# Bound on the distribution of a ratio involving Gaussian distributions

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.

• Are all the matrices here diagonal? It looks that way, especially if $I_K$ is the identity matrix of dimension $K$.
– user44143
Commented Feb 15, 2021 at 14:06
• $I_K$ is indeed the identity matrix of dimension $K$ here. Commented Feb 15, 2021 at 14:16
• 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 :-) Commented Feb 16, 2021 at 4:29

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 $$0=s_0 and consider the random variable $$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.

• 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) Commented Feb 16, 2021 at 12:51