising effective control strategies based on partial information communicated between sensors and actuators across
multiple distributed networks. While the theory of Optimal Control (OC) has become widely used as a framework for studying motor control, the standard
framework of OC neglects many essential attributes of biological control \cite{Izawa2008,Todorov2002,Battaglia2007}.  The classic formulation of closed loop OC
considers a dynamical system (plant) observed through sensors which transmit their output to a controller, which in turn selects a control law that drives actuators to
steer the plant. This standard view, however, ignores the fact that sensors, controllers and actuators are often distributed across multiple sub-systems, and
disregards the communication channels between these sub-systems. While the importance of jointly considering control and communication within a unified framework
was already clear to the pioneers of the field of Cybernetics (e.g., Wiener and Ashby), it is only in recent years that increasing effort is being devoted to the formulation
of a rigorous systems-theoretic framework for control and communication (e.g., \cite{andrievsky2010}). Since the ultimate objective of an agent is to select appropriate
actions, it is clear that sensation and communication must subserve effective control, and should be gauged by their contribution to action selection. In fact, given the
communication constraints that plague biological systems (and many current distributed systems, e.g., cellular networks, sensor arrays, power grids, etc.), a major
concern of a control design is the optimization of sensory information gathering and communication (consistently with theories of active perception). For example,
recent theoretical work demonstrated a sharp communication bandwidth threshold below which control (or even  stabilization) cannot be achieved (for a summary of
such results see \cite{andrievsky2010}). Moreover, when informational constraints exists within a control setting, even simple (linear and Gaussian) problems become
nonlinear and intractable, as exemplified in the famous Witsenhausen counter-example \cite{witsenhausen1968}.\par

The inter-dependence between sensation, communication and control is often overlooked both in control theory and in computational neuroscience, where one
assumes that the overall solution to the control problem consists of first estimating the state of the controlled system (without reference to the control task), followed
by constructing a controller based on the estimated state. This idea, referred to as the \emph{separation principle} in Control Theory, while optimal in certain restricted settings
(e.g., Linear Quadratic Gaussian (LQG) control) is, in general, sub-optimal \cite{Tse1973}. Unfortunately, it is in general very difficult to provide optimal solutions in cases where separation
fails. A special case of the separation principle, referred to as \emph{Certainty Equivalence} (CE), occurs when the controller treats the estimated state as the true state, and
forms a controller assuming full state information. It is generally overlooked, however, that although the optimal control policy does not depend directly on the
observation model at hand, the expected future costs do depend on the specifics of that model \cite{Bar-Shalom1974}. In this sense, even when CE
holds, costs still arise from uncertain estimates of the state and one can optimise the sensory observation model to minimise these costs, leading to sensory
adaptation. At first glance, it might
seem that the observation model that will minimise the expected future cost will be the observation model that minimises the estimation error. We will show,
however, that this is not generally the case.\par

A great deal of the work in computational neuroscience has dealt independently with the problem of sensory adaptation and control, while, as stated above, these two
issues are part and parcel of the same problem. In fact, it is becoming increasingly clear that biological sensory adaptation is task-dependent \cite{gilbert2013,
Huber2012,mattar2013}. In this work we consider a simple setting for control based on spike time sensory coding, and study the optimal coding of sensory information
required in order to perform a well-defined motor task. We show that even if CE holds, the optimal encoder strategy, minimising the control cost, differs from the
optimal encoder required for state estimation. This result demonstrates, consistently with experiments, that neural encoding must be tailored to the task at hand. In
other words, when analyzing sensory neural data, one must pay careful care to the task being performed. Interestingly, work within the distributed control community
dealing with optimal assignment and selection of sensors, leads to similar conclusions and to specific schemes for sensory adaptation.\par

The interplay between information theory and optimal control is a central pillar of modern control theory, and we believe it must be accounted for in the
computational neuroscience community. Though statistical
estimation theory has become central in neural coding issues, often through the Cram\'{e}r-Rao bound, there have been few studies bridging the gap between partially
observed control and neural coding. We hope to narrow this gap by presenting a simple example where control and estimation yield different conclusions.
The remainder of the paper is organised as follows: In \fref{sec:opt_codes} we introduce the notation and concepts; In \fref{sec:control} we derive expressions for the cost-to-go of a linear-quadratic control system observed through spikes from a dense populations of neurons;
in \fref{sec:comparison} we present the results and compare optimal codes for control and estimation with point-process
filtering, Kalman filtering and LQG control; in \fref{sec:conclusion} we discuss the results and their implications.

\subsection{Optimal Codes for Estimation and Control}

\label{sec:opt_codes}

We will deal throughout this paper with a dynamic system with state $X_t$, observed through noisy sensory observations $Z_t$, whose conditional distribution can be
parametrised by a set of parameters $\varphi$, e.g., the widths and locations of the tuning curves of a population of neurons or the noise properties of the observation
process. The conditional distribution is then given by
$P_{\varphi} (Z_t|X_t = x)$. $Z_t$ could stand for a diffusion process dependent on $X_t$ (denoted $Y_t$) or a set of doubly-stochastic Poisson processes
dependent on $X_t$ (denoted $N^m_t$). In that sense, the optimal Bayesian encoder for an estimation problem, based on the Mean Squared Error (MSE) criterion, can
be written as
\[
\varphi^*_e = \argmin_\varphi \boldsymbol{E}_z \left[\boldsymbol{E}_{X_t} \left[\left(X_t - \hat{X_t}(Z_t)\right)^2\middle| Z_t = z\right]\right],
\]
where $\hat X_t(Z_t) = \boldsymbol{E}\left[X_t|Z_t\right]$ is the posterior mean, computable, in the linear Gaussian case, by the Kalman filter. We will throughout this
paper consider the MMSE in the equilibrium, that is, the error in estimating $X_t$ from long sequences of observations $Z_{[0,t]}$.
Similarly, considering a control problem with a cost given by
\[
C(\boldsymbol{X}_0, \boldsymbol{U}_0) = \int_0^T c(X_s,U_s,s) ds + c_T(X_T),
\]
where $\boldsymbol{X}_t = \{X_s |  s \in [t,T] \}, \boldsymbol{U}_t = \{U_s |  s \in [t,T] \}$, and so forth.
We can define
\[
\varphi^*_c = \argmin_\varphi \boldsymbol{E}_z \min_{\boldsymbol{U}_t}  \left[\boldsymbol{E}_{\boldsymbol{X}_t} \left[C(\boldsymbol{X}_0, \boldsymbol{U}_0) \middle| \boldsymbol{Z}_t = z\right]\right].
\]
The certainty equivalence principle states that given a control policy $\gamma^* : \mathcal{X} \to \mathcal{U}$ which minimises the cost $C$,
\[
\gamma^* = \argmin_\gamma C(\boldsymbol{X}_0, \boldsymbol{\gamma}(\boldsymbol{X}_0)),
\]
the optimal control policy for the partially observed problem given by noisy observations $\boldsymbol{Z}_0$ of $\boldsymbol{X}_0$ is given by
\[
\gamma_{CE}(\boldsymbol{Z}_t) =\gamma^*\left( \boldsymbol{E}\left[\boldsymbol{X}_0\middle| \boldsymbol{Z}_t\right]\right).
\]
Note that we have used the notation $\boldsymbol{\gamma}(\boldsymbol{X}_0) = \{\gamma(X_s), s\in [0,T]\}$.
\section{Stochastic Optimal Control}

\label{sec:control}

In stochastic optimal control we seek to minimize the expected future cost incurred by a system with respect to a control variable applied to that system. We will
consider linear stochastic systems governed by the SDE
\begin{subequations}
\begin{equation}
\label{eq:OU}
dX_t = \left(A X_t + B U_t\right) dt + D^{1/2} dW_t,
\end{equation}
with a cost given by
\begin{equation}
C(\boldsymbol{X}_t,\boldsymbol{U}_t,t) = \int_t^T \left(X_s^\top Q X_s + U_s^\top R U_s\right) ds + X_T^\top Q_T X_T.
\end{equation}
\end{subequations}
From Bellman's optimality principle or variational analysis \cite{astrom2006}, it is well known that the optimal control is given by $U^*_t = -R^{-1 }B^\top S_t X_t$, where $S_t$ is the solution of the Riccati equation
\begin{equation}
\label{eq:riccati}
-\dot{S}_t =Q + A S_t + S_t A^\top - S_t B^\top R^{-1} B S_t,
\end{equation}
with boundary condition $S_T = Q_T$. The expected future cost at time $t$ and state $x$ under the optimal control is then given by
$$
J(x,t) = \min_{\boldsymbol{U}_t} \boldsymbol{E}\left[C(\boldsymbol{X}_t,\boldsymbol{U}_t,t)\middle|X_t = x\right] = \frac{1}{2} x^\top S_t x + \int_t^T \Tr\left(D
S_s  \right) ds.
$$

