TITLE: Can a $\gamma$-ray photon give some of his energy to an atom and accelerate it?
QUESTION [3 upvotes]: I know gamma-ray photon can only give its momentum energy to the electrons of an atom.
My question is: Can a photon give some of its momentum to the atom (including its nucleus) to give it heat or speed?
If yes, can you tell me how much energy can it give?

REPLY [2 votes]: Not merely can it transfer its momentum as well as its energy when it interacts, but it must.
If the target atom is in a fluid context (liquid, gas, plasma), then that energy and momentum must end up in the target or some other reaction products(s).
In a solid context the Mössbauer effect can be an issue, allowing the transfer of that momentum to a much more massive collection of mass resulting in lower recoil.
It is also worth noting that this requirement to transfer both energy and momentum is the reason a photon can not be absorbed on a isolated point target in free-space:
$$ \gamma + e \to e \tag{forbidden} \,,$$
though the same reaction is possible in the field of a massive partner
$$ \gamma + e + A \to e + A \tag{allowed}$$
or if the target is a compound object with excited states
$$ \gamma + A \to A^* \tag{allowed} \,.$$


TITLE: How Much More Difficult Is to Guess the Pin in the Second Case with Respect to the First One.
QUESTION [1 upvotes]: I have two probability problems that don't know how to solve. Anyone helps please?
The first problem
A royal flush is a hand consisting of ace, king, queen, jack, and ten all in the same suit. Approximately how many distinct hands showing a royal flush are there in a standard 52 card desk? Hint: the card order is matter.
A. 5
B. 50
C. 500
D. 5000
SECOND PROBLEM
A Cell Phone 4 Key Pin Can Be over the Digits or over the Digits plus the Pound and Star Keys. How Much More Difficult Is to Guess the Pin in the Second Case with Respect to the First One.
A. About twice
B. about 16 times
C. about 100 times
D. much more than 100 times
I don't know how to solve this problem. Anyone helps please?
Thanks

REPLY [0 votes]: For the second problem: In case $1$ there are $10$ possibilities for each digit which means 10 possibilities for each digit which means $10^4$ possible combinations, and in case $2$ there are $12^4$ possible combinations.The difficulty ratio is then given by $\frac{12^{4}}{10^{4}}$ which is roughly $2$, so the answer is $A$


TITLE: Permutations: play a song 3 times before all songs have played at least once.
QUESTION [1 upvotes]: A playlist has $10$ songs. Two musics players ($A$ and $B$) implement the "shuffle and repeat" feature a little differently. In $B$, no song will be played twice in a row, but it is possible in $A$. For each music player, find the probability that at least one song is played $3$ or more times before every song is played at least once.

By the pigeonhole principle, all complete playing of the playlist in which no song repeated $3$ times, lasts $10$ to $19$ songs. The approach for $A$ I'm thinking to consider different lengths of a playlist in which it's possible for playlist to finish before a three-peat. The last song to play must play exactly once, and there are $10$ options for the last song. So using permutation with repeats, and considering the number of combinations of songs of a certain length, the number of plays of length $k$ that finishes without a three-peat is
$$n(k) = 10(k-1)!\frac{^9C_{k-10}}{2^{k-10}}$$
with $k-10$ songs played twice. I can normalize everything to $19$ songs, by using $N(k) = n(k)\cdot(10^{19-k})$. My intent is to use $N(k)$ as the number of complement events, for music player $A$. Incidentally, if a song is to play three times before all songs play once, its three times must be within the first 19 songs.
So my answer to part $A$ is $\displaystyle 1 - \frac{\sum_{k=10}^{19}N(k)}{10^{19}} \approx 0.9786$
But I am not sure if that's sound and if there is better method that scales better with the number of songs. I have no idea how to get started with $B$, but I suspect some inclusion-exclusion can be applied, by modifying the solution to $A$.

REPLY [0 votes]: Here is a solution to the problem for music player $A$, using an exponential probability generating function.  Readers not familiar with generating functions can find many resources in the answers to this question: How can I learn about generating functions?
Consider the situation just before every song has been played at least once.  There is one song that has not been played at all, and all the other songs have been played either one or two times; then on the next play the missing song is played, completing the set.  Let's assume for the sake of simplicity that song number $1$ is the song which completes the set, and let $a_n$ be the probability that on the $n$th play (1) song number $1$ has not been played, and (2) all other songs have been played one or two times.  The probability that song number $1$ is played on the $(n+1)$th play, completing the set, is then $(1/10)a_n$. The exponential generating function for $a_n$ is
$$f(x) = \left( \frac{x}{10} + \frac{1}{2!} \left( \frac{x}{10} \right)^2 \right)^9$$
We would like to know $\sum_{n=0}^{\infty} a_n$ (which is actually a finite sum).  We might expand $f(x)$ to do this, but there is a shortcut based on the identity $$\int_0^{\infty} x^n e^{-x} \;dx = n!$$
Because of this identity,
$$\sum_{n=0}^{\infty}a_n = \int_0^{\infty} f(x) e^{-x} \;dx$$
The integral evaluates to $0.0213966$.  (I admit to using a computer algebra system, but a pencil and paper solution should not be too difficult.) The probability of completing the set with song number $1$ is $(1/10)\sum_{n=0}^{\infty} a_n$.  But the assumption that song number $1$ is the missing song was arbitrary, so the probability of a sequence of plays in which every song is played at least once and no song is played more than two times is
$$10 \cdot \frac{1}{10} \sum_{n=0}^{\infty} a_n = \sum_{n=0}^{\infty} a_n = \boxed{0.0213966}$$


text:
\begin{document}

\maketitle

\begin{abstract}
	This article discusses MLMC estimators with and without weights, applied to nested expectations of the form $\esp{f(\esp{F(Y,Z)|Y})}$. More precisely, we are interested on the assumptions needed to comply with the MLMC framework, depending on whether the payoff function $f$ is smooth or not. A new result to our knowledge is given when $f$ is not smooth in the development of the weak error at an order higher than 1, which is needed for a successful use of MLMC estimators with weights.
\end{abstract}

\noindent \emph{Keywords:} Multilevel Monte Carlo; Weighted Multilevel Monte Carlo; Nested Monte Carlo; Weak error expansion.

\noindent \emph{MSC 2010:} primary 65C05; secondary 65C30.

\section{Introduction}

Multilevel estimators are commonly used when the underlying random variable of interest~--~here~$f\pa{\esp{F(Y,Z)|Y}}$, with $Y$ and $Z$ independent as far as nested simulation is concerned~--~cannot be simulated exactly at a reasonable computational cost. However, such approximations ~--~here  $f\pa{\frac 1N \sum_{k=1}^N F(Y,Z_k)}$~-- induce some bias. Nested simulation is one of the two most popular setting where Multilevel method are implemented, the other being the numerical schemes associated to stochastic dynamics.

The optimal calibration and the resulting performances of Multilevel Monte Carlo estimators depend on the weak and strong error rate of convergence of these simulable proxies
toward  $f\pa{\esp{F(Y,Z)|Y}}$. By weak error, we mean here an expansion of the bias as a function of a given parameter $h$ representative of the (inverse) complexity.

The existence of weak error expansions at order one leads to the (regular and original) Multilevel Monte Carlo (MLMC) method  introduced by M. Giles in~\cite{Gi08}, whereas higher order expansions led naturally  to develop a weighted multilevel framework, called Richardson-Romberg Multilevel  method (ML2R) introduced in~\cite{LePa17}. However, the existence of such an expansion not only depends upon   random variable of interest and its approximations but also on the regularity of the ``payoff'' function $f$, as it has been widely popularized by the analysis of time discretization schemes of Brownian diffusion processes (see~\cite{TaTu90} and~\cite{BaTa196}).

The seminal result concerning the first order weak error expansion  for nested Monte Carlo simulation when $f$ is not regular~--~namely a quantile~--~is due to Gordy and Juneja in~\cite{GoJu10}.

For such indicator function the strong rate of convergence remains slow and reduces the efficiency of regular multilevel estimators since their performances are ruled by this strong convergence rate. In particular they no longer behave  as unbiased or almost unbiased estimators as it is the case for faster strong convergence regimes.

By contrast weighted multilevel estimators are still  almost unbiased in some sense but this performance strongly relies on higher order expansions of the weak error. So the main objective and result of this paper is to establish (see Proposition~\ref{prop:nested_weakII}) such a higher order expansion for non-smooth payoff function $f$ in a nested simulation  framework.

Let us briefly recall the multilevel paradigm (see~\cite{PaNumProb}). Let $Y_0 \!\in L^2(\Omega, {\cal A}, \PP)$ be a random variable and $Y_h$, $h\!\in {\cal H}= \big\{\frac{\mathbf{h}}{n}, \, n\ge1\big\}$ be a family of approximations of $Y_0$ such that $\lim_{h\to 0}\|Y_h-Y_0\|_2 =0$ with a simulation cost of the form
\[
{\rm Cost}(Y_h) = \kappa h^{-1}
\]
so that the parameter $h$ is inverse linear in the complexity.  Its role in the weak expansion error will lead us to call it {\em bias parameter}.

The central idea behind the regular {MLMC} estimator is to consider a $R$-tuple of parameters $h_j = h/M^{j-1}$, $j=1,\ldots,R$ ($h\!\in {\cal H}$) and to write the telescopic sum
\[
\esp{Y_{h_R}} = \esp{Y_{h_1}}  + \sum_{j=2}^R \esp{Y_{h_j}-Y_{h_{j-1}}}
\]
which suggests to introduce the estimator (see~\cite{Gi08})
\begin{equation} \label{def::MLMC}
\widehat I^N_{h,R,q} = \frac{1}{N_1} \sum_{k=1}^{N_1} Y_{h}^{(1),k} + \sum_{j=2}^R \frac{1}{N_j} \sum_{k=1}^{N_j} \pa{Y_{h_j}^{(j),k} - Y_{h_{j-1}}^{(j),k}}
\end{equation}
where $\big( Y_{h_j}^{(j),k} \big)_{k=1,\ldots,N_j}$ are independent copies as $k$ varies of $Y^{(j)}_{h_j}$ itself ``attached''  to $Y^{(j)}_0$ where $(Y^{(1)}_0, \ldots,Y^{(R)}_0)$ are i.i.d. with the same distribution as $Y_0$. The size $N_j$ of each simulation at {\em level} $j$ is of the form $N_j = \lceil q_jN\rceil$, $j=1,\ldots,R$.

