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


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

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!

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Shouldn't this follow from the operator square root being Holder continuous? i.e., $\lVert\sqrt{A}-\sqrt{B}\rVert\le C\lVert A-B\rVert^{1/2}$. –  George Lowther May 14 '13 at 15:43
This argument will do the trick for nonsingular $A$'s. In this situation $f$ need not be Lipschitz. For singular $A$'s things are trickier. The measure $\gamma_A$ is the product between the Dirac measure on $\ker A$ and nonsingular Gaussian on $(\ker A)^\perp$. –  Liviu Nicolaescu May 14 '13 at 17:15
@Liviu: The measure $\gamma_A$ is always the same as the distribution of $\sqrt{A}X$ (singular or not), where $X$ has the standard Gaussian distribution. So, no need to treat the singular case any differently. –  George Lowther May 14 '13 at 17:22
@Duh! Thank you very much. It is embarrassingly obvious now. –  Liviu Nicolaescu May 14 '13 at 17:35
@Liviu: I'll post it as an answer in a few moments –  George Lowther May 14 '13 at 18:19

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.
[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.