This is usually called the optimal cost-to-go. However, the system's state is not always directly accessible and we are often left with noisy observations of it. For a
class of systems e.g. LQG control, CE holds and the optimal control policy for the indirectly observed
control problem is simply the optimal control policy for the original control problem applied to the Bayesian estimate of the system's state. In that sense, if the
CE were to hold for the system above observed through noisy observations $Y_t$ of the state at time $t$, the optimal control would
be given simply by the observation-dependent control  $U^*_t = -R^{-1 }B^\top S_t\boldsymbol{E}\left[X_t\middle| Y_t\right]$ \cite{Bar-Shalom1974}.\par

Though CE, when applicable, gives us a simple way to determine the optimal control, when considering neural systems we are often interested in finding the optimal encoder, or the
optimal observation model for a given system. That is equivalent to finding the optimal tuning function for a given neuron model. Since CE neatly
separates the estimation and control steps, it would be tempting to assume the optimal codes obtained for an estimation problem would also be optimal for an
associated control problem. We will show here that this is not the case.\par
As an illustration, let us consider the case of LQG with incomplete state information. One could, for example, take the observations to be a secondary process
$Y_t$, which itself is a solution to
\[
dY_t = FX_t dt + G^{1/2}dV_t,
\]
the optimal cost-to-go would then be given by \cite{astrom2006}
\begin{flalign}
\label{eq:LQG_incomplete}
J(y,t) =& \min_{\boldsymbol{U}_t} \boldsymbol{E}\left[C(\boldsymbol{X}_t,\boldsymbol{U}_t,t) \middle| Y_{[0,t]} = y\right] \\
=& \nu_t^\top S_t \nu_t+\Tr\left(K_t S_t\right) + \int_t^T \Tr\left(D S_s \right)ds + \int_t^T \Tr\left(S_s B R^{-1} B^\top S_s K_s\right) ds,\nonumber
\end{flalign}
where we have defined $Y_{[0,t]} = \{Y_s, s\in [0,t]\}$, $\nu_t=\boldsymbol{E}[X_t|Y_{[0,t]}]$ and $K_t=\mbox{cov}[X_t|Y_{[0,t]}]$. We give a demonstration of these results in the SI, but for a thorough review see \cite{astrom2006}. Note that through the last term in \fref{eq:LQG_incomplete}
the cost-to-go now depends on the parameters of the $Y_t$ process. More precisely, the variance of the distribution of $X_s$ given $Y_t$, for $s > t$ obeys the
ODE
\begin{equation}
\label{eq:kalman_covar}
\dot{K_t}  = AK_t + K_t A^\top + D - K_t F^\top G^{-1} F K_t.
\end{equation}
One could then choose the matrices $F$ and $G$ in such a way as to minimise the contribution of the rightmost term in \fref{eq:LQG_incomplete}. Note that in
the LQG case this is not particularly interesting, as the conclusion is simply that we should strive to make $K_t$ as small as possible, by making the
term $F^\top G^{-1} F$ as large as possible. This translates to choosing an observation process with very strong steering from the unobserved process (large
$F$) and a very small noise (small $G$). One case that provides some more interesting situations is if we consider a two-dimensional system, where we are
restricted to a noise covariance with constant determinant. That means the hypervolume spanned by the eigenvectors of the covariance matrix is constant. We
will compare this case with the Poisson-coded case below.

\subsection{LQG Control with Dense Gauss-Poisson Codes}

Let us now consider the case of the system given by \fref{eq:OU}, but instead of observing the system directly we observe a set of doubly-stochastic Poisson
processes $\{N^m_t\}$ with rates given by
\begin{equation}
\label{eq:tuning_func}
\lambda^m(x) = \phi \exp\left[-\frac{1}{2} \left(x-\theta_m\right)^\top P^\dagger \left(x-\theta_m\right)\right].
\end{equation}
To clarify, the process $N^m_t$ is a counting process which counts how many spikes the neuron $m$ has
fired up to time $t$. In that sense, the differential of the counting process $dN^m_t$ will give the spike train process, a sum of Dirac delta functions placed at the times
of spikes fired by neuron $m$. Here $P^\dagger$ denotes the pseudo-inverse of $P$, which is used to allow for tuning functions that do not depend on
certain coordinates of the
stimulus $x$.
Furthermore, we will assume that the tuning centre $\theta_m$ are such that the probability of observing a spike of any neuron at a given time
$\hat{\lambda} = \sum_m \lambda^m(x)$ is independent of the specific
value of the world state $x$. This can be a consequence of either a dense packing of the tuning centres $\theta_m$ along a given dimension of $x$, or of an
absolute insensitivity to that aspect of $x$ through a null element in the diagonal of $P^\dagger$. This is often called the dense coding hypothesis \cite{Yaeli2010}.ø
It can be readily be shown that the filtering distribution is given by $P(X_t | \{N_{[0,t)}\}) = \mathcal{N}(\mu_t, \Sigma_t)$, where the mean and covariance are
solutions to the stochastic differential equations (see \cite{Susemihl2013})
\begin{subequations}
\begin{equation}
\label{eq:OU_filter_mean}
d\mu_t = \left(A\mu_t + BU_t\right) dt + \sum_m  \Sigma_t \left(I + P^\dagger \Sigma_t\right)^{-1} P^\dagger \left(\theta_m - \mu_t\right) dN^m_t,
\end{equation}
\begin{equation}
\label{eq:OU_filter_var}
d\Sigma_t =\left(A \Sigma_t + \Sigma_t A^\top + D\right)dt -\Sigma_t P^\dagger \Sigma_t \left(I + P^\dagger\Sigma_t\right)^{-1} \, dN_t,
\end{equation}
\end{subequations}
where we have defined $\mu_t=\boldsymbol{E}[X_t|\{N^m_{[0,t]}\}]$ and $\Sigma_t=\mbox{cov}[X_t|\{N^m_{[0,t]}\}]$.
Note that we have also defined $N^m_{[0,t]} = \{N^m_s | s\in [0,t]\}$, the history of the process $N^m_s$ up to time $t$, and $N_t = \sum_m N^m_t$. Using
Lemma 7.1 from \cite{astrom2006} provides a simple connection between the cost function and the solution of the associated Ricatti equation for a stochastic process.
We have
\begin{flalign*}
C(\boldsymbol{X}_t,\boldsymbol{U}_t,t)=\,&X_T^\top Q_T X_T + \int_t^T \left[X_s^\top Q X_s+U_s^\top R U_s \right] ds\\
=&X_t^\top S_t X_t +\int_t^T (U_s + R^{-1} B^\top S_s X_s)^\top R(U_s + R^{-1} B^\top S_s X_s)ds \\
+& \int_t^T \Tr(D S_s) ds + \int_t^T dW_s^\top D^{\top/2} S_s X_s ds + \int_t^T X_s^\top S_s D^{1/2} dW_s.
\end{flalign*}
We can average over $P(\boldsymbol{X}_t,\boldsymbol{N}_t|\{N_{[0,t)}\})$ to obtain the expected future cost. That gives us
\[
\mu_t^\top S_t \mu_t + \Tr(\Sigma_t S_t) + \boldsymbol{E}\left[\int_t^T (U_s + R^{-1} B^\top S_s X_s)^\top R(U_s + R^{-1} B^\top S_s X_s)ds\middle| \{N_{[0,t)}\}
\right] +\int_t^T \Tr(D S_s) ds
\]
We can evaluate the average over $P(\boldsymbol{X}_t, \{\boldsymbol{N}^m_t\}|\{N^m_{[0,t)}\})$ in two steps, by first averaging over the Gaussian densities
$P(X_s|\{N^m_{[0,s]}\})$ and then over $P(\{N_{[0,s]}\}|\{N_{[0,t)}\})$. The average gives
\[
\boldsymbol{E}\left[\int_t^T (U_s + R^{-1} B^\top S_s \mu_s)^\top R(U_s + R^{-1} B^\top S_s \mu_s) + \Tr\left[S_s BR^{-1}B^\top S_s \Sigma_s(\{N_{[0,s]}\})\right] ds\middle| \{N_{[0,t)}\}
\right],
\]
where $\mu_s$ and $\Sigma_s$ are the mean and variance associated with the distribution $P(X_s|\{N_{[0,s)}\})$. Note that choosing $U_s =- R^{-1} B^\top
S_s \mu_s$ will minimise the expression above, consistently with CE. The optimal cost-to-go is therefore given by
\begin{flalign}
\label{eq:poiss_cost}
J(\{N_{[0,t)}\},t) = \mu_t^\top S_t\mu_t + \Tr(\Sigma_t S_t) +\int_t^T \Tr\left(D S_s\right)ds+ \int_t^T  \Tr\left(S_s BR^{-1}B^\top S_s \boldsymbol{E}\left[\Sigma_s(\{N_{[0,s]}\})|\{N_{[0,t)}\}\right]\right)ds
\end{flalign}
Note that the only term in the cost-to-go function that depends on the parameters of the encoders is the rightmost term and it depends on it only through the
average over future paths of the filtering variance $\Sigma_s$. The average of the future covariance matrix is precisely the MMSE for the filtering problem conditioned
on the belief state at time $t$ \cite{Susemihl2013}. We can therefore analyse the quality of an encoder for a control task by
looking at the values of the term on the right for different encoding parameters. Furthermore,
since the dynamics of $\Sigma_t$ given by \fref{eq:OU_filter_var} is Markovian, we can write the average $\boldsymbol{E}\left[\Sigma_s|\{N_{[0,t)}\}\right]$ as
$\boldsymbol{E}\left[\Sigma_s|\Sigma_t\right]$. We will define then the function $f(\Sigma,t)$ which gives us the uncertainty-related expected future cost for the control problem as
\begin{equation}
\label{eq:f}
f(\Sigma,t)  = \int_t^T  \Tr\left(S_s BR^{-1}B^\top S_s \boldsymbol{E}\left[\Sigma_s\middle|\Sigma_t=\Sigma\right]\right)ds.
\end{equation}