If a first order weak expansion error assumption
\begin{equation} \tag{$WE_{\alpha,1}$}
    \esp{Y_h} = \esp{Y_0} + c_1 h^{\alpha} +o(h^{\alpha})
\end{equation}
is fulfilled for some $\alpha>0$, then
\[
\esp{\widehat I^N_{h,R,q}} = \esp{Y_{h_R}}
 = \esp{Y_0} + c_1\frac{h}{M^{R-1}} + o\big(h/M^{R-1}\big).
 \]
 which dramatically reduces the bias compared to a crude Monte Carlo simulation based on i.i.d. copies of $Y_h$. At this stage the calibration of the allocation parameters $q_1,\ldots,q_{_R}$ across the $R$ levels relies on  a strong error convergence rate assumption
\begin{equation}
\label{strong_error} \tag{$SE_\beta$}
    \forall\, h,h'\!\in \Hr,\quad \normLp[1]{2}{Y_h-Y_{h'}} \le V_1 |h-h'|^{\beta},
\end{equation}
 or its variants (see~$e.g.$~\cite{PaNumProb} among other references where this conditions are discussed). Thus, it happens that $Y_h-Y_{h'}$ is replaced by a random variable $Y_{h,h'}$ satisfying~\eqref{strong_error} and such that $\esp{Y_{h,h'}} = \esp{Y_h-Y_{h'}}$. This calibration aims at minimizing the {\em effort} of the estimator $\widehat I^N_{h,q,R}$, that is the product of its variance by its complexity, given a prescribed Root Mean Square Error ({\emph RMSE}) level $\|\widehat I^N_{h,R,q}-Y_0\|_2 \le \varepsilon$.

If a higher order weak error expansion can be established, namely
\begin{equation}
    \label{weak_error} \tag{$WE_{\alpha,R}$}
\esp{Y_h} = \esp{Y_0} +\sum_{r=1}^R c_r h^{\alpha r} + o(h^{\alpha R}),
\end{equation}
then there exists {\em weights} $(\mathbf{w}_j)_{j=1,\ldots,R}$, {\em only depending on $\alpha$, $M$ and $R$} such that $\sum_{1\le j\le R} \mathbf{w}_j =1$ and satisfying
\[
 \sum_{j=1}^{R} \mathbf{w}_j \esp{Y_{h_j}}  = \esp{Y_0} +\widetilde{ \mathbf{w}}_{_{R+1}} c_R h^{\alpha R} + o\Big(h^{\alpha R}\Big).
\]
These weights, solution to a Vandermonde system (see~\cite{LePa17}), as well as $\widetilde{\mathbf{w}}_{_{R+1}}$ have closed formulas ($ \widetilde{\mathbf{w}}_{_{R+1}}= \sum_{i=1}^R \mathbf{w}_i n_i^{-\alpha R}$). This naturally leads to define the {\em weighted multilevel estimator} (or {\em Richardson-Romberg multilevel estimator}, ML2R) as
\begin{equation}\label{def::ML2R}
\widetilde I^N_{h,R,q} = \frac{1}{N_1} \sum_{k=1}^{N_1} Y_{h}^{(1),k} + \sum_{j=2}^R \frac{\W_j^R}{N_j} \sum_{k=1}^{N_j} \pa{Y_{h_j}^{(j),k} - Y_{h_{j-1}}^{(j),k}},
\end{equation}
where $\mathbf{W}^R_j=  \mathbf{w}_j + \cdots+ \mathbf{w}_{_R}$, $j=1,\ldots,R$ and the $Y^{(j),k}_{h_j}$ are as above. One checks that such an estimator ``kills'' the bias  in a much more efficient manner yet since
\[
\esp{\widetilde I^N_{h,R,q}}  = \esp{Y_0} + \widetilde{ \mathbf{w}}_{_{R+1}} c_R h^{\alpha R} + o\Big(h^{\alpha R}\Big).
\]

Then $\widetilde I^N_{h,R,q} $ can be calibrated like the MLMC estimator to minimize its effort for prescribed \emph{RMSE}. For more precise results on the performances of these two families of estimators, we refer to~\cite{LePa17} or~\cite{Gi17} or~\cite{PaNumProb}. But the important fact to be kept in mind is that, as far as nested Monte Carlo simulations are concerned with $f = \mbox{\bf 1}_{[a,+\infty)}$ (see next section for the specification of the r.v. $Y_h$ for this purpose), the $\beta $ parameter is lower than $1$ (see Proposition~\ref{prop::strong_error_indicator}) so that, as a consequence, the ML2R estimator $\widetilde I^N_{h,R,q} $ behaves ``almost'' like an unbiased estimator, for which the cost is known to be $K \varepsilon^{-2}$, $K>0$ constant.  More precisely, if~\eqref{weak_error} holds for every {\em depth}  $R\ge 1$ and $\lim_{R\to\infty}|c_R|^{\frac 1R} = \widetilde c_\infty \in (0, +\infty)$, then
\[
\mathrm{Cost}\left(\widetilde I^{N(\varepsilon)}_{h(\varepsilon),R(\varepsilon),q(\varepsilon)} \right) \preceq  K_{\alpha, \beta,M}\varepsilon^{-2}\cdot e^{\frac{1-\beta}{\sqrt{\alpha}}\sqrt{2\log(1/\varepsilon)\log(M)} },
\]
where we recall that $f(\varepsilon) \preceq g(\varepsilon)$ if and only if $\limsup_{\varepsilon\to0} g(\varepsilon)/f(\varepsilon) \leq 1$, $K_{\alpha, \beta,M}>0$ is constant and we highlight that $e^{\frac{1-\beta}{\sqrt{\alpha}}\sqrt{2\log(1/\varepsilon)\log(M)} } = o\left( \varepsilon^{-\eta}\right)$ for all $\eta>0$. Note that some numerical experiments carried out in~\cite{LePa17} and in~\cite{Gi17} confirm the fact that weighted multilevel ML2R simulations outperform regular MLMC estimator.

The paper is organized as follows. In Section \ref{sec::nested} we give the description of the nested framework. In Section \ref{sec::useful_results} we give some useful results which will be valid in both frameworks, both $f$ smooth and not. Section \ref{sec::smooth} is devoted to the smooth case, with a particular attention to the antithetic approach, and in Section \ref{sec::non_smooth} we treat the non smooth case and we give a new result concerning the weak error.

\section{Nested Monte Carlo simulation}
\label{sec::nested}

The purpose of the so-called {\em nested} Monte Carlo method is to compute by simulation nested expectations of the form
\begin{equation*}
    \esp{ f\pa[1]{\esp{\Xi|Y}} },
\end{equation*}
where $(\Xi,Y)$ is an $\R\times \R^d$-valued  couple of random variables defined on a probability space $(\Omega,{\cal A}, \PP)$ satisfying $\Xi\!\in L^2$ and $f: \R \to \R$ is a specified function such that $f(\esp{\Xi|Y})\!\in L^2$.

We assume that there exist a Borel function  $F:\R^d\times \R^q \to \R$ and a random vector $Z: (\Omega,{\cal A})\to \R^{q}$ independent of $Y$ such that
\begin{equation*}
	\Xi = F(Y, Z).
\end{equation*}
Let us introduce the Borel function $\phi_0:\R^d \to \R$  defined by $\phi_0(y) = \esp{F(y, Z)}$ so that one may set $\esp{F(Y, Z)|Y} = \phi_0(Y)$.
Then one has the following representation
\begin{equation*}
    \esp{\Xi | Y} = \phi_0(Y) = \int_{\R^{q}} F(Y, z) \PP_{Z}(\d z).
\end{equation*}
To comply with the multilevel framework, we set $K_0\in\N^*$ and  $\Hr = \ac{1/K, \, K\!\in K_0\N^*}$,
\begin{equation*}
	 X_0 := \esp{\Xi|Y}, \quad
	 X_h := \frac 1K \sum_{k=1}^K F(Y, Z_k) \;\text{ with }\; h = \frac{1}{K}\in\Hr,
\end{equation*}
where $(Z_k)_{k\ge 1}$ is an i.i.d. sequence of random vectors with  the same  distribution as $Z$, defined on $(\Omega,{\cal A}, \PP)$  and independent of $Y$ (up to an enlargement of the probability space  if necessary) and
\begin{equation*}
	Y_0 := f(X_0), \quad
	Y_h := f(X_h).
\end{equation*}
To prove that the nested Monte Carlo estimator satisfies the bias error expansion~\eqref{weak_error} and the strong approximation error~\eqref{strong_error}, we introduce the random functions, $\forall y \in \R^d,$
\begin{align}
    D(y) &= F(y, Z) - \esp{F(y, Z)}, \\
    E_h(y) &= \frac{1}{K} \sum_{k=1}^K \Big( F(y, Z_k) - \esp{F(y, Z)} \Big) = \frac{1}{K} \sum_{k=1}^K F(y, Z_k) - \phi_0(y).
\end{align}
Note that $E_h(y)$ is the statistical error of the inner Monte Carlo estimator, which can be rewrited as $\ds E_h(y) = \frac{1}{K} \sum_{k=1}^K D(y)^{(k)}$ where $(D(y)^{(k)})_{k \ge 1}$ is a sequence of \iid copies of $D(y)$, and that $E_h(Y) = X_h-X_0$.

