2
$\begingroup$

Suppose $Z , \epsilon \sim N(0, 1)$ are independent Gaussian random variables. Let $a \ll 1$ be a small positive number. Let $W = aZ + \epsilon$. It can be show that \begin{align} \mathbb{E} [ W^2 (Z^2 - 1)] = 2 a^2. \end{align} Now suppose I truncate random variable $W$. I was wondering what truncation level $R$ should be such that \begin{align} \mathbb{E} [ W^2 \cdot \mathbf{1}\{ | W| > R\} \cdot (Z^2 - 1)] \leq a^2. \end{align} Can we set $R$ to be a constant? This seems an easy problem, but computing the expectation seems not trivial.

$\endgroup$
4
$\begingroup$

There is a simple but often efficient trick to facilitate such computations. Let $H_k$ be the normalized Hermite polynomials and let $\Phi,\Psi$ be any two functions square integrable with respect to the Gaussian measure $\gamma$. Write $\Phi=\sum_k\varphi_kH_k$, $\Psi=\sum_k\psi_kH_k$. Then, if $Y,Z$ are jointly Gaussian with variance $1$ and covariance $\alpha$, we have $$ E(\Phi(Y)\Psi(Z))=\sum_k \varphi_k\psi_k\alpha^k\,. $$ In your case $\Psi$ is essentially $H_2$ (up to normalization) and $\Phi_r(x)=x^2\chi_{[-r,r]^c}(x)$, so putting $Y=(a^2+1)^{-1/2}W$, we see that we just need to drop the integral $2\varphi_2(r)=\int_{\mathbb R} \Phi_r(x)(x^2-1)\,d\gamma(x)$ from its initial value $2$ at $r=0$ to $1$. That corresponds to some fixed level $r_0>0$ and the answer to the original problem is $R=r_0\sqrt{a^2+1}$.

$\endgroup$
1
  • $\begingroup$ fyi: $r_0\approx 2.294417439$. $\endgroup$ – Nawaf Bou-Rabee Aug 14 '16 at 14:37

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.