\subsection{Mutual Information}
\label{sec:mutual_info}
Many results in information theory are formulated in terms of the mutual information of the communication channel $P_\varphi(Y|X)$. For example, the maximum
cost reduction achievable with $R$ bits of information about an unobserved variable $X$ has been shown to be a function of the rate-distortion function with the
cost as the distortion function \cite{kanaya1991}. More recently there
has also been a lot of interest in the so-called I-MMSE relations, which provide connections between the mutual information of a channel and the minimal mean
squared error
of the Bayes estimator derived from the same channel \cite{Merhav2011,guo2005mutual}. The mutual information for the cases we are considering is
not particularly complex, as all distributions are Gaussians.
Let us denote by $\Sigma^0_t$ the covariance of of the unobserved process $X_t$ conditioned on some
initial Gaussian distribution $P_0 = \mathcal{N}(\mu_0,\Sigma_0)$ at time $0$. We can then consider the Mutual Information between the stimulus at time $t$, $X_t$, and
the observations up to time t, $Y_{[0,t]}$ or $N_{[0,t]}$. For the LQG/Kalman case we have simply
\[
I(X_t; Y_{[0,t]} | P_0) = \int dx\, dy P(x,y)\left[\log P(x|y)- \log P(x)\right] = \log | \Sigma^0_t | - \log |\Sigma_t |,
\]
where $\Sigma_t$ is a solution of \fref{eq:kalman_covar}.
For the Dense Gauss-Poisson code, we can also write
\[
I(X_t; N_t | P_0) = \int dx\, dn\, P(x,n)\left[ \log P(x|n) - \log P(x) \right]= \log | \Sigma^0_t| - \boldsymbol{E}_{N_{[0,t]}} \left[\log | \Sigma_t(N_{[0,t]})| \right],
\]
where $\Sigma_t(N_{[0,t]})$ is a solution to the stochastic differential \fref{eq:OU_filter_var} for the given value of $N_{[0,t]}$.

\section{Optimal Neural Codes for Estimation and Control}
\label{sec:comparison}

What could be the reasons for an optimal code for an estimation problem to be sub-optimal for a control problem? We present examples that show two possible
reasons for different optimal coding strategies in estimation and control. First, one should note that control problems are often defined over a finite time horizon. One
set of classical experiments involves reaching for a target under time constraints \cite{Battaglia2007}. If we take the maximal firing
rate of the neurons ($\phi$) to be constant while varying the width of the tuning functions, this will lead the number of observed spikes to be inversely proportional to
the precision of those spikes, forcing a trade-off between the number of observations and their quality. This trade-off can be tilted to either side in
the case of control depending on the information available at the start of the problem. If we are given complete information on the system state at the initial time $0$,
the encoder needs fewer spikes to reliably estimate the system's state throughout the duration of the control experiment,
and the optimal encoder will be tilted towards a lower number of spikes with higher precision.
Conversely, if at the beginning of the experiment we have very little information about the system's state, reflected in a very broad distribution, the encoder will be
forced towards lower precision spikes with higher frequency. These results are discussed in \fref{sec:information}.\par

Secondly, one should note that the optimal encoder for estimation does not take into account the differential weighting of different dimensions of the system's state.
When considering a multidimensional estimation problem, the optimal encoder will generally allocate all its resources equally between the dimensions of the system's
state. In the framework presented we can think of the dimensions as the singular vectors of the tuning matrix $P$ and the resources allocated to it are the
singular values. In this sense, we will consider a set of coding strategies defined by matrices $P$ of constant determinant in \fref{sec:determinant}.
This constrains the overall firing rate of the
population of neurons to be constant, and we can then consider how the population will best allocate its observations between these dimensions. Clearly, if we have an
anisotropic control problem, which places a higher importance in controlling one dimension, the optimal encoder for the control problem will be expected to allocate
more resources to that dimension. This is indeed shown to be the case for the Poisson codes considered, as well as for a simple LQG problem when we constrain the
noise covariance to have the same structure.\par

We do not mean our analysis to be exhaustive as to the factors leading to different optimal codes in estimation and control settings, as the general problem is
intractable, and indeed, is not even separable. We intend this to be a proof of
concept showing two cases in which the analogy between control and estimation breaks down.

\subsection{The Trade-off Between Precision and Frequency of Observations}
\label{sec:information}

In this section we consider populations of neurons with tuning functions as given by \fref{eq:tuning_func} with tuning centers $\theta_m$ distributed along a one-
dimensional line.
In the case of the Ornstein-Uhlenbeck process these will be simply one-dimensional values $\theta_m$ whereas in the case of the stochastic oscillator, we will
consider tuning centres of the form $\theta_m = (\eta_m,0)^\top$, filling only the first dimension of the stimulus space. Note that  in both cases the (dense) population
firing
rate $\hat{\lambda} = \sum_m \lambda_m (x)$ will be given by $\hat{\lambda} = \sqrt{2\pi} p \phi / |\Delta \theta|$, where $\Delta \theta$ is the separation between neighbouring tuning
centres $\theta_m$.

The Ornstein-Uhlenbeck (OU) process controlled by a process $U_t$ is given by the SDE
\[
dX_t = (bU_t-\gamma X_t) dt + D^{1/2}dW_t.
\]
\Fref{eq:poiss_cost} can then be solved by simulating the dynamics of
$\Sigma_s$. This has been considered extensively in \cite{Susemihl2013} and we refer to the results therein. Specifically, it has been found that the dynamics of
the average can be approximated in a mean-field approach yielding surprisingly good results. The evolution of the average posterior variance is given by the average
of \fref{eq:OU_filter_var}, which involves nonlinear averages over the covariances. These are intractable, but a simple mean-field approach yields the approximate
equation for the evolution of the average $\left<\Sigma_s\right> = \boldsymbol{E}\left[\Sigma_s\middle| \Sigma_0\right]$
\[
\frac{d\left<\Sigma_s\right>}{ds} = A \left<\Sigma_s\right> + \left<\Sigma_s \right>^\top A^\top + D - \hat{\lambda} \left<\Sigma_s\right>P^\dagger  \left<\Sigma_s\right> \left(I+P^\dagger  \left<\Sigma_s\right>\right)^{-1}.
\]
The alternative is to simulate the stochastic dynamics of $\Sigma_t$ for a large number of samples and compute numerical averages. These results can be directly
employed to evaluate the optimal cost-to-go in the control problem $f(\Sigma,t)$.\par
Alternatively, we can look at a system with more complex dynamics, and we take as an example the stochastic damped harmonic oscillator given by the system of equations
\begin{equation}
\label{eq:stoch_osc}
\dot{X}_t = V_t,\quad dV_t=\left(b U_t-\gamma V_t  - \omega^2 X_t \right)dt + \eta^{1/2} dW_t.
\end{equation}
Furthermore, we assume that
the tuning functions only depend on the position of the oscillator, therefore not giving us any information about the velocity. The controller in turn seeks to keep the
oscillator close to the origin while steering only the velocity. This can be achieved by the choice of matrices
$A = (0, 1; -\omega^2, -\gamma)$, $B= (0,0;0, b)$, $D = (0,0;0, \eta^2)$, $R= (0,0;0,r)$, $Q = (q, 0;0 , 0)$ and $P = (p^2,0;0,0)$.\par