We distinguish between two main frameworks, depending on whether or not $f$ is smooth, a classical example of non-smoothness being $f= \un_{(a,b)}$ (see~\cite{DeLo09}).
When the function $f$ is smooth enough, say $f \in {\cal C}^{1+\rho}(\R,\R)$ with $\rho\!\in (0,1]$, a variant of the former Multilevel \emph{nested} estimator has been proposed in~\cite{BuHaRe15},~\cite{HaAl12} and~\cite{ChLi12} (see also~\cite{Gi15}) to improve the  rate of strong convergence in order to attain the asymptotically unbiased setting, namely~\eqref{strong_error} with $\beta>1$.
A root $M \ge 2$ being given, the idea is  to replace in the successive refined levels of the MLMC and ML2R estimators (see~\eqref{def::MLMC} and~\eqref{def::ML2R}) the difference $Y_{\frac hM}-Y_h$ (where $h= \frac{1}{K}$, $K\!\in K_0\N^*$)  by an antithetic type as follows
\[
Y_{h,\frac hM} :=  f\left(\frac{1}{MK} \sum_{k=1}^{MK} F\big(Y, Z_{k}\big)\right)- \frac 1M \sum_{m=1}^M f\left(\frac 1K \sum_{k=1}^K F\big(Y, Z_{(m-1)K+k}\big)\right) .
\]
It is clear that $\esp{Y_{h,\frac hM} } = \esp{Y_{\frac hM}-Y_{h} }$.

Before getting into the smooth and non smooth case, we give some useful results that will be valid in both frameworks and will be used to establish the higher order of weak error expansion.

\section{Useful results}
\label{sec::useful_results}

Following Comtet~\cite{Co74}, we introduce the partial Bell polynomials $\BB_{n, k}$ for $n \ge 1$ and $k =1,\dots,n$ defined by
\begin{equation} \label{def:partialBell}
	\BB_{n, k}(x_1, \dots, x_{n-k+1}) = \sum \frac{n !}{\ell_1! \cdots \ell_{n-k+1}!} \pa[2]{\frac{x_1}{1!}}^{\ell_1} \cdots\pa[2]{\frac{x_{n-k+1}}{({n-k+1})!}}^{\ell_{n-k+1}}
\end{equation}
where the summation takes place over all integers $\ell_1, \dots, \ell_n \ge 0$, such that $\ell_1 + 2 \ell_2 + \cdots + (n-k+1) \ell_{n-k+1} = n$ and $\ell_1 + \cdots + \ell_{n-k+1} = k$. Note that $\deg (B_{n,k})=k$.
The complete Bell polynomials $\BB_n$ are defined by
\begin{equation*}
	\BB_n(x_1,\dots,x_n) = \sum_{k=1}^n \BB_{n, k}(x_1, \dots, x_{n-k+1}).
\end{equation*}
The first statement is a formal Taylor expansion with integral remainder of $\esp{g( X_h)}$ around $\esp{g( X_0)}$, with $g:\RR\to\RR$ a test function.

\begin{Lemma}[Taylor expansion] \label{lem:taylor} Let $R \ge 0$ and let $g:\R \to \mathbf{C}$ be a $2R+1$ times differentiable function.

Assume $\Xi = F(Y, Z) \in L^{2R+1}$ and let $\kappa_j(\xi)$ be  the $j$--th cumulant (a.k.a. semi-invariant) of a random variable $\xi$. We set $\kappa_{j,y}:= \kappa_j(D(y))$ with $D(y) = F(y, Z)-\esp{F(y, Z)}$ for $y \in \R^d$ and $j \in \ac{1,\dots,R}$. Let $(\BB_{n,k})_{1\le k\le n}$ be  the {\em partial Bell polynomials} defined by~\eqref{def:partialBell}. We then define for $r \in \N$, $r+1 \le n \le 2r$ and every $y \in \R^d$,
$$
b_{r,n-r}(y) = \BB_{r,n-r}\pa{\frac{\kappa_{2,y}}{2}, \dots, \frac{\kappa_{2r-n+2,y}}{2r-n+2}}.
$$
Then
\begin{equation} \label{eq:taylor_nested}
    \forall h \in \Hr, \quad \esp{g( X_h)} = \esp{g( X_0)}
+ \sum_{r=1}^{2R-1} c(r,(2r+1)\wedge 2R) h^r + \calig{R}_{2R+1},
\end{equation}
with
\begin{equation} \label{eq:taylor_coeff}
 c(r,k) = \frac{1}{r!} \sum_{\ell = r+1}^{k} \esp{g^{(\ell)}( X_0) b_{r, \ell-r}(Y)}, \;  1 \le r < k \le 2r+1,
\end{equation}
and
\begin{equation} \label{eq:taylor_reste}
    \calig{R}_{2R+1} = \frac{1}{(2R)!} \esp{\int_0^{ X_h- X_0} g^{(2R+1)}(t+ X_0) ( X_h- X_0-t)^{2R} \d t}.
\end{equation}
\end{Lemma}

\begin{proof}[Proof]
    The case $R=0$ is trivial, since it is a direct application of the fundamental theorem of calculus.

Let $R \ge 1$ be an integer. The Taylor formula at order $2R$ applied to $g$ at $\phi_0(y)$ reads
\begin{equation} \label{eq:taylorBell}
    \esp{g\pa[3]{\frac{1}{K} \sum_{k=1}^K F(y, Z_k)}} = g(\phi_0(y)) + \sum_{n = 1}^{2R} \frac{g^{(n)}(\phi_0(y))}{n !} \esp{(E_h(y))^n} + \calig{R}_{2R+1}(y),
\end{equation}
where $\calig{R}_{2R+1}(y) = \frac{1}{(2R)!} \esp{\int_{0}^{E_h(y)} g^{(2R+1)}(t + \phi_0(y)) (E_h(y) - t)^{2R} \d t}$.

The Bell polynomials allow us to explicitly compute the moments $\esp[1]{(E_h(y))^n}$, $n =1,\dots,R$ of $E_h(y)$ as follows. Let $\kappa_{j,y} = \kappa_j(D(y))$, $j = 1,\dots,R$, $y\in\R^d$. Additivity and homogeneity of cumulants give
\begin{equation*}
    \forall j = 1, \dots, 2R-1, \quad \kappa_j\pa[1]{E_h(y)} = h^{j-1} \kappa_{j,y}.
\end{equation*}
Moments of $E_h(y)$ can be expressed in terms of cumulants using complete Bell polynomials (see~\cite{Co74} p.160 Equation$(2)$) as:
\begin{equation*}
	\esp[1]{(E_h(y))^n} = \BB_n\pa[2]{\kappa_1\pa[1]{E_h(y)}, \dots, \kappa_n\pa[1]{E_h(y)}}.
\end{equation*}
First note that $\kappa_{1,y} = 0$ so that $\kappa_1\pa[1]{E_h(y)} = 0$. Moreover, it follows from the definition~\eqref{def:partialBell} that $\BB_{n, k}$ is $k$--homogeneous, consequently
\begin{align*}
    \esp[1]{(E_h(y))^n} = h^{n} \sum_{k=1}^n h^{-k} \BB_{n, k}\pa[1]{0, \kappa_{2,y}, \dots, \kappa_{n-k+1,y}}.
\end{align*}
We again derive from~\eqref{def:partialBell} that $\BB_{n,n} (0) = 0$, hence the last term in the above sum is null. In particular the sum in~\eqref{eq:taylorBell} starts from $n=2$.
Note now that
\begin{equation*}
    \BB_{n, k}\pa{0, \kappa_{2,y}, \dots, \kappa_{n-k+1,y}}
    = \begin{cases}
        \frac{n!}{(n-k)!} b_{n-k, k}(y) & \text{if $1 \le k \le \lceil n / 2 \rceil$}, \\
        0 & \text{if $k > \lceil n / 2 \rceil$},
    \end{cases}
\end{equation*}
with $\ds b_{n-k, k}(y) = \BB_{n-k, k}\pa{\frac{\kappa_{2,y}}{2},\dots, \frac{\kappa_{n-2k+2,y}}{n-2k+2}}$ which implies that

\begin{align} \label{eq:momentsBell}
    \esp[1]{(E_h(y))^n} = h^{n} \sum_{k=1}^{\lceil{n/2\rceil}} h^{-k}
    \frac{n!}{(n-k)!} b_{n-k, k}(y).
\end{align}
Plugging~\eqref{eq:momentsBell} in~\eqref{eq:taylorBell} gives, since the sum starts at $n=2$ as mentioned above,
\begin{equation*}
    \esp{g\pa[3]{\frac{1}{K} \sum_{k=1}^K F(y, Z_k)}}
    = g(\phi_0(y)) + \sum_{n = 2}^{2R} g^{(n)}(\phi_0(y))
    \sum_{k=1}^{\lceil n/2 \rceil} \frac{h^{n-k}}{(n-k)!} b_{n-k,k}(y)
	+ \calig{R}_{2R+1}(y).
\end{equation*}
Setting $r = n-k$ in the above expression, noting that $\lceil n/2\rceil + \lfloor n/2 \rfloor = n$ and that $\lfloor n/2 \rfloor\le r$ if and only if $n\leq 2r+1$, on derives by interchanging the sums that
\begin{equation*}
\esp{g\pa[3]{\frac{1}{K} \sum_{k=1}^K F(y, Z_k)}} = g(\phi_0(y)) + \sum_{r = 1}^{2R-1} \frac{h^{r}}{r!} \pa{\sum_{n = r+1}^{(2r+1) \wedge 2R} g^{(n)}(\phi_0(y)) b_{r, n-r}(y)}
	+ \calig{R}_{2R+1}(y).
\end{equation*}
We conclude by integrating with respect to $\PP_Y(\d y)$.
\end{proof}

Taking advantage of this expansion we will derive two results. First a bias error expansion for smooth enough payoff functions, in which no regularity is required on the law of $( X_0,  X_h)$ (see Subsection~\ref{sec:smooth_bias_error}). Conversely a second result will be established relying on the regularity of the distribution of $(X_0,  X_h)$ when the payoff function is not smooth (see Subsection~\ref{sec:non_smooth_bias_error}).

As concerns the strong error, elementary computations show that, if $\Xi\!\in L^2$, then
\begin{equation}
\label{eq::nested_X_L2_bound}
    \normLp[1]{2}{X_h- X_0}^2 = \frac{1}{K} \int \PP_{Y}(\d y) {\rm var}\big(F(y,Z)\big) =  h \esp{(F(Y,Z)-\phi_0(Y))^2}  \leq  h\,{\rm var}\big(F(Y,Z)\big),
\end{equation}
since $\phi_0(Y) = \esp{F(Y,Z)|Y}$. To prove~\eqref{strong_error} we extend this result to $\normLp[1]{p}{X_h - X_{h'}}$ when $\Xi \in L^p$, $p > 1$, as described in Lemma~\ref{lemma::strong_conv_useful}.

