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](https://www.sciencedirect.com/science/article/pii/S0047259X99918602).



Bound on the distribution of a ratio involving Gaussian distributions
Royen's proof of the Gaussian correlation conjecture (see [here](https://arxiv.org/pdf/1512.08776.pdf) 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.