In \fref{fig:comparison_uni} we provide the uncertainty-dependent costs for LQG control, for the Poisson observed control, as well as the MMSE for the Poisson
filtering problem and for a Kalman-Bucy filter with the same noise covariance matrix $P$. This
illustrates nicely the difference between Kalman filtering and the Gauss-Poisson filtering considered here. The Kalman filter MSE has a simple, monotonically
increasing dependence on the noise
covariance, and one should simply strive to design sensors with the highest possible precision ($p=0$) to minimise the MMSE and control costs. The Poisson case
leads to optimal performance at a non-zero value of $p$. Importantly the optimal values of $p$ for estimation and control differ. Furthermore, in view of
\fref{sec:mutual_info}, we also
plotted the mutual information between the process $X_t$ and the observation process $N_t$, to illustrate that information-based arguments would lead to the same
optimal encoder as MMSE-based arguments.\par
\begin{figure}
\includegraphics[width=.5\columnwidth]{figure-1-1.pdf}
\includegraphics[width=.5\columnwidth]{figure-1-2.pdf}
\caption{\small{The trade-off between the precision and the frequency of spikes is illustrated for the OU process (a) and the stochastic oscillator (b). In both figures, the initial condition has a very uncertain estimate of the system's state, biasing the optimal tuning width towards higher values. This forces the encoder to amass the maximum number of observations within the duration of the control experiment. Parameters for figure (a)
were: $T =2,\, \gamma = 1.0,\, \eta = 0.6,\, b = 0.2,\, \phi = 0.1,\, \Delta\theta = 0.05,\, Q = 0.1,\, Q_T = 0.001,\, R = 0.1$.  Parameters for figure (b) were $T=5,\,
\gamma=0.4,\,\omega=0.8,\,\eta = 0.4,\, r=0.4,\, q=0.4,\, Q_T = 0,\, \phi = 0.5,\, \Delta\theta=0.1$.   }}
\label{fig:comparison_uni}
\end{figure}

\subsection{Allocating Observation Resources in Anisotropic Control Problems}
\label{sec:determinant}

A second factor that could lead to different optimal encoders in estimation and control is the structure of the cost function $C$. Specifically, if the cost functions
depends more strongly on a certain coordinate of the system's state, uncertainty in that particular coordinate will have a higher impact on expected future costs than
uncertainty in other coordinates. We will here consider two simple linear control systems observed by a population of neurons restricted to a certain firing rate. This
can be thought of as a metabolic constraint, since the regeneration of membrane potential necessary for action potential generation is one of the most significant
metabolic expenditures for neurons \cite{attwell2001energy}. This will lead to a trade-off, where an increase in precision in one coordinate will result in a decrease in
precision in the other coordinate.\par

We consider a
population of neurons whose tuning functions cover a two-dimensional space. Taking a two-dimensional isotropic OU system with state $X_t=(X_{1,t},X_{2,t})^\top$
where both dimensions are uncoupled,
we can consider a population with tuning centres $\theta_m = (\eta^m_1,\eta^m_2)^\top$
densely covering the stimulus space. To consider a smoother class of stochastic systems we will also consider a two-dimensional stochastic oscillator with state
$X_t = (X_{1,t},V_{1,t},X_{2,t},V_{2,t})^\top$, where again, both dimensions are uncoupled, and the tuning centres of the form
$\theta_m = (\eta^m_1, 0, \eta^m_2,0)^\top$, covering densely the position space, but not the velocity space.\par
Since we are interested in the case of limited resources, we will restrict ourselves to populations with a tuning matrix $P$ yielding a constant population firing rate. We
can parametrise these simply as
$P_{OU}(\zeta) = p^2\Diag(\tan(\zeta),\cotan(\zeta))$, for the OU case and $P_{Osc}(\zeta) = p^2\Diag(\tan(\zeta), 0, \cotan(\zeta),0)$
for the stochastic oscillator, where $\zeta \in (0,\pi/2)$.
Note that this will yield the firing rate $\hat{\lambda} = 2\pi p \phi/(\Delta \theta)^2$, independent of the specifics of the matrix $P$.\par

We can then compare the performance of all observers with the same firing rate in both control and estimation tasks.
As mentioned, we are interested in control problems where the cost functions are anisotropic, that is, one dimension of the system's state vector contributes more
heavily to the cost function. To study this case we consider cost functions of the type
\[
c(X_t,U_t) = Q_1 X_{1,t}^2+ Q_2 X_{2,t}^2 + R_1 U_{1,t}^2 +R_2 U_{2,t}^2.
\]
This again, can be readily cast into the formalism introduced above, with a suitable choice of matrices $Q$ and $R$ for both the OU process as for the stochastic
oscillator. We will also consider the case where the first dimension of $X_t$ contributes more strongly to the state costs (i.e., $Q_1 > Q_2$).

The filtering error can be obtained from the formalism developed in \cite{Susemihl2013} in the case of Poisson observations and directly from the Kalman-Bucy
equations in the case of Kalman filtering \cite{Bucy1965}. For LQG control, one can simply solve the control problem for the system mentioned using the
standard methods (see e.g. \cite{astrom2006}). The Poisson-coded version of the control problem can be solved using either direct simulation of the
dynamics of $\Sigma_s$ or by a mean-field approach which has been shown to yield excellent results for the system at hand. These results are summarised in
\fref{fig:comparison_radial}, with similar notation to that in \fref{fig:comparison_uni}. Note the extreme example of the stochastic oscillator, where the optimal encoder
is concentrating all the resources in one dimension, essentially ignoring the second dimension.
\begin{figure}
\includegraphics[width=.5\columnwidth]{comparison_multi_new.pdf}
\includegraphics[width=.5\columnwidth]{figure-2-2.pdf}
\caption{\small{The differential allocation of resources in control and estimation for the OU process (left) and the stochastic oscillator (right). Even though the estimation MMSE leads to a symmetric optimal encoder both in the Poisson and in the Kalman filtering problem, the optimal encoders for the control problem are asymmetric, allocating more resources to the first coordinate of the stimulus.
}}
\label{fig:comparison_radial}
\end{figure}

\section{Conclusion and Discussion}

\label{sec:conclusion}

We have here shown that the optimal encoding strategies for a partially observed control problem is not the same as the optimal encoding strategy for the
associated state estimation problem. Note that this is a natural consequence of considering noise covariances with a constant determinant in the case of Kalman
filtering
and LQG control, but it is by no means trivial in the case of Poisson-coded processes. For a class of stochastic processes for which the certainty
equivalence principle holds we have provided an exact expression for the optimal cost-to-go and have shown that minimising this cost provides us with an encoder that
in fact minimises the incurred cost in the control problem.\par
Optimality arguments are central to many parts of computational neuroscience, but it seems that partial observability and the importance of combining adaptive
state estimation and control have rarely been considered in this literature, although supported by recent experiments.
We believe the present work, while treating only a small subset of the formalisms used in neuroscience, provides a first
insight into the differences between estimation and control. Much emphasis has been placed on tracing the parallels between the two (see \cite{Kalman1960,Todorov2008}, for example), but one must not forget to take into account the differences as well.\par

\bibliography{library}{}
\bibliographystyle{unsrt}

\end{document}


TITLE: Spline with varying tension, selection of tension factor
QUESTION [1 upvotes]: I need to perform a special interpolation, using that kind of basis :
$$\varphi_{i,j}(x) =  a_i + b_ix + c_i(\cosh(\tau\ x) - 1) + d_i(\sinh(\tau\ x) - \tau\ x)$$
where the $a_i$, $b_i$, $c_i$ and $d_i$ are determined with the following constraints :
$$
\begin{array}{|c|c|c|c|}
\hline
&\varphi_{i,j}(x_i)&\varphi_{i,j}(x_{i+1})&\varphi_{i,j}^\prime(x_i)&\varphi_{i,j}^\prime(x_{i+1})\\
\hline
j=0&1&0&0&0\\
\hline
j=1&0&1&0&0\\
\hline
j=2&0&0&1&0\\
\hline
j=3&0&0&0&1\\
\hline
\end{array}
$$
I need to determinate the minimum tension factor $\tau$, on each interval. I've read this paper, but I don't really understood. Does anyone ever used that kind of interpolation ?
Ps. I apologize for my English.

REPLY [3 votes]: We used these "splines with tension" in a CAD system for many years. As J.M. said, the idea is that the user adjusts the tension values until he/she is happy with the shape of the curve. I read the paper quickly, and didn't see how "optimal" tension values are selected. But, clearly, you first have to decide what you're going to optimise. There is no such thing as a universally "best" tension value.
In answer to your question, the minimum tension value is normally taken to be zero, in which case these splines are just regular cubic splines.
We eventually ditched the idea. Splines with exponential terms don't fit nicely into the CAD world, and there are other ways to get "tension" kinds of shape control, using splines that are simply piecewise polynomials.