Note that from now on we give the results for a generic $p>1$ instead of $p=2$, because this wider assumption can be useful to establish a condition of uniform  integrability needed to prove a Central Limit Theorem (and strong law of large numbers) for Multilevel Monte Carlo estimators, see Lemma 5.2 in~\cite{GiLePa17}.

The proof of Lemma~\ref{lemma::strong_conv_useful} relies on the Marcinkiewicz-Zygmund inequality that we recall for clarity. If $(\xi_n)_{n \ge 1}$ is a sequence of centered independent random variables such that $\esp{\abs{\xi_n}^p} < +\infty$, $1 < p < +\infty$, then
\begin{equation} \label{eq:marcinkZ}
    \normLp[4]{p}{\sum_{k=1}^K \xi_k} \le (B_p)^{\frac{1}{p}} \normLp[4]{\frac{p}{2}}{\sum_{k=1}^K \xi_k^2}^{\frac{1}{2}},
\end{equation}
where $B_{p} = \frac{18p^{\frac 32}}{(p-1)^{\frac 12}}$ (see~\cite{Sh96} p.499). If moreover $(\xi_n)_{n \ge 1}$ are identically distributed we have
\begin{equation} \label{eq:marcinkZ_iid}
    \normLp[4]{p}{\sum_{k=1}^K \xi_k} \le (B_p)^{\frac{1}{p}} \sqrt{K} \normLp[1]{p}{\xi_1}
\end{equation}
We make an intensive use of this inequality in Section~\ref{sec::smooth}.

\begin{Lemma}\label{lemma::strong_conv_useful}
Assume $\Xi\!\in L^p$, $p > 1$.  Then, for every $h$, $h'\!\in \Hr$,
\begin{equation}\label{eq:Yh-Yhprime}
    \normLp[1]{p}{X_h- X_{h'}} \le 2B_p \normLp[1]{p}{\Xi - \esp{\Xi|Y}} |h-h'|^{\frac 12}.
\end{equation}
\end{Lemma}

\begin{proof}[Proof] Assume first that $h' \le h$, and set $K=\frac 1h$, $K'=\frac{1}{h'} \ge K$.
First note that by Fubini's theorem
\[
    \normLp[1]{p}{X_h- X_{h'}}^p = \int_{\R^d} \PP_{_Y}(\d y) \esp{\left|E_h(y) - E_{h'}(y)
\right|^p}.
\]
Setting $\widetilde F(y,z) = F(y,z)-\phi_0(y)$, we write
\begin{equation*}
    E_h(y) - E_{h'}(y) = \pa{\frac{1}{K} - \frac{1}{K'}} \sum_{k=1}^K \widetilde F(y,Z_k) + \frac{1}{K'} \sum_{k=K+1}^{K'} \widetilde F(y,Z_k).
\end{equation*}
Then, for every $y\!\in \R^d$, it follows from Minkowski's Inequality,
\begin{equation*}
    \normLp{p}{E_h(y) - E_{h'}(y)}
        \le |h-h'| \normLp[4]{p}{\sum_{k=1}^K \widetilde F(y,Z_k)}
        + h'\normLp[4]{p}{\sum_{k=K+1}^{K'} \widetilde F(y,Z_k)}.
\end{equation*}
Applying Marcinkiewicz-Zygmund Inequality to both terms on the right hand side of the  above inequality yields
\begin{align*}
    \normLp{p}{E_h(y) - E_{h'}(y)}
    &\le |h-h'| B_p \normLp[4]{\frac{p}{2}}{\sum_{k=1}^K \widetilde F(y,Z_k)^2}^{\frac 12}
    + h' B_p \normLp[4]{\frac{p}{2}}{\sum_{k=K+1}^{K'}\widetilde F(y,Z_k)^2}, \\
    &\le |h-h'|B_p K^{\frac 12} \normLp[1]{p}{\widetilde F(y,Z)}
    + h'B_p(K'-K)^{\frac 12} \normLp[1]{p}{\widetilde F(y,Z)}.
\end{align*}
Finally, for every $y\!\in \R^d$,
\begin{align*}
    \normLp{p}{E_h(y) - E_{h'}(y)}
    &\le B_p \normLp[1]{p}{\widetilde F(y,Z)} \left( (h-h') \frac{1}{\sqrt{h}} +h'\Big(\frac{1}{h'}-\frac{1}{h}\Big)^{\frac 12}\right)\\
    &= B_p \normLp[1]{p}{\widetilde F(y,Z)} (h-h')^{\frac 12}\left( \Big(1-\frac{h'}{h}\Big)^{\frac12}  +\Big( \frac{h'}{h}\Big)^{\frac 12}\right)\\
    &\le 2 B_p \normLp[1]{p}{\widetilde F(y,Z)} (h-h')^{\frac 12}.
\end{align*}
Plugging this bound in the above equality yields, owing to Minkowski's Inequality and Jensen's Inequality for conditional expectations, the announced result
\begin{align*}
    \normLp{p}{X_h - X_{h'}}^p & \le (2 B_p)^p  \int_{\R^d} \PP_{_Y}(\d y)
    \normLp[1]{p}{\widetilde F(y,Z)}^p (h-h')^{\frac p2}\\
    &= (2 B_p)^p \normLp[1]{p}{\Xi-\esp{\Xi|Y}}^p (h-h')^{\frac p2}.

\end{align*}
\end{proof}

\section{Smooth payoff function}
\label{sec::smooth}

We first focus on the smooth case, where we give a bias error expansion and a strong convergence rate when the payoff function $f$ is smooth. This result beyond its direct application will be an important step when dealing with indicator functions.

\subsection{Weak error}
\label{sec:smooth_bias_error}
The bias error expansion of the nested Monte Carlo estimator when $f$ is smooth is a consequence of Lemma~\ref{lem:taylor}, as we emphasized in the proof of the following Proposition.

\begin{Proposition} [Bias error (I): smooth functions] \label{prop:nested_weakI} Let $R \in \N^*$ and let $f:\R \to \R$ be a $2R+1$ times differentiable payoff function with bounded derivatives $f^{(k)}$, $k=R+1,\dots,2R+1$.  Assume $X \in L^{2R+1}$. Then there exists $c_1, \dots, c_R$ such that
\begin{equation}
    \forall h \in \Hr, \quad \esp{f( X_h)} = \esp{f( X_0)} + \sum_{r=1}^{R} c_r h^r + \calig{O}(h^{R+1/2}).
\end{equation}
\end{Proposition}
\begin{proof}[Proof]
    Applying Lemma~\ref{lem:taylor} with the function $g=f$ we get, for every $h \in \Hr$,
\begin{equation}
    \esp{f( X_h)} = \esp{f( X_0)} + \sum_{r=1}^{R-1} c(r,2r+1) h^r + c(R,2R) h^R + \sum_{r=R+1}^{2R-1} c(r,2R) h^r + \calig{R}_{2R+1},
\end{equation}
with $c(r,k)$ defined in~\eqref{eq:taylor_coeff} and $\calig{R}_{2R+1}$ in~\eqref{eq:taylor_reste}. Establishing  the proposition amounts  to proving that the remainder term $\calig{R}_{2R+1}$ is well controlled. Using that $f^{(2R+1)}$ is bounded, we have
\begin{equation*}
	\abs{\calig{R}_{2R+1}} \le \frac{\norm{f^{(2R+1)}}_{\infty}}{(2R+1)!} \esp{\abs{ X_h -  X_0}^{2R+1}}.
\end{equation*}
Using successively  the Marcinkiewicz-Zygmund Inequality and the Minkowski Inequality for the  $L^{R+\frac 12}(\PP)$-norm, we get, keeping in mind that $h= \frac 1K$,
\begin{align*}
    \esp{\abs{E_h(y)}^{2R+1}}
    & \le (B_{2R+1})^{2R+1} h^{2R+1} \esp{\abs[3]{
\sum_{k=1}^K (F(y, Z_k) - \phi_0(y))^2}^{R + 1/2}} \\
& \le (B_{2R+1})^{2R+1} h^{R+1/2} \esp{\abs{F(y, Z) - \phi_0(y)}^{2R+1}}
\end{align*}
Integrating with respect to $\PP_{Y}$ finally yields
\begin{align}
\nonumber    \esp{\abs{X_h - X_0}^{2R+1}}
    &\le (B_{2R+1})^{2R+1} h^{R+1/2} \esp{\abs{F(Y, Z) - \phi_0(Y)}^{2R+1}}\\
\nonumber    &\le 2^{R+\frac 12} (B_{2R+1})^{2R+1} h^{R+1/2}\esp{|\Xi|^{2R+1} + |\esp{\Xi|Y}|^{2R+1}} \\
    &\le 2^{R+\frac 32} (B_{2R+1})^{2R+1} h^{R+1/2}\esp{|\Xi|^{2R+1}}\label{eq:Yh-Y_0:2R+1}
\end{align}
so that $\abs{\calig{R}_{2R+1}} = \calig{O}(h^{R+1/2})$.
\end{proof}

\subsection{Strong convergence rate}
If we assume that $f$ is Lipschitz continuous, Lemma~\ref{lemma::strong_conv_useful} straightforwardly shows that the standard \emph{nested} Monte Carlo satisfies a strong convergence at a rate $h^\beta$ with $\beta=1$. More precisely, if $\Xi \in L^2$, we have
\begin{equation*}
    \normLp[1]{2}{Y_h - Y_{h'}} \le 2 B_2 [f]_{\Lip} \normLp[1]{2}{\Xi - \esp{\Xi|Y}} \abs{h - h'}^{\frac{1}{2}},
\end{equation*}
where $[f]_{\Lip}$ denotes the Lipschitz coefficient of $f$.

