3
$\begingroup$

Let $U \sim \mathcal{N}(0, I_K)$ be a Gaussian vector of dimension $K$ and $V \sim \mathcal{N}(0,1)$, independent of $U$. Let $\Delta$ be a diagonal matrix with non-negative diagonal elements, $c\in\mathbb{R}^K$ and $\sigma^2\geq 0$. Consider the ratio $$R= \frac{c^T U + (\sigma^2 + U^T\Delta U)^{1/2} V}{(c^Tc + \sigma^2 + U^T\Delta U)^{1/2}}.$$ Is it the case that $R$ is ``less dispersed'' than a standard normal, namely, for all $x\in \mathbb{R}^+$, $$\Pr(|R|\leq x)\geq \Pr(|V|\leq x)? \quad (1).$$ Note that if $\Delta c = 0$, then $c^T U$ is independent of $U^T\Delta U$, by Cochran's theorem. As a result, $$R | U^T \Delta U \sim \mathcal{N}(0,1),$$ and therefore (1) holds (with an equality sign). Extensive simulations suggest that (1) holds when $\Delta c\neq 0$ but its proof remains elusive to me. I have not been able to use the exact distribution of $(c^T U, U^T \Delta U)$ as displayed in this paper.

$\endgroup$
3
  • $\begingroup$ Are all the matrices here diagonal? It looks that way, especially if $I_K$ is the identity matrix of dimension $K$. $\endgroup$
    – user44143
    Commented Feb 15, 2021 at 14:06
  • $\begingroup$ $I_K$ is indeed the identity matrix of dimension $K$ here. $\endgroup$
    – bdx77
    Commented Feb 15, 2021 at 14:16
  • $\begingroup$ It is certainly true if $K=1$ and, if the proof of the Gaussian correlation conjecture goes the way I believe it does, also true in general but my memory is rather rusty, so I have to check a few details in the literature before committing to anything more definite :-) $\endgroup$
    – fedja
    Commented Feb 16, 2021 at 4:29

1 Answer 1

4
$\begingroup$

Royen's proof of the Gaussian correlation conjecture (see here yields the following general statement:

Let $(W_1(t),W_2(t))\in \mathbb R^{n_1+n_2}$ be a Gaussian vector for every fixed $t\in[0,1]$ with the correlation matrix $C(t)=\begin{bmatrix}C_{11}&tC_{12}\\ tC_{21}&C_{22}\end{bmatrix}$ where $C_{11}$ and $C_{ij}$ are $n_i\times n_j$ blocks. Fix any two origin symmetric convex bodies $K_1,K_2$. Then the probability $P[W_1(t)\in K_1 \& W_2(t)\in K_2]$ is a non-decreasing function of $t$.

Thereby, under the same assumptions, the probability $P[W_1(t)\in K_1 \& W_2(t)\notin K_2]=P[W_1(t)\in K_1]-P[W_1(t)\in K_1 \& W_2(t)\in K_2]$ is a non-increasing function of $t$.

Now take any increasing positive step function $F:[0,+\infty)\to[0,+\infty)$, $F(s)=f_j$ on $[s_j,s_{j+1})$ with some $0=s_0<s_1<\dots<s_{q+1}=+\infty$ and consider the random variable $$ R=\frac{c^TU+F(U^T\Delta U)V}{\sqrt{c^Tc+F(U^T\Delta U)^2}} $$ Put $K_1=[-x,x], K_2=\{y\in\mathbb R^k: y^T\Delta y\le 1\}$ Then $$ P[R\in[-x,x]]=\sum_j P[W_j\in K_1\& U\in\sqrt{s_{j+1}}K_2\setminus \sqrt{s_{j}}K_2] $$ where $W_j,U$ is a Gaussian vector in $\mathbb R^{1+k}$ with the correlation matrix $\begin{bmatrix}1&\frac{c^T}{\sqrt{c^Tc+f_j^2}}\\\frac{c}{\sqrt{c^Tc+f_j^2}}&I_k\end{bmatrix}$.

Now take the last term $P[W_{q}\in K_1\& U\notin\sqrt{s_{q}}K_2]$ in this sum and replace in it $W_{q}$ by $W_{q-1}$. Since the cross-correlations went up proportionally, the probability went down. But that is equivalent to moving the last (infinite) interval on the graph of $F$ down to join it with the previous one (i.e., replacing $f_q$ by $f_{q-1}$), thus reducing the complexity of $F$. Repeating this procedure $q$ times, we come to the constant $F$ for which the statement is trivial (with equality instead of inequality).

The case of continuously changing $F$ follows by approximation.

$\endgroup$
1
  • 1
    $\begingroup$ Amazing! I very much doubt I would have found the answer by myself, even though I did look at Latala's paper on Royen's proof. So this was extremely useful :-) (Note: because I am new on this forum, my upvote does not appear yet) $\endgroup$
    – bdx77
    Commented Feb 16, 2021 at 12:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.