TITLE: Show that $f=g$ when $\int_0^\varepsilon f dx = \int_0^\varepsilon g dx$ for all $\varepsilon\in\mathbb{R}$
QUESTION [0 upvotes]: I know that this should be pretty straightforward to prove, but I'm trying to find a fairly clean proof for this.  I have one that seems to be far too complicated, and I'm wondering if this is actually easy to prove.
Let $f$ and $g$ be continuous functions on $\mathbb{R}$.  Show that if $\int_0^\varepsilon f \,dx = \int_0^\varepsilon g \,dx$ for all $\varepsilon\in\mathbb{R}$, then $f=g$

REPLY [0 votes]: As an alternate, and more elementary, method to the using the Fundamental Theorem of Calculus:
Let $h=f-g$. We need to show $h$ is identically zero.
Note that $h$ is continuous and that $\int_0^x h(t)\,dt=0$ for all $x$. Show that this implies that $\int_a^b h(t)\,dt=0$ for any numbers $a,b$.  Now assume $h$ is not identically zero. Then by the continuity of $h$  there is an $\alpha>0$ and an interval $(a,b)$ such that either $h(x)>\alpha$ for all $x\in(a,b)$ or $h(x)<-\alpha$ for all $x\in(a,b)$. Consider now the integral $\int_a^b h(t)\,dt$ and derive a contradiction.


TITLE: Is there a generalized upper bound for $\sum_{i = 1}^n | X \cap D_i |$ for $n > 2$?
QUESTION [2 upvotes]: Let $D_1, ..., D_n$ be arbitrary $n$ sets where $D_i \cap D_j \neq \emptyset$. In the simplified case where $n = 2$, we have that
$$
\begin{split}
| X \cap D_1 | + | X \cap D_2 | = &| X \cap (D_1 \setminus D_2) | + |  X \cap (D_1 \cap D_2) | \\
&+ | X \cap (D_2 \setminus D_1) | + |  X \cap (D_1 \cap D_2) | \\
= & |X| + |  X \cap (D_1 \cap D_2) | \\
\leq & |X| + | D_1 \cap D_2 |.
\end{split}
$$
My question is that, can we generalize the above upper bound to something like
$$
\sum_{i = 1}^n | X \cap D_i | \leq |X| + c,
$$
where $c$ is dependent on $(D_1, D_2, ..., D_n)$? It is self-evident that, if $D_1, ..., D_n$ is a disjoint partition of a universe, then we have $\sum_{i = 1}^n | X \cap D_i | = |X|$. However, it seems difficult for me to bound $c$ when $D_1, ..., D_n$ are not disjoint.
It would be appreciated if you could give me any hint.

REPLY [1 votes]: Following Brandons answer, what I would use for an upper bound of similar efficiency (In the spirit of sieve techniques) would be:
$$\sum_{i = 1}^n | X \cap D_i | \leq |X| + n/2\cdot(n-1)\cdot \max_{i\neq j}(|D_i\cap D_j|),$$
This is the truncated sum, and if you can control the size of intersections, and $n$ is relatively small compared to $|X|$ this can yield a good estimate.


TITLE: Functional analysis, Schwartz space, Integration
QUESTION [0 upvotes]: The following question is related to this: Tempered distribution, Schwartz space, functional analysis

For $f:\mathbb{R}\to\mathbb{C}$, $f(x)=|x|$ and $h:\mathbb{R}\to\mathbb{C}$, $h(x)=\begin{cases} 1, ~\text{for}~ x\geq 0\\ -1,~\text{else}\end{cases}$
holds $(S_f)'=S_h$. Calculate the derivative of $S_h$ (therefore $(S_h)'$).

I am not sure what I have to do.
I have to show, that the functions $(S_f)'$ and $S_h$ are equal.
Hence for every $g\in\mathcal{S}(\mathbb{R})$ we have $(S_f(g))'=S_h(g)$. Right?
I tried to evaluate:
$S_h(g)=\int_{\mathbb{R}} h(x)g(x)\, dx=\int_0^\infty g(x)\, dx-\int_{-\infty}^0 g(x)\, dx$
Using integration by parts, I get to
$xg(x)\vert_{x=0}^{x=\infty}-\int_0^\infty xg'(x)\, dx-(xg(x)\vert_{x=0}^{x=\infty}-\int_{-\infty}^0 xg'(x)\, dx=-\int_{\mathbb{R}} xg'(x)\, dx$
But how do I calculate $(S_f(g))'=\left(\int_{\mathbb{R}} |x|g(x)\, dx\right)'$, since this is not even a function, and the derivative should be 0?
Thanks in advance for your comments.

REPLY [1 votes]: $\displaystyle\int_{0}^{\infty}g(x)dx-\int_{-\infty}^{0}g(x)dx=-\int_{0}^{\infty}xg'(x)dx+\int_{-\infty}^{0}xg'(x)dx=-\int_{\bf{R}}xh(x)g'(x)dx$, note that $xh(x)=|x|$.
So $\left<g,S_{h}\right>=-\displaystyle\int_{\bf{R}}f(x)g'(x)dx=-\left<g',S_{f}\right>=\left<g,S_{f}'\right>$.
And $\left<g,S_{h}'\right>=-\left<g',S_{h}\right>=-\displaystyle\int_{0}^{\infty}g'(x)dx+\displaystyle\int_{-\infty}^{0}g'(x)dx=g(0)+g(0)=2g(0)$.


TITLE: Co-coercivity of gradient
QUESTION [4 upvotes]: If $f$ is convex with $dom$ $f$ $= R^{n}$ and $g(x) = x^{T}x - f(x)$ is convex,
how to prove the Co-coercivity of gradient?
$$(\nabla f(x) - \nabla f(y))^{T}(x - y) \geq 1/L \parallel \nabla f(x) - \nabla f(y) \parallel ^{2}_{2}$$

REPLY [4 votes]: I find the answer in ucla lecture Page 1-16.


text:
\begin{document}

\subsection{Study of the approximation rank}
\label{subsection:approximation_rank}
As the rank $\mathbf{r} = \left(r_t\right)_{t \in \mathcal{T}}$ in a tree tensor network is a tuple depending on the underlying tree $\mathcal{T}$, we consider the \emph{maximum rank} $\rmax = \max_{t} r_t$ and the \emph{effective rank} $\reff$. The effective rank $\reff$ of a tensor representation is defined such that the storage cost for this representation equals the cost to store one with rank $\mathbf{r} = \left(\reff \right)_{t \in \mathcal{T}}$. Since intuitively a rank should be an integer, $\reff$ is rounded up to the nearest integer.

We compute low-rank approximations of the marginal distribution $\pPM$ using~\cref{alg:uniformtization} as described in~\cref{subsection:settings}.
In the plots we only show the mean values of the ranks, since their variance among the $100$ realizations of $\PM$ was very small.~\Cref{figure:max_rank_b,figure:eff_rank_b} show the maximum and effective approximation rank of $\pPM$ as a function of the number $d$ of automata for blocks of size $b \in \{2, 4, \frac{d}{2}\}$.
\begin{figure}[tbhp]
	\begin{minipage}[tbhp]{0.45\textwidth}
		\vspace{0pt}
		\centering

		\includegraphics[width =\textwidth]{tikz/mean_maximum_rank}

		\subcaption{\footnotesize{Mean of maximum rank.}}
		\label{figure:max_rank_b}
	\end{minipage}
	\hfill
	\begin{minipage}[tbhp]{0.45\textwidth}
		\vspace{0pt}
		\centering

		\includegraphics[width =\textwidth]{tikz/mean_effective_rank}

		\subcaption{\footnotesize{Mean of effective rank.}}
		\label{figure:eff_rank_b}
	\end{minipage}
	\caption{\footnotesize{Mean of maximum and effective approximation rank of $\pPM$ as a function of the number $d$ of automata using~\cref{alg:uniformtization} with tolerance $\tol=10^{-4}$ and maximum relative truncation error $\eps = 10^{-8}$ for $100$ sample parameters each.}}
\end{figure}
We observe that $\pPM$ is approximated to a tolerance of $\tol = 10^{-4}$ with low rank in all cases.
In particular, for all block sizes $b$ considered, the effective rank increases less than linearly in the number $d$ of automata and is bounded by $\reff < 20$.
As expected, the maximum rank reacts much more sensitively to changes in $d$ and $b$.

