Show that $E\left[\|V\|^2| (V+U,U) \in C \times C \right] < E\left[\|V\|^2 \right]=3$ where $V$ and $U$ are standard normal

Question

Let $C\in \mathbb{R}^3$ be a cone. Specifically, assume that $C$ is given by \begin{align} C=\{(x_1,x_2,x_3): x_1 \le x_2 \le x_3 \}. \end{align}

I am interested in the quantity \begin{align} E\left[\|V\|^2| (V+U,U) \in C \times C \right] \end{align} where $V\in \mathbb{R}^3$ and $U\in \mathbb{R}^3$ are independent and standard normal.

Question: Can we show that \begin{align} E\left[\|V\|^2| (V+U,U) \in C \times C \right] < E\left[\|V\|^2 \right]=3. \end{align}

Things that I have tried:

Cauchy-Schwarz: \begin{align} E\left[\|V\|^2| (V+U,U) \in C \times C \right] &=\frac{E\left[\|V\|^2 1_{ (V+U,U) \in C \times C} \right] }{ E[1_{ (V+U,U) \in C \times C}]}\\ & \le \frac{ \sqrt{E\left[\|V\|^4 \right] E\left[ 1_{ (V+U,U) \in C \times C} \right]} }{ E[1_{ (V+U,U) \in C \times C}]}\\ &=\frac{\sqrt{15}}{\sqrt{ E[1_{ (V+U,U) \in C \times C}]}}. \end{align}

However, the above is large than $3$.

Re-writing : I was thinking that we can define $W=V+U$ in which case we have that \begin{align} E\left[\|W-U\|^2| (W,U) \in C \times C \right], \end{align} but this also didn't lead anywhere.

I think the approach has to use the fact that $C$ is a cone, but I am not sure how to use it. I also tried to move the problem into spherical coordinates but didn't get anywhere.

Answer 1

OK, here are the details. Note that $$ E\{|V|^2:U,U+V\in C\}= P(C)^{-1}\int_C E\{|V|^2:u+V\in C\}\,dP(u) $$ so it suffices to show that $E\{|V|^2:u+V\in C\}\le E|V|^2$ for every $u\in C$.

Now we have $E|V|^2=\int_0^\infty r^2 p(r)\,dr$ where $p(r)$ is the density of the distribution of $|V|$. Similarly, $E\{|V|^2:u+V\in C\}=\int_0^\infty r^2 q(r)\,dr$ where $q(r)$ is the probability density proportional to $p(r)\omega(r)$ and $\omega(r)$ is the portion of the sphere of radius $r$ contained in $C$.

Now, if $u+re\in C$, then $u+r'e\in C$ for all $r'\in[0,r]$ by the convexity of $C$, so $\omega=q/p$ is a non-increasing function of $r$. Let $r_0$ be a point such that $p(r_0)=q(r_0)$ (the area under both graphs is $1$, so they must intersect somewhere). Then, by the decreasing property of $\omega$, we have $p(r)\ge q(r)$ for $r\ge r_0$ and $p(r)\le q(r)$ for $r\le r_0$, so $\int_0^\infty (p(r)-q(r))(r^2-r_0^2)\,dr\ge 0$ as the integral of a nonnegative function. Cancelling $\int_0^\infty p(r)r_0^2=r_0^2=\int_0^\infty q(r)r_0^2$, we get exactly what we need.

That's all :-)