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

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 $$B\subset \mathbb{R}^n$$.

Assumptions about the set $$B$$: 1) Assume that $$1>P(Z\in B)>0$$; 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.

## 1 Answer

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] \\&= E\left[Z^2 1\{X \in [-1,1]\} 1\left\{Z \in \left[\frac{-1 - X}{\sqrt{t}},\frac{1 - X}{\sqrt{t}} \right] \right\}\right] \\ &= E[Z^2 1_{[-1,1]}(X)] - E\left[Z^2 1\{X \in [-1,1]\} 1\left\{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}\cdot\phi(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]\} 1\left\{Z \notin \left[\frac{-1 - X}{\sqrt{t}},\frac{1 - X}{\sqrt{t}} \right] \right\}\right] \\ &\geq E\left[Z^2 1\{X \in [1-\sqrt{t},1]\} 1\left\{Z \notin \left[\frac{-1 - X}{\sqrt{t}},\frac{1 - X}{\sqrt{t}} \right] \right\}\right] \\ &\geq E\left[Z^2 1\{X \in [1-\sqrt{t},1]\} 1\left\{Z > 0 \right\}\right] \end{align*}

• Thank you for your answer. I also think it can be a negative infinity. I followed everything except the lower bound can tell what you dropped in there?
– Boby
Jul 28, 2020 at 23:55
• Sure I fleshed it out a bit more. Jul 29, 2020 at 0:13
• Do you think your argument would still work if you somehow approximate $B$ by a ball of radius $r$? I guess the problem with this approach is that $B$ might be such that the ball is not a good approximation.
– Boby
Jul 29, 2020 at 0:26
• I'm not sure. I'd guess the error contribution still comes from when $X$ is near the boundary of $B$. Jul 30, 2020 at 0:59