For constant block sizes $b = 2$ and $4$ no smooth increase in the number $d$ of automata can be detected.
Instead, maximum and effective rank oscillate in the number $d$ of automata.
Particularly low ranks typically occur when the strongly correlated automata are not separated by the tree structure.
For example, in case of $d = 32$ automata with a block size of $b = 4$ the strongly correlated automata are split only in the lower levels of the tree, and we observe a very low rank.
For $d = 28$ automata with a block size of $b = 4$, however, the strongly correlated automata $\{\A_{13}, \A_{14}, \A_{15}, \A_{16}\}$, for example, are separated already in the first level of the tree, and we observe a much higher maximum rank.
A closer look at the rank components reveals an increase especially at the points where the blocks are split up.

In the case of $b=\frac{d}{2}$ the block size increases linearly in the number $d$ of automata.
In~\cref{figure:max_rank_b} we see a strong increase of the maximum rank for $d=24, 28, 32$ automata.
However, the effective rank grows smoothly and logarithmically in the number $d$ of automata, see~\cref{figure:eff_rank_b}.
This indicates that the maximum rank occurs only sporadically.
A closer look shows that the maximum ranks again occur only in isolated cases at the vertices where highly correlated automata are separated.
This observation and the fact that the effective rank is small implies that most rank components are small as well.

These results suggest that neither the number $d$ of automata nor the size of the blocks $b$ alone are responsible for an increase in rank, but that especially the distribution of the automata in the tree has a large effect.
How to construct an appropriate tree in order to keep ranks low using a-priori information on the parameters is a topic of ongoing research.
\\

We also  analyze how the approximation rank is influenced by the maximum relative truncation error $\eps$ allowed within the algorithm.
To this end, we compute a low-rank approximation of the marginal distribution $\pPM$ for $d = 32$ automata and blocks of size $b = 4$ using~\cref{alg:uniformtization}. The iteration stops when the norm of the relative residual for~\cref{eq:linear_system} is smaller than $\tol \in \{10^{-1}, 10^{-2}, 10^{-3}, 10^{-4}\}$.
In order to achieve this for all $100$ sample parameters, $\eps$ has to be chosen smaller than $10^{-6}$. If we allow for a higher truncation error, the iteration stagnates in some cases.
We again restrict ourselves to plotting mean values since the variances are small.

~\Cref{figure:max_rank_truncation,figure:eff_rank_truncation} show semi-logarithmic plots of the mean of the maximum and effective rank of the approximation of $\pPM$ with $d = 32$ automata and block size $b = 4$ as a function of the maximum relative truncation error $\eps$ for several values of the tolerance $\tol$.
\begin{figure}[tbhp]
	\begin{minipage}[tbhp]{0.45\textwidth}
		\vspace{0pt}
		\centering

		\includegraphics[width =\textwidth]{tikz/mean_maximum_rank_eps}

		\subcaption{\footnotesize{Mean of the maximum rank.}}
		\label{figure:max_rank_truncation}
	\end{minipage}
	\hfill
	\begin{minipage}[tbhp]{0.45\textwidth}
		\vspace{0pt}
		\centering

		\includegraphics[width =\textwidth]{tikz/mean_effective_rank_eps}

		\subcaption{\footnotesize{Mean of the effective rank.}}
		\label{figure:eff_rank_truncation}
	\end{minipage}
		\caption{\footnotesize{Mean of maximum and effective approximation rank of $\pPM$ as a function of the maximum relative truncation error $\eps$ using~\cref{alg:uniformtization} for $100$ sample parameters with $d=32$ automata and block size $b = 4$. The number of iterations necessary to achieve a relative residual smaller than $\tol$ increases logarithmically in $\tol$ independent of $\eps$.}}
\end{figure}
For $\tol = 10^{-2}$ maximum and effective rank are nearly constant with $\rmax \approx 17$ and $\reff \approx 7$.
For $\tol \neq 10^{-2}$, we observe that there are only small differences in maximum and effective rank, respectively.
In these cases, the maximum rank increases logarithmically in the maximum relative truncation error $\eps$, see~\cref{figure:max_rank_truncation}, and the effective rank is nearly constant for all truncation errors, see~\cref{figure:eff_rank_truncation}.
For very small $\eps$ all curves converge.
This observation suggests, on the one hand, that $\pPM$ can be accurately approximated with a tensor of maximum rank $\rmax \approx 17$ and effective rank $\reff \approx 7$, since a more accurate truncation, i.e., smaller $\eps$, has only very small impact on the approximation ranks.
On the other hand, the conclusion that the ranks are low and nearly independent of the tolerance value indicates that the ranks during the iteration are also low.
This allows not only for efficient storage of the resulting approximation but also for efficient computation using~\cref{alg:uniformtization}.

\end{document}

config: arxiv


text:
\section{A combinatorial extended \(S_{n+k}\) action on the poset of the boundary strata of   $Y_{\Fc_{A_{n-1}}}$} \label{sec:combinatorialaction}
This section and the next one are devoted to point out that another hidden  extended action of the symmetric group appears in the geometry of   the minimal models  $Y_{\Fc_{A_{n-1}}}$. This action   is different from the  one,  described in Section \ref{sec:action1},  that motivated the construction of supemaximal models.
More precisely, we are going to deal with a purely combinatorial action  on the poset  \(\Bc(n-1)\) that indexes  the strata of $Y_{\Fc_{A_{n-1}}}$.
This does not correspond to an action on the variety $Y_{\Fc_{A_{n-1}}}$, but it gives rise, as we will see, to an interesting permutation action on the monomials of the Yuzvinski basis of  $H^*(Y_{\Fc_{A_{n-1}}}, \Z)$.

Let us denote by \(F^k(\Bc(n-1))\) the subset of \(\Bc(n-1)\) made by the elements of cardinality \(k+1\).
These elements indicize the \(k\)-codimensional strata of $Y_{\Fc_{A_{n-1}}}$ (the only element in \(F^0(\Bc(n-1))\) is \(\{V\}\) that  corresponds to the big open part).
In \cite{gaifficayley} it has been described an explicit bijection between \(F^k(\Bc(n-1))\) and the set of unordered partitions of \(\{1,2,...,n+k\}\) into \(k+1\) parts of cardinality greater than or equal to 2.
To recall this bijection, we identify the elements of \(\Fc_{A_{n-1}}\) with subsets of \(\{1,2,...n\}\), as in Section \ref{secexamplean}.

\begin{definition}
\label{ordering}
We fix the following  (strict) partial ordering on  \(\Fc_{A_{n-1}}\): given  \(I\) and \(J\) in \(\Fc_{A_{n-1}}\)  we put \(I< J\)  if the minimal number  in \(I\) is less than the minimal number  in \(J\).
\end{definition}

Let us  consider a nested set   \(\MS\)   that belongs to \(F^k(\Bc(n-1))\).
 It can be represented by an oriented  rooted tree on \(n\) leaves as in Section \ref{secexamplean}. The leaves are the sets \(\{1\},\{2\},\cdots,\{n\}\).
 Now we put labels on the vertices of this tree. We start by labelling  the vertices \(\{1\},\{2\},\cdots,\{n\}\) respectively by the labels \(1,2,...,n\).

Then we can partition the set of vertices of the tree into  levels
with the same criterion as in the proof of Theorem
\ref{teo:chiusoazionebuilding}:  level $0$ is made by the leaves, and in general,   level \(j\)  is made by the vertices \(v\) such that the maximal length of an oriented  path that connects \(v\) to a leaf is \(j\).

Now  we  label the internal vertices of the tree in the following way. Let us suppose that there are \(q\) vertices in level 1. These  vertices correspond, by the nested property, to pairwise disjoint  elements of  \(\Fc_{A_{n-1}}\), therefore   we can totally order them using the ordering of Definition \ref{ordering} and  we  label them with  the numbers from \(n+1\) to \(n+q\) (the label  \(n+1\)  goes to the minimum, while \(n+q\) goes to the maximum).

At the same way, if there are   \(t\) vertices in level 2, we can label them with the numbers from \(n+q+1\) to \(n+q+t\) , and so on. At the end of the process, the root is labelled with the number \(n+k+1\).