When asking for more smoothness, more precisely that $f'$ is $\rho$--Hölder, we can build an antithetic version of the \emph{nested} Monte Carlo which attains a strong convergence at a rate $h^\beta$ with $\beta>1$. As we saw, this corresponds to the optimal unbiased setting in terms of minimization of the computational cost. This antithetic multilevel estimator is obtained by replacing each difference $Y_{\frac hM}-Y_h$, $h \in h_1,\dots,h_{R-1}$, in the MLMC~\eqref{def::MLMC} and ML2R~\eqref{def::ML2R} estimators by the following random variable
\[
    Y_{h,\frac hM} :=  f\left(\frac{1}{MK} \sum_{k=1}^{MK} F\big(Y, Z_{k}\big)\right)- \frac 1M \sum_{m=1}^M f\left(\frac 1K \sum_{k=1}^K F\big(Y, Z_{(m-1)K+k}\big)\right),
\]
satisfying $\esp{Y_{h,\frac hM} } = \esp{Y_{\frac hM}-Y_{h} }$.
We set
\begin{equation*}
\bar X_{K,m} = \frac 1K \sum_{k=1}^K F(Y, Z_{K(m-1)+k}) \quad \mbox{and} \quad \bar X_{MK} = \frac 1M \sum_{m=1}^M \bar X_{K,m} = \frac 1{MK} \sum_{k=1}^{MK} F(Y,Z_k),
\end{equation*}
so that the nested antithetic MLMC estimator~\eqref{def::MLMC} then reads
\begin{equation*}
    \widehat I^N_{h,R,q} = \frac{1}{N_1} \sum_{i=1}^{N_1} f(\bar X_{K,1}^{(i)}) + \sum_{j=2}^R \frac 1{N_j} \sum_{i=1}^{N_j} \left( f(\bar X_{MK}^{(i)}) - \frac 1M \sum_{m=1}^M f(\bar X_{K,m}^{(i)})\right),
\end{equation*}
with $(\bar X_{K,m}^{(i)})_{i\geq 1}$ independent copies of $\bar X_{K,m}$, and similarly for the ML2R estimator~\eqref{def::ML2R} with the weights $(W_j)_{2 \le j \le R}$.

\begin{Proposition}
Let $p > 1$ and $0 < \rho \le 1$. Assume $\Xi \in L^{p(1+\rho)}$ and $f'$ $\rho$--Hölder, i.e.
\begin{equation} \label{hp::rho_holder}
    \forall x, y \in \R, \quad \left| f'(x) - f'(y) \right| \leq \left[ f'\right]_{\rho} |x-y|^\rho.
\end{equation}
Then $Y_{h, \frac hM}$ satisfies a strong approximation error control similar as~\eqref{strong_error} with $\beta = 1+\rho>1$. More precisely, we prove that there exists $\widetilde V_1 > 0$ depending only on $p, \rho, [f']_\rho$ and $M$ such that
\begin{equation} \label{eq:prop_strong_antithetic}
    \normLp[2]{p}{Y_{h, \frac{h}{M}}} \le \widetilde V_1 \pa{h - \frac{h}{M}}^{\frac{1 + \rho}{2}}.
\end{equation}
\end{Proposition}

\begin{proof}[Proof]
Owing to Taylor's formula, for all $m=1, \ldots, M$, there exists $x_m$ in the geometric segment $\left(\bar X_{K,m}, \bar X_{MK} \right)$ such that
$$f(\bar X_{K,m}) = f(\bar X_{MK}) + f'(\bar X_{MK}) (\bar X_{K,m} -\bar X_{MK}) + \left( f'(x_m)-f'(\bar X_{MK}) \right) (\bar X_{K,m}-\bar X_{MK}).$$
Hence, using the definition of $\bar X_{MK}$,
\begin{equation} \label{eq::f_a_i_sum}
 \frac 1M \sum_{m=1}^M f(\bar X_{K,m}) = f(\bar X_{MK}) + \frac 1M \sum_{m=1}^M \left( f'(x_m) - f'(\bar X_{MK})\right)(\bar X_{K,m}-\bar X_{MK}).
\end{equation}
We aim at computing $\ds\normLp[2]{p}{Y_{h,\frac hM}} = \normLp[4]{p}{\frac 1M \sum_{m=1}^M f(\bar X_{K,m}) - f(\bar X_{MK})}$. Owing to the decomposition~\eqref{eq::f_a_i_sum}, to Minkowski's Inequality and to the $\rho$--Hölder assumption~\eqref{hp::rho_holder} on $f'$, we get
\begin{align} \label{eq::rho_inequality}
    \normLp[4]{p}{\frac 1M \sum_{m=1}^M f(\bar X_{K,m}) - f(\bar X_{MK})}
    &= \normLp[4]{p}{\frac 1M \sum_{m=1}^M \left( f'(x_m) - f'(\bar X_{MK})\right)(\bar X_{K,m}-\bar X_{MK})} \nonumber\\
    &\leq [f']_\rho \frac 1M \sum_{m=1}^M \normLp[2]{p}{X_{K,m}-\bar X_{MK}|^{1+\rho}}.
\end{align}
We first notice by an exchangeability argument that the variables $(\bar X_{K,m}-\bar X_{MK})_{m=1\ldots,M}$ are identically distributed with $\bar X_{K,m}-\bar X_{MK} \sim \bar X_{K,1}-\bar X_{MK}$. Moreover we write
$$\bar X_{K,1} - \bar X_{MK} = \bar X_{K,1} - \frac 1M \sum_{m=1}^M \bar X_{K,m} = \frac 1M \sum_{m=1}^M (\bar X_{K,1} -\bar X_{K,m}) = \frac 1M \sum_{m=2}^M (\bar X_{K,1} -\bar X_{K,m}).$$
Hence, we get
\begin{equation} \label{eq::main-inequality}
    \normLp[4]{p}{\frac 1M \sum_{m=1}^M f(\bar X_{K,m}) - f(\bar X_{MK})}
    \leq [f']_\rho \frac 1{M^{1+\rho}} \normLp[4]{p}{\left|\sum_{m=2}^M (\bar X_{K,1} -\bar X_{K,m})\right|^{1+\rho}}.
\end{equation}
Owing to the independence of $Y$ and $(Z_k)_{k\geq1}$, we may write
\begin{align*}
 &\esp{\left|\sum_{m=2}^M (\bar X_{K,1} -\bar X_{K,m})\right|^{(1+\rho)p}} \\
 &= \int \PP_Y(\d y) \esp{\left|\sum_{m=2}^M \left(\frac 1K \sum_{k=1}^K F(y, Z_{k}) -\frac 1K \sum_{k=1}^K F(y, Z_{K(m-1) + k})\right)\right|^{(1+\rho)p}} \\
 &=  \int \PP_Y(\d y) \frac{1}{K^{(1+\rho)p}} \esp{\left| \sum_{k=1}^K  \left( (M-1)F(y, Z_{k}) - \sum_{m=2}^M  F(y, Z_{K(m-1) + k})\right)\right|^{(1+\rho)p}}.
\end{align*}
We notice that, for each fixed $y\in\RR^d$, the random variables $\xi_k = (M-1)F(Z_{k},y) - \sum_{m=2}^M  F(y, Z_{K(m-1) + k})$, $k \ge 1$, are centered and \iid. Moreover $(1+\rho)p > 1$ hence, owing to Marcinkiewicz-Zygmund inequality~\eqref{eq:marcinkZ_iid}, we have
\begin{multline*}
    \esp{\left|\sum_{m=2}^M (\bar X_{K,1} -\bar X_{K,m})\right|^{(1+\rho)p}} \\
    \leq B_{(1+\rho)p} \frac{1}{K^{\frac{(1+\rho)p}{2}}} \int \PP_Y(\d y) \esp{\left|  (M-1)F(y, Z_{1}) - \sum_{m=2}^M  F(y, Z_{K(m-1) + 1})\right|^{(1+\rho)p}}.
\end{multline*}
Applying twice Minkowski's Inequality yields
\begin{equation*}
    \esp{\left|\sum_{m=2}^M (\bar X_{K,1} -\bar X_{K,m})\right|^{(1+\rho)p}}
    \leq C_{p,\rho} \frac{1}{K^{\frac{(1+\rho)p}{2}}} (M-1)^{(1+\rho)p} 2^{(1+\rho)p} \esp{\left|  F(Y, Z_{1}) \right|^{(1+\rho)p}}.
\end{equation*}
Plugging this in~\eqref{eq::main-inequality} we get
\begin{equation*}
    \normLp[4]{p}{\frac 1M \sum_{m=1}^M f(\bar X_{K,m}) - f(\bar{x})}^p \leq  \widetilde V_1 \pa{1 - \frac{1}{M}}^{\frac{p(1+\rho)}{2}} \frac 1{K^{\frac{p(1+\rho)}{2}}},
\end{equation*}
with $\ds \widetilde V_1 = [f']_\rho^p C_{p,\rho} 2^{p(1+\rho)} \left( 1-\frac1M\right)^{\frac{(1+\rho)p}{2}} \normLp{(1+\rho) p}{\Xi}^{(1+\rho) p}$, and~\eqref{eq:prop_strong_antithetic} is proved.
\end{proof}

