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

Mathematics Asked on December 18, 2021

Let $Cin mathbb{R}^3$ 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}
Eleft[|V|^2| (V+U,U) in C times C right]
end{align}

where $Vin mathbb{R}^3$ and $Uin mathbb{R}^3$ are independent and standard normal.

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

Things that I have tried:

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

One Answer

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

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, $E{|V|^2:u+Vin C}=int_0^infty r^2 q(r),dr$ 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 $u+rein C$, then $u+r'ein C$ 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 $rge r_0$ and $p(r)le q(r)$ for $rle r_0$, so $int_0^infty (p(r)-q(r))(r^2-r_0^2),drge 0$ 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 :-)

Answered by fedja on December 18, 2021

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