# Is the probability that the norm of a Gaussian N(mu,I_n) exceeds some q conditionally that the Gaussian is in a cone is maximized when mu=0?

Assume that $Y\sim \mathcal{N}(\mu,I_n)$. Let $C$ be some polyhedral cone in $\mathbb{R}^n$, that is $C = \{x\in\mathbb{R}^n, Ax\geq 0\}$ for some matrix $A$ which we could take to be a full rank square matrix.

I assume that $\mu\notin C$. I'm studying the following conditional probability $$\mathbb{P}(\|Y\|^2>q\left|Y\in C\right.)$$ I want to prove that $$\mathbb{P}(\|Y\|^2>q\left|Y\in C\right.)\leq \mathbb{P}(\|Y-\mu\|^2>q\left|Y-\mu\in C\right.)$$ (which is practically $\mathbb{P}(\|Y-\mu\|^2>q)$ since the direction and the norm of a Gaussian $\mathcal{N}(0,I_n)$ are independent).

I was able to prove this when the polyhedral cone $C$ is a quadrant (or orthant in higher dimensions) by proving that the conditional probability is increasing in each coordinate of $\mu$ (assuming that $\mu<0$).

For example, let $\mu\in(-\infty,0)$. Let $C=\{x, x>0\}$. I can prove that $$\mathbb{P}(Y^2>q|Y>0)$$ is increasing in $\mu$. See the figure attached.

The case of a general polyhedral cone $C$ seems to follow the same way, but I still could not prove it.

Any help ?

• Please clarify what you mean by a "cone": are you discussing one lobe or two? If it's just one, as suggested by your quadrant example, then there are obvious counterexamples for $n=1$. – whuber May 24 '18 at 15:26
• Sorry, I meant the polyhedral cone $C$ is a quadrant (or orthant in higher dimensions). When $n=1$ and $\mu<0$ and let the polyhedral cone be $\{x, x>0\}$, then my claim writes $\mathbb{P}(Y^2>q|Y>0)$ is increasing in $\mu$. – Diaa Al mohamad May 24 '18 at 21:48
• Note that when $n=1$ (or 2), a cone is a polyhedral cone and vice versa. Moreover, the type of polyhedral cones I consider have their the origin as the point of intersection of all the polyhedron faces. When $n=1$, the only cones are $(-\infty,0]$ and $[0,\infty)$. – Diaa Al mohamad May 25 '18 at 10:17
• I'm just a lazy bum, so I haven't even checked this for a simple counterexample. If the following stronger statement is true (I'm not saying it is), your conjecture immediately follows as a consequence. If the minimum to the optimization problem: minimize w.r.t. $x$, $\mathbb{P}(\|Y - x\| <= q \| A(Y-x) \ge 0)$ is achieved at $x = \mu$, you're done. – Mark L. Stone May 25 '18 at 11:49
• Yes Mark, it is indeed a valid approach (actually it is the title of my question :) ). The problem resides in working with the derivatives. – Diaa Al mohamad May 25 '18 at 20:47