If we replace the $\rho$--Hölder assumption on $f'$ by a weaker assumption $f'$ locally $\rho$--Hölder, $i.e.$
$$\forall x,y\in\R, \quad \left| f'(x) - f'(y) \right| \leq C |x-y|^\rho \left( 1 + |x|^q + |y|^q \right),$$
a strong convergence assumption with $\beta = 1 + \rho > 1$ similar to~\eqref{eq:prop_strong_antithetic} can still be  proved. Since $|x_m|^q\leq\max(|\bar X_{K,m}|^q,|\bar X_{MK}|^q)\leq |\bar X_{K,m}|^q+|\bar X_{MK}|^q$, Inequality~\eqref{eq::rho_inequality} must be replaced by
\begin{multline} \label{eq::locally_rho_inequality}
 \normLp[4]{p}{\frac 1M \sum_{m=1}^M f(\bar X_{K,m}) - f(\bar X_{MK})} = \normLp[4]{p}{\frac 1M \sum_{m=1}^M \left( f'(x_m) - f'(\bar X_{MK})\right)(\bar X_{K,m}-\bar X_{MK})} \\
  \leq [f']_\rho \frac 1M \sum_{m=1}^M \normLp[2]{p}{|\bar X_{K,m}-\bar X_{MK}|^{1+\rho}(1+|\bar X_{K,m}|^q+2|\bar X_{MK}|^q)}.
\end{multline}
Owing to Hölder's Inequality with $r,s>1$ such that $\frac1r +\frac 1s = 1$ and  Minkowski's Inequality, we get
\begin{multline*}
    \normLp[2]{p}{|\bar X_{K,m}-\bar X_{MK}|^{1+\rho}(1+|\bar X_{K,m}|^q+2|\bar X_{MK}|^q)}
    \\ \leq  \normLp[2]{pr}{|\bar X_{K,m}-\bar X_{MK}|^{1+\rho}}
    \left( 1+ \normLp[2]{ps}{|\bar X_{K,m}|^q} + 2 \normLp[2]{ps}{|\bar X_{MK}|^q} \right).
\end{multline*}
Since the variables $(\bar X_{K,m})_{m=1,\ldots,M}$ are identically distributed, Inequality ~\eqref{eq::locally_rho_inequality} yields
\begin{multline*}
    \normLp[4]{p}{\frac 1M \sum_{m=1}^M f(\bar X_{K,m}) - f(\bar X_{MK})} \\
    \leq [f']_\rho \normLp[2]{pr}{|\bar X_{K,1}-\bar X_{MK}|^{1+\rho}}
    \left( 1+ \normLp[2]{ps}{|\bar X_{K,m}|^q} + 2 \normLp[2]{ps}{|\bar X_{MK}|^q} \right).
\end{multline*}

The analysis of the term $\normLp[2]{pr}{|\bar X_{K,1}-\bar X_{MK}|^{1+\rho}}$  does not change, except for the condition $\Xi\in L^{(1+\rho)pr}$. Under the assumption $\Xi \in L^{qps}$, the term $\normLp[2]{ps}{|\bar X_{K,m}|^q} + 2 \normLp[2]{ps}{|\bar X_{MK}|^q}$ is bounded, since
\begin{equation*}
    \normLp[2]{ps}{X_{K,1}|^q} = \normLp[4]{ps}{\left|\frac 1K \sum_{k=1}^K F(Y, Z_k) \right|^q}
    \leq \normLp[1]{qps \vee 1}{\Xi}^q.
\end{equation*}
Keeping in mind that $r = s/(s-1)$, the optimal choice for $s$ which minimizes both $(1+\rho)pr$ and $qps$ is given by $s = (1+\rho+q)/q$ (hence $r=(1+\rho+q)/(1+\rho)$). This leads to  the additional condition $\Xi\in L^{p(1+\rho+q)}$.
In conclusion, if $f'$ is locally $\rho$--Hölder, under the assumption $\Xi \in L^{p(1+\rho+q)}$, $Y_{h, \frac hM}$ satisfies the $L^p$ version of the strong convergence assumption with $\beta = 1+\rho>1$, similarly to~\eqref{eq:prop_strong_antithetic}.

\section{Indicator function and smooth density}
\label{sec::non_smooth}
There are many situations where we need to consider  non smooth payoff functions of the type $f=\un_{\{g(\esp{\Xi|Y})\in I\}}$, with $g:\R\to\R$ and $I\subset \R$ interval.
Among them we can cite the  computation of loss thresholds, \ie when we search, a threshold $q\in\RR$ being fixed, for the corresponding $\alpha_q\in[0,1]$ such that
$$1-\alpha_q = \PP (g(\esp{\Xi|Y})\geq q) = \esp{\un_{\{ g(\esp{\Xi|Y})\geq q\}}},$$
or the inverse problem, which consists in computing the quantile $q_\alpha$ such that for a fixed $\alpha\in[0,1]$,
$$1-\alpha = \PP (g(\esp{\Xi|Y})\geq q_\alpha) = \esp{\un_{\{ g(\esp{\Xi|Y})\geq q_\alpha\}}}.$$
Another situation of interest is the approximation of density functions (see the seminal paper of Bally and Talay~\cite{BaTa196} and~\cite{BaTa296}, treating the law of the Euler scheme for distributions).

The payoff function $f$ being non smooth, the regularity assumptions on $f$ that we needed to prove the weak and the strong convergence of the estimator in the smooth case, will be replaced by some regularity assumptions on the density functions, as we detail in the next two Subsections.

\subsection{Weak error}
\label{sec:non_smooth_bias_error}
We recall the notation that $X_0 = \esp{\Xi|Y}$ and $X_h = \frac{1}{K} \sum_{k=1}^K F(Y, Z_k)$ with $h = \frac{1}{K} \in \Hr$ and we introduce the notation
\begin{equation*}
    \Delta_h = X_h - X_0.
\end{equation*}

The following result on the weak error derives from Lemma~\ref{lem:taylor} and gives a bias error expansion relying on the density of the joint distribution of $(X_0,  \Delta_h)$ and of $(X_0,  Y)$. More precisely, assume that $(X_0,  \Delta_h)$ is a random vector with smooth density with respect to the Lebesgue measure on $\R^2$. Let $f_{X_0}$ be the density of $X_0$, let $f_{X_0, Y}$ be the  density of $(X_0, Y)$ and let $f_{X_0, \Delta_h}$ be the  density of $(X_0, \Delta_h)$.
Moreover let $F_{ X_h}(x)$ and $F_{ X_0}(x)$ be the cumulative distribution functions of $ \Delta_h $ and $ X_0$.

\begin{Proposition} [Bias error (II): smooth density] \label{prop:nested_weakII}  $(a)$ Let $R \ge 0$. Assume that the partial derivatives $\partial_x^{(\ell)} f_{X_0,Y}(x,y)$ exist for $\ell=1,\ldots, 2R$, that the partial derivatives $\partial_x^{(\ell)} f_{X_0,\Delta_h}(x,y)$ exist for $\ell=1,\ldots, 2R+1$ and that $\partial_x^{(2R+1)} f_{X_0,\Delta_h}(x,y)$ is continuous.
    Assume that $\Xi \in L^{2R+1}$.

    Let
    \begin{equation*}
     P_{r}(x) =\frac{1}{ f_{ X_0}(x)} \sum_{\ell=r+1}^{(2r+1)\wedge 2R}(-1)^{\ell} \int_{\R} b_{r,\ell-r} (y) \partial_x^{(\ell)} f_{X_0,Y}(x,y) \d y.
    \end{equation*}

    \noindent $(a)$ If $\sup_{h\in \Hr, x, v \in \R}\left|\partial_x^{(2R+1)} f_{X_0 | \Delta_h = v}(x)\right|<+\infty$, then
    \begin{equation}\label{eq:fY_0}
        f_{ X_h}(x) = f_{ X_0}(x) + f_{ X_0}(x) \sum_{r=1}^{R} \frac{h^r}{r!} P_{r}(x)  + \calig{O}(h^{R+\frac{1}{2}})
    \end{equation}
    uniformly with respect to $x \in \R$.

    \noindent $(b)$   If furthermore $\sup_{h\in \Hr, x, v\in \R}\left|\partial_x^{(2R)} f_{X_0|\Delta_h = v}(x) \right|<+\infty$ and $\lim_{x\to -\infty}\partial_x^{(2R)} f_{X_0|\Delta_h = v}(x)=0$ for every $v\!\in \R$, then
  \begin{equation}
        F_{ X_h}(x) = F_{ X_0}(x) + \sum_{r=1}^{R} \frac{h^r}{r!} \esp{P_{r}( X_0)\mbox{\bf 1}_{\{ X_0\le x\}} } + \calig{O}(h^{R+\frac{1}{2}})
    \end{equation}
     uniformly with respect to $x \in \R$.
\end{Proposition}

\begin{proof}[Proof]
The case $R=0$ is trivial, using the expansion~\eqref{eq:taylor_nested} and the convention $\sum_{r=1}^0 = 0$.
Let $g:\R\to \R$ be an infinitely differentiable test function with compact support. We apply the expansion~\eqref{eq:taylor_nested} to the smooth function $g$  where coefficients $c(r,2r+1)$, $r=1,\ldots,R-1$ and $c(r, 2R)$, $r=R,\ldots,2R-1$ are given by~\eqref{eq:taylor_coeff} and the remainder term $\calig{R}^{g}_{2R+1}$ is given by~\eqref{eq:taylor_reste}.