We can now  associate to such a tree an unordered  partition of \(\{1,2,...,n+k\}\) into \(k+1\) parts  by assigning  to every internal vertex \(v\) the set of the labels of the vertices covered by \(v\) (see Figure \ref{labellednestedtree}).

 \begin{figure}[ht]

 \center
\includegraphics[scale=0.5]{labellednestedtree2}
\caption{On top of the picture there is a  nested set \(S\) with 6 elements  in \(\Fc_{A_{8}}\).  In the middle there is  its representation by an oriented labelled rooted tree. At the bottom one can read  the resulting partition  of \(\{1,2,...,14\}\) into 6 parts.}
 \label{labellednestedtree}
\end{figure}

This bijection allows us to consider new actions of the symmetric group on \(\Bc(n-1)\): every subset \(F^k(\Bc(n-1))\) is equipped with  an action of \(S_{n+k}\).
\begin{remark}\label{rem:comparing}
We notice that when \(k>2\), if we first embed \(S_n\) into \(S_{n+k}\) in the standard way and then restrict  the  \(S_{n+k}\) action to \(S_n\) we do not obtain the natural \(S_n\) action on \(\Bc(n-1)\). For instance, let us consider \(n=5\) and \(k=3\), and the following nested set \(\MS\) in \(\Bc(4)\):
\[\MS=\{\{1,2\}, \{3,4\}, \{3,4,5\}, \{1,2,3,4,5\}\}\]
On one hand, the  natural action of the transposition \((1,3)\) sends \(\MS\) to
\[\MS'= \{\{2,3\}, \{1,4\}, \{1,4,5\}, \{1,2,3,4,5\}\}\]
On the other hand,  the partition of \(\{1,2,3,..,8\}\) associated with \(\MS\) is
\[\{1,2\}, \{3,4\}, \{5,7\}, \{6,8\}  \]
that is sent by the transposition \((1,3)\) to
\( \{2,3\}, \{1,4\}, \{5,7\}, \{6,8\}.  \)

This last partition corresponds to the nested set
\[\MS''= \{\{1,4\}, \{2,3\}, \{2,3,5\}, \{1,2,3,4,5\}\}\]
and we notice that \(\MS'\neq \MS''\).

Moreover, we observe that the  natural \(S_5\) action  on \(F^3(\Bc(4))\) and the \(S_5\) action  restricted from \(S_8\)  differ in the number of orbits, therefore   when we consider the associated permutation representations they differ in the multiplicity of the trivial representation, which is 3 for the natural \(S_5\) representation and 4 for the  restricted one.  \end{remark}

\section{The  \(S_{n+k}\) action on the  basis of $H^*(Y_{\Fc_{A_{n-1}}}, \Z)$} \label{ss:action3}

As we observed in the preceding section, the combinatorial action of \(S_{n+k}\) on  \(F^k(\Bc(n-1))\) can  be read as an action on the \(k\)-codimensional strata of $Y_{\Fc_{A_{n-1}}}$.
Moreover we notice  that this action can in turn be extended to the Yuzvinski basis of $H^*(Y_{\Fc_{A_{n-1}}}, \Z)$ described in Section \ref{constru}.
In fact we can represent the elements of the Yuzvinski basis by {\em labelled partitions} in the way illustrated by the following example.
\begin{example}
\label{esempiobasi}
Let \(n=7\) and let us consider the monomial \(c_{A_1}^2c_{A_2}\) in the Yuzvinski basis of  \(H^6(Y_{\Fc_{A_{6}}}, \Z)\), where \(\{A_1, A_2\}\) is the nested set given by the subspaces \(A_1=\{1,2,3,5\}\), \(A_2=\{4,6,7\}\). Since  \(V\) does not belong to this nested set, we write this monomial as \(c_{A_1}^2c_{A_2}c_V^0\). Now,  according to  the  bijection described in Section \ref{sec:combinatorialaction}, we can associate to the nested set \(\{V, A_1, A_2\}\) the following partition of the set \(\{1,2,..,9\}\):
\[\{1,2,3,5\}\{4,6,7\}\{8,9\}\]
where,  \(A_1\) corresponds to  \(\{1,2,3,5\}\), \(A_2\) corresponds to \(\{4,6,7\}\) and \(V\) corresponds to \(\{8,9\}\).
Finally  we associate to \(c_{A_1}^2c_{A_2}\) the following {\em labelled partition} of \(\{1,2,..,9\}\):
\[\{1,2,3,5\}^2\{4,6,7\}^1\{8,9\}^0\]
As another example,  we represent the monomial \(c_{A_1}^2c_{V}^2\) of \(H^8(Y_{\Fc_{A_{6}}}, \Z)\) by the labelled partition of \(\{1,2,..,8\}\):
\[\{1,2,3,5\}^2\{4,6,7,8\}^2\]
\end{example}
We notice that this representation provides  us with an easy way to `read' the bounds
 for the exponents in the Yuzvinski basis (see the end of Section \ref{constru}). More in detail, the bounds \(d_{(supp\, f)_A,A}^{\{V\}}\) can be translated in this language in the following way. Let \(I\) be a part of  a  labelled partition of \(\{1,....,n+k\}\)  that represents a monomial in the Yuzvinski basis: then  the   exponent (i.e. the label)  \(\alpha_I\) of   \(I\)  satisfies \(0\leq \alpha_I \leq |I|-2\).
 Moreover, it  may be equal to 0 only if \(I\) contains the number \(n+k\), i.e. when \(I\) represents \(V\), and in the monomial the variable \(c_V\) does not appear (that is, according to the convention established before, it  appears with exponent 0). In particular all the sets in the partition have cardinality \(\geq 3\) except possibly for the set containing \(n+k\), that may have cardinality equal to 2.

Now we observe that  \(S_{n+k}\) acts on the labelled partitions of \(\{1,....,n+k\}\) into \(k+1\) parts, and this provides us with a permutation  action on the monomials of the Yuzvinski basis of $H^*(Y_{\Fc_{A_{n-1}}}, \Z)$.
More in detail:
\begin{itemize}
\item \(S_{n+k}\) acts on the set of all the monomials that are represented by a labelled partition of \(\{1,....,n+k\}\) into \(k+1\) parts  with  all the labels \(> 0\).
\item \(S_{n+k-1}\) acts on the set of all the monomials that are represented by a labelled partition of \(\{1,....,n+k\}\) into \(k+1\) parts with  one of  the labels  equal to  \(  0\). In fact if  there is a part labelled by 0, it must contain the number \(n+k\), and \(S_{n+k-1}\) is embedded into  \(S_{n+k}\) as the subgroup that keeps \(n+k\) fixed.
\end{itemize}

This representation, once restricted in the standard way to $S_n$, is not isomorphic to the natural $S_n$
representation\footnote{One can see this for instance by counting the multiplicity of the trivial representation in  \(H^6(Y_{A_7})\). The key point is provided by the monomials of type \(c_{A_1}^1c_{A_2}^1c_{A_3}^1c_V^0\) that span an invariant subspace \(H\) for both the natural \(S_8\) action and the extended \(S_{10}\) action.  By an argument similar to that of Remark \ref{rem:comparing},
i.e. by counting the number of orbits, one can check that on \(H\) the natural \(S_8\) representation and the \(S_8\) representation restricted from \(S_{10}\) differ  in the  the multiplicity of the trivial representation (that is respectively 3 and 4). }.
Therefore it  is also not compatible with  the action on cohomology produced by the extended geometric \(S_{n+1}\) action described in Section \ref{sec:action1}.
Nevertheless it is interesting since it  splits  the  cohomology module  into a sum of   induced representations of the form  \(Ind_G^{S_{n+k}}Id\) or \(Ind_G^{S_{n+k-1}}Id\)  where \(G\) is the stabilizer of a partition of \(\{1,....,n+k\}\).

Moreover, the  orbits of this  action can  be used to write  a generating formula  for the Poincar\'e polynomials of the models \(Y_{\Fc_{A_{n-1}}}\) that is different from the recursive formula for the Poincar\'e series recalled at the beginning of Section \ref{sec:poincaresupermodels}.

Let us denote by \(\Psi(q,t,z)\)  the following exponential generating series:
\[\Psi(q,t,z)= 1+ \sum_{n\geq 2, \; \MS\in \Nc(\Fc_{A_{n-1}})}P(S) z^{|\MS|} \frac{t^{n+|\MS|-1}}{(n+|\MS|-1)!}\]
where, for every \(n\geq 2\),
\begin{itemize}
\item  \(\MS\) ranges over all the nested sets of the building set \(\Fc_{A_{n-1}}\) (i.e., \(\MS\) may not contain \(\{V\}\));
\item  \(P(\MS)\) is the polynomial, in the variable \(q\),  that expresses the contribution to $H^*(Y_{\Fc_{A_{n-1}}}, \Z)$ provided by all the monomials \(m_f\) in the Yuzvinski basis such that \(supp \ f=\MS\).

