Expectation of product of two inner products

Question

Suppose $a,w \sim \mathcal{N}(0,I_n)$ where the entries of $a$ and $w$ are independent. Define the following random vectors: $w_+$ and $w_-$ whose $i$-th entry is $$(w_+)_i = w_i 1_{\{w_i > 0\}}\ \text{and}\ (w_-)_i = |w_i| 1_{\{w_i < 0\}}.$$ That is, $w_+$ keeps the positive entries of $w$ and $w_-$ keeps the magnitude of the negative entries of $w$. I am interested in calculating $$\mathbb{E}|\langle a,w_+ \rangle \langle a,w_- \rangle|.$$ How could one go about this calculation? Since $a$ and $w$ are independent, is there a way to use conditional expectation to simplify this?

Answer 1

You might be able to simplify this:

Conditional on $w$ the random variables $U=\langle a,w_+\rangle$ and $V=\langle a,w_-\rangle$ are independent mean $0$ gaussians with variances $\|w_+\|^2$ and $\|w_-\|^2$, respectively. (Independent because $\langle w_+,w_-\rangle=0$.) So your desired expectation is $(E|Z|)^2 E(\|w_+\| \|w_-\|)$, where $Z\sim N(0,1)$, that is, $$\left( \frac2{\sqrt{2\pi}}\int_0^\infty z e^{-z^2/2}\,dz\right)^2 E(\|w_+\| \|w_-\|)=\frac 2 \pi \,E(\|w_+\| \|w_-\|).$$ We can write a formula for $E(\|w_+\| \|w_-\|)$ as follows: $$E(\|w_+\| \|w_-\|) = \sum_{k=0}^n \binom n k 2^{-n} E\sqrt{X_k}\,E\sqrt{X_{n-k}}\tag{*}$$ where $X_k$ is a $\chi^2$ random variable with $k$ degrees of freedom, and $X_{n-k}$ is another, with $n-k$ degrees of freedom. To see this, introduce $n$ iid Bernoulli random variables $b_1,\dots,b_n$ and note that the joint distribution of $(\|w_+\|^2, \|w_-\|^2)$ is the same as that of $$(R,S)=(\sum_{i=1}^n b_i w_i^2,\sum_{i=1}^n (1-b_i) w_i^2).$$ The idea here is that the $b_i$ decide for the $w_i$ whether they are positive or not. Conditional on $\sum_i b_i = k$, $R$ and $S$ are independent $\chi^2$ rvs, with $k$ and $n-k$ degrees of freedom, respectively. Finally, using the fact that the expectation of the square root of a $\chi^2$ rv with $\nu$ degrees of freedom is $0$ if $\nu=0$ and $$E\sqrt{X_\nu} = \sqrt 2 \frac{\Gamma(\frac{\nu+1}2)}{\Gamma(\frac \nu 2 )}$$ otherwise, we have an answer to the original question: $$ E|\langle a,w_+\rangle\langle a,w_-\rangle| = \frac {2^{2-n}} \pi \sum_{k=1}^{n-1}\binom n k \frac{\Gamma(\frac{k+1}2)}{\Gamma(\frac k 2 )}\frac{\Gamma(\frac{n-k+1}2)}{\Gamma(\frac {n-k} 2 )}.$$