We first note that, for every $\ell\!\in \{1,\ldots, 2R\}$,
\[
 \esp{g^{(\ell)}( X_0) b_{r, \ell-r}(Y)} = \int_{\R^2} g^{(\ell)}(x) b_{r, \ell-r}(y) f_{X_0,Y}(x,y)\d x \d y.
\]
Then, performing successively $\ell$ integrations by parts yields
\[
 \esp{g^{(\ell)}( X_0) b_{r, \ell-r}(Y)} =  \int_{\R} g(x)  \int_{\R} (-1)^{\ell} b_{r, \ell-r}(y) \partial_x^{(\ell)}f_{X_0,Y}(x,y) \d y \d x.
 \]
 As for the remainder term,
\begin{align*}
 \calig{R}^{g}_{2R+1} &= \frac{1}{(2R)!} \esp{\int_0^{X_h-X_0} g^{(2R+1)}(t+ X_0) ( X_h- X_0-t)^{2R} \d t}\\
 &= \frac{1}{(2R)!}  \esp{\int_0^1 g^{(2R+1)}(X_0 + s \Delta_h) (\Delta_h)^{2R+1} (1-s)^{2R} \d s}\\
 &= \frac{1}{(2R)!}  \int_0^1 \int_{\R^2} g^{(2R+1)}(x) v^{2R+1} f_{X_0, \Delta_h}(x - sv, v) \d x \d v (1-s)^{2R} \d s.
\end{align*}
Performing successively $2R+1$ integrations by parts yields
\begin{align*}
    \calig{R}^{g}_{2R+1} &= \frac{1}{(2R)!} \int_0^1 \int_{\R} \pa{\int_{\R} g(x) \partial_x^{(2R+1)} f_{X_0, \Delta_h}(x - sv, v) \d x} v^{2R+1} \d v (1-s)^{2R} \d s, \\
    &= \int_{\R} g(x) r(h,x) \d x,
\end{align*}
where
\begin{equation}\label{eq::rest_def}
    \begin{aligned}
    r(h, x) &= \frac{1}{(2R)!}  \int_{0}^1 \int_{\R} \partial_x^{(2R+1)} f_{X_0, \Delta_h}(x - sv, v) v^{2R+1} \d v (1-s)^{2R} \d s, \\
    &= \frac{1}{(2R)!}  \int_{0}^1 \int_{\R} \partial_x^{(2R+1)} f_{X_0|\Delta_h = v}(x - sv) f_{\Delta_h}(v) v^{2R+1} \d v (1-s)^{2R} \d s. \\
    \end{aligned}
\end{equation}
 Plugging these identities in~\eqref{eq:taylor_nested}, we get  that, for every test-function $g$,
 \[
 \esp{g( X_h)} = \int_{\R^d} g(x)\left[f_{ X_0}(x) + f_{ X_0}(x) \sum_{r=1}^{R} \frac{h^r}{r!} P_{r}(x) + \widetilde r(h,x)\right]\d x,
 \]
 where
 \[
  \widetilde r(h,x) = f_{ X_0}(x) \sum_{r=R+1}^{2R-1} \frac{h^r}{r!} P_{r}(x) + r(h,x),
 \]
 Hence
 \begin{equation}
  \label{eq::first_density}
  f_{ X_h}(x) = f_{ X_0}(x) + f_{ X_0}(x) \sum_{r=1}^{R} \frac{h^r}{r!} P_{r}(x)    + \widetilde r(h,x).
 \end{equation}
 The continuity of the function on the right hand side of the above equality will establish the announced expansion, provided we show that $r(h,x)= \calig{O}\big(h^{R+\frac 12}\big)$ uniformly with respect to $x\!\in \R$.  It follows from the boundedness assumption made on $\partial_x^{(2R+1)} f_{X_0|\Delta_h = v}(x)$ that
\begin{equation}
\label{eq::rest_upper_bound}
    \abs{r(h, x)} \le \frac{1}{(2R)!}  \sup_{x,v \in \R} \Big|\partial_x^{(2R+1)} f_{X_0|\Delta_h = v}(x) \Big|\esp{\frac{\abs{\Delta_h}^{2R+1}}{2R+1}}\le\frac{C_{_{\Xi,R}}}{(2R+1)!}h^{R+\frac12},
\end{equation}
owing to the upper-bound established in~\eqref{eq:Yh-Y_0:2R+1} for $\esp{| X_h- X_0|^{2R+1}}$, since $\Xi \in L^{2R+1}$.

\noindent $(b)$ The claim amounts to integrating Equation~\eqref{eq::first_density}, provided we show that the integrals of $P_{r}(x) f_{ X_0} (x)$, for $r=1, \ldots,2R-1$, are at least semi-convergent and that, for all $b\in\RR$,  $\int_{-\infty}^b r(h,x) \d x = \calig{O}(h^{R+\frac12})$.  Owing to Fubini's Theorem, using the definition~\eqref{eq::rest_def} of $r(h,x)$, we have for all $a < b \in \R$,
\begin{multline*}
    \int_a^b r(h, x) \d x = \frac{1}{(2R)!}  \int_{0}^1 \int_{\R} \pa{\partial_x^{(2R)} f_{X_0|\Delta_h = v}(b - sv) - \partial_x^{(2R)} f_{X_0|\Delta_h = v}(a-sv)} \\ \times f_{\Delta_h}(v) v^{2R+1} \d v (1-s)^{2R} \d s
\end{multline*}

The assumption $\sup_{h\in \Hr,x,v\in \R}| f^{(2R)}_{ X_0| \Delta_h=v}(x)|<+\infty$ and the upper bound~\eqref{eq:Yh-Y_0:2R+1}, yield that $\int_a^b \abs{r(h, x)} \d x < +\infty$.

Hence, owing to Lebesgue's Dominated Convergence Theorem and to the assumption \newline $\lim_{x\to -\infty}f^{(2R)}_{ X_0| \Delta_h=v}(x)=0$, we get
\begin{equation}\label{eq::finite_rest}
\begin{aligned}
    \int_{-\infty}^b r(h,x) \d x &= \frac{1}{(2R)!}  \int_{0}^1 \int_{\R} \partial_x^{(2R)} f_{X_0|\Delta_h = v}(b - sv) f_{\Delta_h}(v) v^{2R+1} \d v (1-s)^{2R} \d s, \\
    &= \frac{1}{(2R)!} \int_0^1 \esp{\partial_x^{(2R)} f_{X_0|\Delta_h}(b - s \Delta_h) (\Delta_h)^{2R+1}} (1-s)^{2R} \d s.
\end{aligned}

\end{equation}
and then, likewise~\eqref{eq::rest_upper_bound}, using the boundedness of $f^{(2R)}_{_{ X_0\,|\, \Delta_h=v}}(x) $,
\begin{equation}\label{eq::rest_semiconvergent}
   \int_{-\infty}^b r(h,x) \d x  = \calig{O}(h^{R+\frac12}).
\end{equation}
Owing to Equation~\eqref{eq::first_density}, if we take $h_1,\ldots, h_{2R-1} \in\H$ pairwise distinct,  we get, for all $i=1,\ldots, 2R-1$,
\begin{equation}\label{eq::p_vandermond_system}
 \sum_{r=1}^{2R-1} h_i^{r-1} \left( \frac{ P_r(x)}{r!} f_{ X_0} (x) \right) = \rho_i(x),
\end{equation}
with
\begin{equation}\label{eq::Xi_def}
\rho_i(x) = \frac{f_{ X_{h_i}}(x) - f_{ X_{0}}(x) - r(h_i, x)}{h_i}, \quad i=1,\ldots,2R-1.
\end{equation}
Hence we get a Vandermonde system, $V u(x) = \rho(x)$ with
$$V = V(h_1,\ldots, h_{2R-1}) = \left(
\begin{array}{ccccc}
 1 & h_1 & h_1^2 & \ldots & h_1^{2R-2} \\
 1 & h_2 & h_2^2 & \ldots & h_2^{2R-2} \\
 \vdots & \vdots & \vdots & \ldots & \vdots \\
 1 & h_{2R-1} & h_{2R-1}^2 & \ldots & h_{2R-1}^{2R-2}
\end{array} \right),
$$
$$u(x)= (u_1(x), \ldots,u_{2R-1}(x)) \quad \mbox{with} \quad u_r(x) =  \frac{ P_r(x)}{r!} f_{ X_0} (x)$$
and
$$\rho(x) = (\rho_1(x),\ldots,\rho_{2R-1}(x)).$$
We set, for all $j=1, \ldots,2R-1$,
$$\widetilde V_j(x) = \widetilde V_j(h_1,\ldots,h_{2R-1}, \rho(x)) = \left(
\begin{array}{ccccccc}
 1 	& \ldots & h_1^{j-2} 	& \rho_1(x) 	& h_1^j 	& \ldots & h_1^{2R-2} \\
 1 	& \ldots & h_2^{j-2} 	& \rho_2(x) 	& h_2^j 	& \ldots & h_2^{2R-2} \\
 \vdots & \ldots & \vdots 	& \vdots 	& \vdots 	& \ldots & \vdots \\
 1 	& \ldots & h_{2R-1}^{j-2} & \rho_{2R-1}(x) 	& h_{2R-1}^j 	& \ldots & h_{2R-1}^{2R-2}
\end{array} \right).$$
By expanding along the $j^{th}$ column, the determinant of $\widetilde V_j$ writes
$$\det(\widetilde V_j(x)) = \sum_{i=1}^{2R-1} (-1)^{i+j} d_{ij} \rho_i(x), $$
where $d_{ij} := d_{ij}(h_1,\ldots,h_{i-1},h_{i+1},\ldots,h_{2R})\in\RR$ with
$$d_{ij}(h_1,\ldots,h_{i-1},h_{i+1},\ldots,h_{2R-1}) = \det\left(
\begin{array}{cccccc}
 1 	& \ldots & h_1^{j-2} 	&  h_1^j 	& \ldots & h_1^{2R-2} \\
 \vdots & \ldots & \vdots 	&  \vdots 	& \ldots & \vdots \\
 1 	& \ldots & h_{i-1}^{j-2}&  h_{i-1}^j 	& \ldots & h_{i-1}^{2R-2} \\
 1 	& \ldots & h_{i+1}^{j-2}&  h_{i+1}^j 	& \ldots & h_{i+1}^{2R-2} \\
 \vdots & \ldots & \vdots 	&  \vdots 	& \ldots & \vdots \\
 1 	& \ldots & h_{2R-1}^{j-2} &  h_{2R-1}^j 	& \ldots & h_{2R-1}^{2R-2}
\end{array} \right).$$
Hence, owing to Cramer's rule, the solution of the Vandermonde system writes
$$u_j(x) = \frac{\det\left( \widetilde V_j(x)\right)}{\det(V)} = \frac{1}{\det(V)}\sum_{i=1}^{2R-1}  (-1)^{i+j} d_{ij} \rho_i(x).$$
Finally, since we saw that for all $i = 1, \ldots, 2R-1$, the integral of $\rho_i(x)$ is semi-convergent, we deduce the semi-convergence of the integral
$$\int_{-\infty}^b  P_r(x) f_{ X_0} (x) \d x  =  \frac{r!}{\det(V)}\sum_{i=1}^{2R-1} (-1)^{i+r} d_{ir} \int_{-\infty}^b \rho_i(x) \d x,$$
where, owing to the expression~\eqref{eq::Xi_def} and to~\eqref{eq::rest_semiconvergent},  $\int_{-\infty}^b \rho_i(x) \d x$ is finite,
which concludes the proof.

