# How fast does $lim_{ t to 0} E left[ |Z|^2 1_{B}(X,X+sqrt{t} Z) right]= E left[ |Z|^2 right] E[1_B(X)]$

Mathematics Asked on January 3, 2022

Let $$X in mathbb{R}^n$$ and $$Z in mathbb{R}^n$$ be two independent standard normal random vectors.

We are interested in the following quantity:
begin{align} E left[ |Z|^2 1_{B times B}(X,X+sqrt{t} Z) right] end{align}
for some set $$Bsubset mathbb{R}^n$$.

Assumptions about the set $$B$$: 1) Assume that $$1>P(Zin B)>0$$; 2) (Optional) $$B$$ is convex.

Concretely, we are interested in how this quantity behaves as $$t to 0$$.

First, it is easy to show that
begin{align} lim_{ t to 0} E left[ |Z|^2 1_{B times B}(X,X+sqrt{t} Z) right]= E left[ |Z|^2 right] E[1_B(X)], end{align}
where we have used the dominated convergence theorem and the bound $$|Z|^2 1_{B times B}(X,X+sqrt{t} Z) le |Z|^2$$.

My question is: Can we say something about how fast does this approach the limit? Specificaly, can we say something about
$$lim_{ t to 0} frac{d}{dt} E left[ |Z|^2 1_{B times B}(X,X+sqrt{t} Z) right]= ???$$

Edit: The derivative is given by
begin{align} &2 frac{d}{dt} E left[ |Z|^2 1_{B times B}(X,X+sqrt{t} Z) right]\ &=frac{E[|Z|^4 1_{B times B}(X,X+sqrt{t} Z)]- (n+2) E[|Z|^2 1_{B times B}(X+sqrt{t} Z ,X) ]}{t} end{align}

Now, if take the limit as $$t to 0$$ of the numerator than we get
begin{align} &lim_{n to infty} E[|Z|^4 1_{B times B}(X,X+sqrt{t} Z)]- (n+2) E[|Z|^2 1_{B times B}(X+sqrt{t} Z ,X) ]\ &= E left[ |Z|^4 right] E[1_B(X)] – (n+2) E left[ |Z|^2 right] E[1_B(X)]\ &=0 end{align}
where we have used that the fourth moment is given by $$E left[ |Z|^4 right]=n(n+2)$$.

Therefore, we have zero over zero.
I tried using L’hospital rule more times, but we keep getting zero over zero no matter how many times we apply L’hospital rule.

This isn't a full answer, but it does give some more info: I believe the derivative you seek can be negative infinity. For instance, take the example of $$n = 1$$ and $$B = [-1,1]$$.
For legibility, I'll write $$1{A}$$ for the indicator of an event $$A$$. Then begin{align*} E&left[Z^2 1{(X,X+sqrt{t}Z) in [-1,1]^2} right] \&= Eleft[Z^2 1{X in [-1,1]} 1left{Z in left[frac{-1 - X}{sqrt{t}},frac{1 - X}{sqrt{t}} right] right}right] \ &= E[Z^2 1_{[-1,1]}(X)] - Eleft[Z^2 1{X in [-1,1]} 1left{Z notin left[frac{-1 - X}{sqrt{t}},frac{1 - X}{sqrt{t}} right] right}right],. end{align*}

I claim that this second term in absolute value is $$Omega(sqrt{t})$$ as $$t to 0$$. To see this, we may bound it below in absolute value by begin{align*} E[Z^2 1{ X in [1 - sqrt{t},1] 1{Z > 0} ] &sim sqrt{t}cdotphi(1) E[Z^2 1{Z > 0}] \ &= sqrt{t} cdot phi(1) / 2 end{align*} where $$phi(1)$$ is the standard normal density evaluated at $$1$$. I suspect an upper bound (in this case) of $$sqrt{t}$$ is possible to achieve as well.

EDIT: Some more details on the LB: begin{align*} E&left[Z^2 1{X in [-1,1]} 1left{Z notin left[frac{-1 - X}{sqrt{t}},frac{1 - X}{sqrt{t}} right] right}right] \ &geq Eleft[Z^2 1{X in [1-sqrt{t},1]} 1left{Z notin left[frac{-1 - X}{sqrt{t}},frac{1 - X}{sqrt{t}} right] right}right] \ &geq Eleft[Z^2 1{X in [1-sqrt{t},1]} 1left{Z > 0 right}right] end{align*}

Answered by Marcus M on January 3, 2022