For instance, with reference to the Example \ref{esempiobasi}, if \(\MS\) is the nested set \(\{A_1,A_2\}\), then \(P(\MS)=(q+q^2)q\) since we have to take into account all the possible ways to label the partition
\[ \{1,2,3,5\}\{4,6,7\}\{8,9\},\]
while if \(\MS\) is \(\{A_1, V\}\) then \(P(\MS)=(q+q^2)^2\) since we are dealing with the possible labellings of  the partition
\[\{1,2,3,5\}\{4,6,7,8\}.\]
\end{itemize}
We observe that the series \(\Psi(q,t,z)\) encodes the same information that is encoded by the Poincar\'e series. In particular, for a fixed \(n\), the Poincar\'e polynomial  of the model \(Y_{\Fc_{A_{n-1}}}\) can be read from the coefficients of the  monomials whose \(z,t\) component is \(t^kz^s\) with \(k-s=n-1\) (see the Example \ref{esempiopolpoincare} at the end of this section).
 \begin{theorem}
 \label{teo:formulapoicareminimale}
 We have the following formula for the series \(\Psi(q,t,z)\):
 \begin{equation}
 \label{formulapoicareminimale} \Psi(q,t,z) = e^t\prod_{i\geq 3}e^{zq[i-2]_q\frac{t^i}{i!}}
 \end{equation}
 where \([j]_q\) denotes the \(q\)-analog of \(j\): \([j]_q=1+q+\cdots+q^{j-1}\).

 \end{theorem}
 \begin{proof}
 We think of the monomials of the Yuzvinsky bases as labelled partitions. Then  we single out the contribution given to \(\Psi\) by all the parts    represented by subsets with cardinality \(i\geq 3\) and with non trivial label.
  If in a partition there is only one such part  its contribution is \(z(q+q^2+\cdots+ q^{i-2})\frac{t^i}{i!}\), if there are \(j\) such parts their contribution is \(z^j(q+q^2+\cdots+ q^{i-2})^j \frac{(\frac{t^i}{i!})^j}{j!}\).
  In conclusion the contribution  of all the parts    represented by subsets with cardinality \(i\geq 3\) and with non trivial label is provided by
 \[e^{z(q+q^2+\cdots+ q^{i-2})\frac{t^i}{i!}}-1\]

 Let us now focus on the contribution to \(\Psi\) that comes  from the parts with cardinality \(i\geq 2\) and with label equal to 0. For every monomial in the basis there is  at most one such  part, and its contribution is \(\frac{t^{i-1}}{(i-1)!}\).  The exponent \(i-1\) (instead of  \(i\)) takes into account that such part does not contribute to the cardinality \(|\MS|\).
The total contribution of the elements with label equal to 0 is therefore     \(\sum_{i\geq 2}\frac{t^{i-1}}{(i-1)!}\). Summing up, we observe that the expression
  \[ e^t\prod_{i\geq 3}e^{zq[i-2]_q\frac{t^i}{i!}}  \]
  allows us to take into account the contribution to \(\Psi\) of all the possible monomials in the Yuzvinski bases.
   \end{proof}

  \begin{example}
 \label{esempiopolpoincare}
 If one wants to compute the Poincar\'e polynomial of \(Y_{\Fc_{A_{4}}}\) one has to single out all the monomials in \(\Psi\) whose  \(z,t\) component is \(t^kz^s\) with \(k-s=4\).
 A product of the exponential functions that appear in the formula (\ref{formulapoicareminimale}) gives:
 \[\frac{t^4}{4!} [1]+ \frac{t^5}{5!}z [16q+6q^2+q^3]+ \frac{t^6}{6!}z^2 [10q^2]\]
Therefore the  Poincar\'e polynomial is \(1+16q+16q^2+q^3\).

 \end{example}


TITLE: CDF from a probability function
QUESTION [0 upvotes]: Hi I have this exercise where I must calculate the CDF and I wanted to see if what I have done is correct:
The exercise goes as follows:
Determine the distribution function F of the probability measure $P$ on $(\mathbb{R}, B(\mathbb{R}))$ with the
following density:
$f(x)=\left\{\begin{matrix}
&0 & x<m-5\\
&\frac{1}{25}(x-m+5) &  m-5 \leq x \leq m\\
&-\frac{1}{25}(x-m-5) & m\leq x \leq m+5\\
& 0 & m + 5 < x
\end{matrix}\right.$
Then, to calculate the CDF I simply calculated the respective integrals:
$F(x)=\left\{\begin{matrix}
& F_1(t) & x<m-5\\
& F_2(t) &  m-5 \leq x \leq m\\
&F_3(t) & m\leq x \leq m+5\\
& F_4(t) & m + 5 < x
\end{matrix}\right.$
I know that $F_1(x)=0$ and $F_4(x)=1$
$F_2(t)=\int_{m-5}^{t}\frac{1}{25}(x-m+5)dx=\dfrac{x\cdot\left(x-2m+10\right)}{50}$
$F_3(t)=\int_{m}^{t}\frac{-1}{25}(x-m-5)dx=\dfrac{x-m-5}{25}$
So at the end we have:
$F(x) =
\left\{\begin{matrix}
 &0  &   x<m-5\\
 & \dfrac{x\cdot\left(x-2m+10\right)}{50} &  m-5 \leq x \leq m \\
 &\dfrac{x-m-5}{25} & m\leq x \leq m+5\\
 & 1 & m + 5 < x
\end{matrix}\right.
$
Have I done any errors or is everything allright?

REPLY [1 votes]: Just extending a little bit the suggestion from @Rik93, using your notation and indicator functions:
$$F(x)=\mathbf{1}_{(m-5\le x \le m)}(x)F_2(x)+\mathbf{1}_{(m< x \le m+5)}(x)[F_2(m)+F_3(x)]+\mathbf{1}_{(m+5<x)}(x)$$


TITLE: Feynman Diagram in QED non-relativistic limit
QUESTION [2 upvotes]: I am following Peskin & Schroeder to read a Feynman diagram. But in this given image, they used the non-relativistic limit to write the incoming fermion gamma mu product. How did he derive the bottom formula? Any explanations in detail?

REPLY [2 votes]: How did he get it?

How did he get what? Which part of the calculation it is that's confusing you. You have listed multiple equalities, which one is at issue?
Eq 3.55 of the pdf online version of Peskin and Schroder shows that:
$$
u^\dagger u=2E_p\xi^\dagger \xi
$$
and since
$$
\gamma_0^2=1
$$
we know that
$$
\bar u\gamma_0u=u^\dagger u
$$
and, in the non-relativistic limit (i.e. $p<<mc$)
$$
E_p\to m\;.
$$
(N.b., $c=1$)


TITLE: What Algebra steps, do I need to take in the following proof (induction proof: Recurrences)
QUESTION [3 upvotes]: Edit: Apparently my question wasn't clear enough so I will go step by step:
I will start from the beginning:
Claim : Let $a_n$ be the sequence defined by $a_1=1, a_2=8,$ and $a_n= a_{n-1} + 2a_{n-2} $ , for $ n \ge 3$
Proof: Prove by strong induction that, for all $ n \in N$,
(*) $a_n= 3 \cdot2^{n-1} + 2(-1)^n$
Base case: When $n= 1$, the left side is $a_1=1$, and the right side is $3 \cdot2^0 + 2(-1)^1 =1 $ . So both sides are equal and (*) is true for $n=1$.
Same is the case with $n=2$. Now I move to induction step.
Induction step: Let $k \in N $ with $k \ge 2 $ be given and suppose (*) is true for n = 1, 2 ..... k. Then:
$a_{k+1} = a_k + 2a_{k-1}$ (by recurrence of $a_n$ )
$=3*2^{k-1} + 2(-1)^k + 2(3*2^{k-2} + 2(-1)^{k-1})  $ (by $n=k$ and $n= k-1$)

This is as far as I got in solving the problem. My question is how do I get to final solution: $3*2^k + 2(-1)^{k+1}$ from the last step I made.
I tried myself but I have gaps in my Algebra  knowledge, so I always got different outcome. In other words, my question is can someone solve a problem above from where I left it, step by step, pointing out why and what he used to get to the solution.
P.S I hope this is clear enough now.If not,  please let me know  what you didn't understand like @ctst did. Also this is my first time, I ever asked online for a help to solve a math problems or used Math Stackexchange.

REPLY [2 votes]: Yes it is clear enaugh, indeed. I don't think you still deserve downvotes.
So, you found that $$a_{k+1}=3\times 2^{k-1}+2(-1)^k+2\times \left(3\times 2^{k-2}+2(-1)^{k-1}\right)\tag{1}$$ Your final result should be $$3\times2^k+2(-1)^{k+1}\tag{2}$$ Notice that this final expression is a sum of two terms: $3$ times a power of $2$ and $2$ times a power of $(-1)$. So why not try to regroup powers of $2$ and powers of $(-1)$ in expression $(1)$, like this:$$a_{k+1}=\left(3\times 2^{k-1}+2\times 3\times 2^{k-2}\right)+\left( 2(-1)^k+2\times 2(-1)^{k-1}\right)$$
All you need now is to simplify this expression. You can do it!