\end{proof}

\subsection{Strong convergence rate}
We conclude by showing the strong convergence rate~\eqref{strong_error} for the nested Monte Carlo estimator.
The following Lemma  is more or less standard (see for instance~\cite{Av09}).
\begin{Lemma}\label{lem:StrEr1}
 Let $\xi$ and $\xi'$ be two  real valued random variables lying in $L^p$, $p\ge 1$, with densities $f_{\xi}$ and $f_{\xi'}$ respectively. Then, for every $x \!\in \R$,
\begin{equation}\label{eq:StrErQuantile}
\Big\|   \mbox{\bf 1}_{\{\xi\le x \}}-\mbox{\bf 1}_{\{\xi' \le x\}} \Big\|_2^2\le \Big(p^{\frac{p}{p+1}} +p^{\frac{1}{p+1}}\Big)\left(\big\|f_{\xi}\big\|_{\sup} +\big\|f_{\xi'}\big\|_{\sup}\right)^{\frac{p}{p+1}}\big\|\xi-\xi'\big\|_p^{\frac{p}{p+1}}.
\end{equation}
\end{Lemma}

\begin{proof}[Proof] Let $L>0$. Note that
\begin{eqnarray*}
\Big\|   \mbox{\bf 1}_{\{\xi\le x \}}-\mbox{\bf 1}_{\{\xi'\le x \}} \Big\|_2^2 &=& \PP\big(\xi\le x\le \xi'\big)+ \PP\big(\xi'\le x\le \xi\big)\\
&\le& \PP\big(\xi \le x,\, \xi'\ge x+L\big)+\PP\big(\xi \le x \le \xi'\le x+L\big)\\
&& + \PP\big(\xi'\le x,  \xi\ge x+L\big)+\PP\big(\xi' \le x \le \xi\le x+L\big)\\
&\le &  \PP\big(\xi'-\xi\ge L) + \PP\big(\xi-\xi'\ge L)\\
&& + \PP(\xi'\!\in [x, x+L]) +  \PP(\xi\!\in [x, x+L]) \\
&=& \PP\big(|\xi'-\xi|\ge L) + \PP(\xi\!\in [x, x+L]) +  \PP(\xi'\!\in [x, x+L]) \\
&\le& \frac{\esp{|\xi'-\xi|^p}}{L^p} +L \Big( \big\|f_{\xi}\big\|_{\sup}+ \big\|f_{\xi'}\big\|_{\sup}   \Big).
\end{eqnarray*}

A straightforward optimization in $L$ yields the announced result.
\end{proof}

The strong convergence is a consequence of the Proposition~\ref{prop:nested_weakII} combined with the previous Lemma and Lemma~\ref{lemma::strong_conv_useful}.

\begin{Proposition}
\label{prop::strong_error_indicator}
Assume $X\!\in L^p$, $p\ge 2$. Under the assumptions of Proposition~\ref{prop:nested_weakII} $(a)$ with $R=0$  and if  the density $f_{ X_0}$ is bounded, then there exists  $h_0= \frac{1}{K_0}\!\in \Hr\setminus\{0\}$ such that, for every $h, h'\!\in (0, h_0]$
\[
\Big\|   \mbox{\bf 1}_{\{ X_h\le x \}}-\mbox{\bf 1}_{\{ X_{h'}\le x \}} \Big\|_2^2\le C |h-h'|^{\frac{p}{2(p+1)} },
\]
where $C =     2^{\frac{p}{p+1}}  \Big(p^{\frac{p}{p+1}} +p^{\frac{1}{p+1}}\Big) (\big\|f_{_{ X_0}}\big\|_{\sup} +1)^{\frac{p}{p+1}} \big (4\, B_p \big\|X\big\|_p \big)^{\frac{p}{p+1}}$.  This means that the strong approximation error assumption holds with $\beta =\frac{p}{2(p+1)}\!\in (0,\frac 12)$.
\end{Proposition}
\begin{proof}[Proof] It follows from Proposition~\ref{prop:nested_weakII} $(a)$ that
\[
f_{ X_h}(x) \le f_{ X_0}(x) + o(h^{\frac 12})\quad \mbox{ uniformly with respect to $x\!\in \R$}.
\]
Consequently, there exists an $h_0= \frac{1}{K_0}  \!\in \Hr\setminus\{0\}$ such that, for every $h\!\in (0, h_0]$,
\[
\forall\, x\!\in \R, \; f_{ X_h}(x) \le f_{ X_0} (x)+1.
\]
Plugging the above  bound and~\eqref{eq:Yh-Yhprime} in Inequality~\eqref{eq:StrErQuantile} of Lemma~\ref{lem:StrEr1} applied with $\xi= X_h$ and $\xi'=  X_{h'}$ completes the proof.
\end{proof}
\printbibliography

\end{document}

config: arxiv


TITLE: Does the weak force get transmitted at speeds less than $c$?
QUESTION [7 upvotes]: The force carrier of EM is the photon (or off-shell photons at least). These are massless field disturbances. However the force carriers in Weak interactions are the $W$ and $Z$ bosons. Having mass, do they transmit information at subluminal speeds then?

REPLY [3 votes]: The weak interaction is mediated by a field, not by particles, despite the poetic/lazy language that physicists sometimes use. The W and Z particles are special manifestations of the W and Z fields that mediate the weak interaction, but the fields are not made of particles. Analogy: a tornado is a special manifestation of the dynamics of the atmosphere, but the dynamics of the atmosphere is not made of tornadoes.
With that clarification in mind, we can address the question using a simplified model in which the mediating field is a single scalar field $\phi$ whose equation of motion is
$$
 \partial^2\phi+m^2\phi\propto J,
$$
where:

$m$ is the mass of the corresponding particle ($m\sim 90\,\times$ the mass of a proton),

$J$ is the appropriate "charge density" of the quarks/leptons (using the word "charge" in a generalized sense, not referring to electrostatics at all),

the units are such that $\hbar=c=1$.

This equation of motion has two properties relevant to the question:

For simplicity, temporarily think of $\phi$ as a classical field (instead of as a quantum field). Unlike waves in the electromagnetic field (which would have $m=0$ in this analogy), waves in the field $\phi$ are dispersive: an initially localized wavepacket spreads out along the direction of motion as it propagates, so the leading edge travels faster than the trailing edge. The group velocity (speed of the wavepacket's centroid) is less than the speed of light, but the leading edge still moves as the speed of light.

For simplicity, temporarily think of the "charge density" $J$ as a static classical quantity. The "force field" of such a source depends on the distance $r$ from the source like $\phi\sim (e^{-mr})/r$. Since $m\sim 90\,\times$ the mass of a proton, this "force" is completely negligible on scales larger than the size of a proton.

In reality, $\phi$ is a quantum field, and the quarks/leptons that define $J$ are quantum entities, so the idea of the "speed of transmission of the weak force" is even fuzzier than point 1 already emphasized, especially since its extremely short range (point 2) keeps it firmly on the quantum side of the spectrum of behaviors. We can still say that the influence cannot be transmitted any faster than the speed of light, but to say much beyond that would require reframing the question in terms that make more sense for a quantum field that behaves in a very non-classical way — not like classical particles or classical fields.


TITLE: Explain Kerr-Newmann Black Hole Spins in SI Units
QUESTION [4 upvotes]: I'm trying to run some calculations on Kerr-Newman black holes, but I'm having two major difficulties.  First, most equations I've been able to find are only for Kerr black holes.  Second, essentially all equations are in one of several incompatible varieties of "natural units" (this is a big problem for me, since my background is not in Physics, and so I don't magically know when to multiply things by $c$, $G$, $k_B$, $\hslash$, $M$, etc.).
In approximately increasing order of my confusion:

"$J$": I assume this is angular momentum, with the usual $\text{kg}\cdot\text{m}^2\cdot\text{s}^{-1}$ units?
"$J_{max}(M)$": Presumably the maximum allowable value of $J$ for a given mass $M$ (in $\text{kg}$).  I found an SI formula for a Kerr black hole.  Would it be different for a Kerr-Newman black hole?
"$\Omega$": Angular velocity (from an observer at infinity, with units $\text{rad}\cdot\text{s}^{-1}$, I assume).  I found this formula for Kerr, not Kerr-Newman, and I couldn't get the units to make sense.
"$M$": I assume this is mass in $\text{kg}$.  But, I've also seen it rescaled to $\text{m}$, somehow.
"$a$": Some kind of spin parameter?  Unclear how exactly this relates to $J$.  AFAICT, there are multiple definitions, with dimensionless units, meters (e.g. as $J/(M c)$ from Abramowicz and Fragile 2013 $\S 3$), and possibly other units.  Is the dimensionless version supposed to be $a/M$, with $M$ rescaled to length (by $GM/c^2$) or something?  I also found a dimensionless version here as a ratio of $J$ to $J_{max}(M)$.  How does that tie in?
"$a/M$": Specific angular momentum?  Seen in e.g. Bardeen et al. 1972 (fig 1.).  Also seems to be called $\chi$, "dimensionless spin".  Does this have anything to do with spin?

For each of these quantities (and anything else relevant), I'm looking for (1) their scientific names (so I can research them), (2) their SI definitions (so I can do dimensional analysis on them), and (3) their SI equations for Kerr-Newman black holes (so I can calculate them).  At least a self-consistent explanation of $a$, $a/M$ and how they relate to $J$, as treated in the literature, so that I'm not completely lost when reading it anymore.

REPLY [1 votes]: To get the dimensionless spin parameter use
$$\rm \bar a=\frac{J c}{G M^2}$$
where J is the angular momentum of the black hole (the angular momentum J can be expressed in terms of GM²/c, so divide by that to get rid of the dimensions). This always gives a number between 0 and 1 for black holes (regular objects also can have spin parameters larger than 1, but those have to eject some of their angular momentum before collapsing to a black hole). If for some reasons you want to keep the M in the equations instead of setting G=M=c=K=1, just divide by M instead of M².


text:
{\bf Problem.} What is the sum of the prime factors of 91?
{\bf Level.} Level 2
{\bf Type.} Prealgebra
{\bf Solution.} Since 91 is not divisible by the first three prime numbers, we try dividing $91\div 7$.  We find a quotient of 13, so $91=7\times 13$ and the sum of these prime factors is $7+13=\boxed{20}$.

set_name: MATH