Continuous dependence of the expectation of a r.v. on the probability measure

Question

$\newcommand{\bsV}{\boldsymbol{V}}$ $\newcommand{\bsE}{\boldsymbol{E}}$ $\newcommand{\bR}{\mathbb{R}}$ Suppose that $\bsV$ is an $N$-dimensional real Euclidean space. Denote by $\newcommand{\eA}{\mathscr{A}}$ $\eA$ the space of symmetric positive semidefinite operators $A:\bsV\to \bsV$. To each $A\in \eA$ we can associate in a canonical fashion a centered Gaussian measure $\gamma_A$ on $\bsV$ which is concentrated on $(\ker A)^\perp$. For example, if $A$ is nondegenerate, then

$$ \gamma_A(dv)= \frac{1}{\sqrt{\det 2\pi A}} e^{-\frac{1}{2} (A^{-1}v,v)}dv, $$

while if $A=0$, then $\gamma_0$ is the Dirac measure concentrated at the origin.

Fix a locally Lipschitz function $f:\bsV\to\bR$ which is positively homogeneous of degree $\alpha \geq 2$. For any $A\in \eA$ we denote by $\bsE_A(f)$ the expectation of $f$ with respect to the probability measure $\gamma_A$ on $\bsV$. Consider the function

$$ \eA\ni A\mapsto \bsE_A(f)\in \bR. $$

This function is continuous and positively homogeneous of degree $\frac{\alpha}{2}$, i.e.,

$$ \bsE_{tA}(f)=t^{\frac{\alpha}{2}} \bsE_A(f),\;\;\forall t>0,\;\;A\in\eA. $$

I am interested in its modulus of uniform continuity on the ball

$$\eA_1:=\bigl\lbrace A\in\eA;\;\;\Vert A\Vert\leq 1\bigr\rbrace, $$

I was able to prove that on this ball the above function is Holder continuous, with Holder exponent $\frac{1}{2N+3}$. This suffices for the applications I have in mind, but I strongly suspect that it is far from optimal. I believe that the Holder exponent $\frac{1}{2}$ is uniformly optimal in the following sense: there exist $C, r>0$ so that for any $A, B\in\eA_1$ satisfying

$$\Vert A- B\Vert \leq r, $$

we have

$$\bigl\vert \bsE_A(f)-\bsE_B(f)\bigr\vert\leq C\Vert A-B\Vert^{\frac{1}{2}}. \tag{1} $$

Remark. To see that the exponent $\frac{1}{2}$ is the best one can hope for consider the case $\bsV=\bR^2$, $f(x,y)=|xy|$ and $\newcommand{\ve}{{\varepsilon}}$ and the Gaussian measures

$$ \gamma_{A_\ve}=\frac{1}{2\pi\ve} e^{-\frac{1}{2\ve^2}x^2-\frac{1}{2}y^2} |dxdy| $$

Then $\Vert A_\ve-A_0\Vert =\ve^2$,

$$\bsE_{A_0}(f)=0,\;\; \bsE_{A_\ve}(f)=\left(\int_{\bR}|x|e^{-\frac{1}{2}x^2} |dx|\right)^2 \ve. $$

My question is the following: have you encountered continuity results of this sort, and if so, can you indicate some references that deal with this? Thanks!

Answer 1

This problem reduces quickly to Holder continuity of the operator square root. That is, there exists a $C > 0$ such that $$ \begin{align} \lVert\sqrt{A}-\sqrt{B}\rVert\le C\lVert A-B\rVert^{1/2}&&{\rm(1)} \end{align} $$ for any positive semidefinite operators $A,B$.^[1]

Assuming (1), the proof of continuity with Holder exponent $1/2$ as mentioned in the question is quite direct. If $X$ is an $\mathbb{R}^N$-valued standard normal random variable, then $\sqrt{A}X$ has distribution $\gamma_A$. So, if $f$ has Lipschitz constant $K$ on the unit ball then, writing $\hat X=X/\lVert X\rVert$, $$ \begin{align} \left\lvert\mathbb{E}_A(f)-\mathbb{E}_B(f)\right\rvert &=\left\lvert\mathbb{E}[f(\sqrt{A}X)-f(\sqrt{B}X)]\right\rvert\cr &=\left\lvert\mathbb{E}[\lVert X\rVert^\alpha(f(\sqrt{A}\hat{X})-f(\sqrt{B}\hat{X}))]\right\rvert\cr &\le K\mathbb{E}[\lVert X\rVert^\alpha]\lVert\sqrt{A}-\sqrt{B}\rVert\cr &\le CK\mathbb{E}[\lVert X\rVert^\alpha]\lVert A-B\rVert^{1/2}, \end{align} $$ as required.

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^[1] Holder continuity of the square root looks quite obvious, so I would expect it to be standard. I don't have a reference for this though, but I can give a proof now using the Taylor expansion of the square root (maybe there is a quicker proof). It is tempting to suggest that it holds with $C=1$ as in the 1-dimensional case, but this is possibly rather too strong when $A$ and $B$ do not commute.

Multiplying through by a scalar, we can assume that $\lVert A\rVert$ and $\lVert B\rVert$ are bounded by $1$. Then, write $A=1-X$, $B=1-Y$ for positive semidefinite operators $X,Y$ with $\lVert X\rVert,\lVert Y\rVert\le1$. By Taylor expansion, $$ \sqrt{1-X}=1-\sum_{n=1}^\infty a_n X^n $$ where $$ \begin{align} a_n&=\frac12\prod_{k=2}^n\left(1-\frac3{2k}\right) \le\frac12\prod_{k=2}^n\exp\left(-\frac3{2k}\right)\cr &\le\exp\left(-\frac32\log n\right) =n^{-3/2}. \end{align} $$ Then, $$ \sqrt{A}-\sqrt{B}=\sum_{n=1}^\infty a_n(Y^n-X^n). $$ As $X,Y$ have norm bounded by 1, the term $Y^n-X^n$ has norm bounded by $n\lVert Y-X\rVert=n\lVert A-B\rVert$. As $X,Y$ are also positive semidefinite, $Y^n-X^n$ is bounded by 1 in norm. So, $$ \begin{align} \lVert\sqrt{A}-\sqrt{B}\rVert&\le\sum_{n=1}^\infty n^{-3/2}\min\left(n\lVert A-B\rVert,1\right)\cr &=\lVert A-B\rVert\sum_{n\le\lVert A-B\rVert^{-1}}n^{-1/2}+\sum_{n > \lVert A-B\rVert^{-1}}n^{-3/2}. \end{align} $$ As $\sum_{n\le x}n^{-1/2}$ and $\sum_{n > x}n^{-3/2}$ are bounded by fixed multiples of $\sqrt x$ and $1/\sqrt x$ respectively (for $x\ge1$), both terms on the right hand side of the inequality above are bounded by a multiple of $\lVert A-B\rVert^{1/2}$. This gives (1) as required. Note that the constant $C$ does not depend on the dimension $N$, and (1) also holds in infinite dimensions.