10
$\begingroup$

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.

$\endgroup$

1 Answer 1

1
$\begingroup$

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*}

$\endgroup$
4
  • $\begingroup$ 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? $\endgroup$
    – Boby
    Jul 28, 2020 at 23:55
  • $\begingroup$ Sure I fleshed it out a bit more. $\endgroup$
    – Marcus M
    Jul 29, 2020 at 0:13
  • $\begingroup$ 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. $\endgroup$
    – Boby
    Jul 29, 2020 at 0:26
  • $\begingroup$ I'm not sure. I'd guess the error contribution still comes from when $X$ is near the boundary of $B$. $\endgroup$
    – Marcus M
    Jul 30, 2020 at 0:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.