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 nonnegative 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.

$\begingroup$ Are all the matrices here diagonal? It looks that way, especially if $I_K$ is the identity matrix of dimension $K$. $\endgroup$– user44143Commented Feb 15, 2021 at 14:06

$\begingroup$ $I_K$ is indeed the identity matrix of dimension $K$ here. $\endgroup$– bdx77Commented 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 :) $\endgroup$– fedjaCommented Feb 16, 2021 at 4:29
1 Answer
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 nondecreasing 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 nonincreasing 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 $W_{q1}$. Since the crosscorrelations 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 $f_{q1}$), 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.

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) $\endgroup$– bdx77Commented Feb 16, 2021 at 12:51