$ \lim_{ \|x\| \to 0 } \frac{E \left[ \log \left(1+ \left(\|x\|^2+ \langle x,Z \rangle \right)^2 \right) \right]}{\| x\|^2} = 1$

Question

How to compute the limit \begin{align} \lim_{ \|x\| \to 0 } \frac{E \left[ \log \left(1+ \left(\|x\|^2+ \langle x,Z \rangle \right)^2 \right) \right]}{\| x\|^2} \end{align} where $Z$ is i.i.d. Gaussian vector of lenght $n$.

Upper bound:

\begin{align} \lim_{ \|x\| \to 0 } \frac{E \left[ \log \left(1+ \left(\|x\|^2+ \langle x,Z \rangle \right)^2 \right) \right]}{\| x\|^2} \le \lim_{ \|x\| \to 0 } \frac{E \left[ \left(\|x\|^2+ \langle x,Z \rangle \right)^2 \right]}{\| x\|^2} \end{align}

here we used that $\log(x) \le x-1$

Next, observe that \begin{align} E[\langle x, Z \rangle]=0,\\ E[\langle x, Z \rangle^2 ]= \|x\|^2 \end{align} where the last two stament follow from i.i.d. assumption.

So, in conclusion, we have

\begin{align} \lim_{ \|x\| \to 0 } \frac{E \left[ \log \left(1+ \left(\|x\|^2+ \langle x,Z \rangle \right)^2 \right) \right]}{\| x\|^2} \le 1. \end{align}

I am stuck with showing the other direction.

For the lower bound one can use $\log(1+x) \ge \frac{x}{x+1}$ which result in \begin{align} \lim_{ \|x\| \to 0}\frac{1}{\|x\|}E \left[ \frac{\left(\|x\|^2+ \langle x,Z \rangle \right)^2 }{1+\left(\|x\|^2+ \langle x,Z \rangle \right)^2 } \right] \end{align}

However, not sure how to proceed next.

Answer 1

I think the following is an outline where you can get the solution.

First, consider the random variable $Y :=\langle x, Z \rangle/||x||$. Obviously, $Y$ is standard normal. Now let $a:= ||x||$, and the problem reduced to

$\lim_{a\downarrow 0} E\frac{[\log(1+(a^2+aY)^2]}{a^2}$.

Now if you want, you can calculate the expectation, however, since we are concerned with limit, I want to interchange the order or limit and expectation. Your previous solution shows that this is allowed because of dominated convergence theorem.

After interchanging the limit and the expectation, we arrived at

$E \lim_{a \downarrow 0} \frac{[\log(1+(a^2+aY)^2]}{a^2}$

To calculate the limit, use L'Hospital's Rule, and take derivative both in the denominator and numerator, you have something like

$E \lim_{a \downarrow 0} \frac{(a+Y)(2a+Y)}{1+(a^2+aY)^2} = EY^2 =1$