Let $Z$ be a Gaussian random vector $\mathcal{N}(0,I_n)$ for some positive integer $n$. Let $q>0$, $F\subset\mathbb{R}^n$ be a cone, and let $F^o$ be its polar cone. Then, $$\mathbb{P}\left(\|Z+\mu\|>q|Z+\mu\in F\right)\leq \mathbb{P}\left(\|Z\|>q|Z\in F\right), \quad \forall \mu\in F^o.$$ Proof:
We rewrite the conditional probability in polar coordinates. Let $S=\{\omega\in\mathbb{R}^n| \; \|\omega\|=1\}$ be the unit sphere in $\mathbb{R}^n$. Denote $\nu$ the surface measure on $S$. We have $$\mathbb{P}\left(\|Z+\mu\|>q|Z+\mu\in F\right) = \frac{\int_{r=q}^{\infty}\int_{\omega\in S \cap F}t^{n-1}e^{-\frac{1}{2}\|tw-\mu\|^2}\nu(d\omega)dt}{\int_{r=0}^{\infty}\int_{\omega\in S \cap F}s^{n-1}e^{-\frac{1}{2}\|sw-\mu\|^2}\nu(d\omega)ds}$$ This means that the conditional density of $\|Z+\mu\|$ provided that $\{Z+\mu\in F\}$ is given by $$g_{\mu}(t) = \frac{t^{n-1}e^{-\frac{1}{2}t^2}\int_{\omega\in S \cap F}e^{tw^T\mu}\nu(d\omega)}{\int_{r=0}^{\infty}\int_{\omega\in S \cap F}s^{n-1}e^{-\frac{1}{2}s^2+sw^T\mu}\nu(d\omega)ds}.$$ Denote $A(\mu)$ the normalization constant in the previous display, that is $$A(\mu) = \int_{r=0}^{\infty}\int_{\omega\in S \cap F}s^{n-1}e^{-\frac{1}{2}s^2+sw^T\mu}\nu(d\omega)ds.$$ Define $$G_{\mu}(t) = \frac{g_0(t)}{g_{\mu}(t)} = \frac{A(\mu)}{A(0)}\frac{\int_{\omega\in S \cap F}\nu(d\omega)}{\int_{\omega\in S \cap F}e^{t\omega^T\mu}\nu(d\omega)}.$$ Since $\mu\in F^o$, then $$\forall\omega\in S \cap F, \quad \omega^T\mu \leq 0.$$ Thus, function $t\mapsto G_{\mu}(t)$ is nondecreasing over $(0,\infty)$ whatever the value of $\mu\in F^o$.
Let $h$ be some measurable nondecreasing function defined on $\mathbb{R}_+$. Note that for any couple of nonnegative real numbers $(r_1,r_2)$, since both $G_{\mu}(t)$ and $h(t)$ are nondecreasing functions in $t$ over $(0,\infty)$, we have $$(h(r_1)-h(r_2))(G_{\mu}(r_1)-G_{\mu}(r_2))\geq 0 \qquad (1)$$ Let $R_1$ and $R_2$ be two i.i.d. random variables with a common density defined on $(0,\infty)$. We have, due to (1), $$\mathbb{E}\left[\left(h(R_1)-h(R_2)\right)\left(G(R_1)-G(R_2)\right)\right]\geq 0.$$ We deduce that $$\mathbb{E}\left[h(R_1)G_{\mu}(R_1)\right]\geq \mathbb{E}\left[h(R_1)\right]\mathbb{E}\left[G_{\mu}(R_1)\right]. \qquad (2)$$ Assume now that $R_1$ has the density $g_{\mu}$, and denote $\mathbb{E}_{\mu}$ (resp. $\mathbb{E}_0$) the expectation under $g_{\mu}$ (resp. $g_0$). We have now \begin{align*} \mathbb{E}_{\mu}\left[h(R_1)G_{\mu}(R_1)\right] & = \mathbb{E}_{0}\left[h(R_1)\right] \\ \mathbb{E}_{\mu}\left[G_{\mu}(R)\right] & = 1. \end{align*} The second line in the previous display comes from the fact that $g_0$ is a density over $(0,\infty)$. Hence, due to (2), we may write $$\mathbb{E}_{\mu}\left[h(R)\right] \leq \mathbb{E}_{0}\left[h(R)\right]. \qquad (3)$$ Set $h(t) = 1_{t>q}$ (the indicator function of the set $(q,\infty)$). Function $h$ is nondecreasing over $(0,\infty)$, therefore we could apply (3) on it and get $$\frac{\int_{r=q}^{\infty}\int_{\omega\in S \cap F}t^{n-1}e^{-\frac{1}{2}\|tw^T\mu\|^2}\nu(d\omega)dt}{\int_{r=0}^{\infty}\int_{\omega\in S \cap F}s^{n-1}e^{-\frac{1}{2}\|sw^T\mu\|^2}\nu(d\omega)ds} \leq \frac{\int_{r=q}^{\infty}\int_{\omega\in S \cap F}t^{n-1}\nu(d\omega)dt}{\int_{r=0}^{\infty}\int_{\omega\in S \cap F}s^{n-1}\nu(d\omega)ds}.$$ This is exactly what we claim.