1

This is an interview question I found online.

Let $X$ and $Y$ be independent standard Gaussians. What is $E[X|XY]$?

I don't know how the expectation function distributes over conditionals. Any tips?

Gregory Grant
  • 15,324
  • 5
  • 38
  • 63
user1742188
  • 191

3 Answers3

2

Here is a hint: Note that $X$ and $Y$ are symmetric distributions. Suppose you are told that the product of $X$ and $Y$ is, say, $1$. Does that give you preferential information about the sign of $X$; that is to say, is the value of $XY$ informative about whether $X$ is positive or negative?

Do you have reason to believe the conditional distribution of $X \mid XY$ would be asymmetric?

heropup
  • 143,828
  • If XY is positive, both X and Y have the same sign. Similarly, if XY is negative, X and Y have opposite signs. Thus, X can be positive or negative with equal probability. The value of XY gives no infromation, so E[X|XY] = E[X]? – user1742188 May 10 '15 at 14:16
  • You mean that this equation holds for every symmetric distribution? – Mehdi Jafarnia Jahromi May 10 '15 at 16:40
1

Consider a Borel set of the form $A=(a, b)$. Then \begin{align*} \int_{\{XY \in A\}} E(X\mid XY) dP &= \int_{\{XY \in A\}} X dP\\ &=\int_{\Omega} X \pmb{1}_{\{XY \in A\}} dP\\ &=\iint_{\mathbb{R}^2} x \pmb{1}_{\{xy \in A\}} dF(x) dF(y)\\ &=\int_{-\infty}^0\int_{\frac{b}{x}}^{\frac{a}{x}} xdF(x) dF(y) + \int_0^{\infty}\int_{\frac{a}{x}}^{\frac{b}{x}} xdF(x) dF(y)\\ &=\int_{-\infty}^0 x\left(F(a/x) - F(b/x) \right) dF(x) + \int_0^{\infty}x\left(F(b/x) - F(a/x) \right) dF(x)\\ &=\int_{\infty}^0x\left(F(b/x) - F(a/x) \right) dF(x) + \int_0^{\infty}x\left(F(b/x) - F(a/x) \right) dF(x)\\ &=0. \end{align*} By the generalization machinery, we obtain that \begin{align*} E(X\mid XY) = 0. \end{align*}

Alternative Solution.

By the density transformation technique, see also this question, the joint density of $X$ and $Z=XY$ is given by \begin{align*} f_{X, Z}(x, z) = f(x)f\left(\frac{z}{x}\right)\frac{1}{|x|}, \end{align*} and the marginal density of $Z$ is given by \begin{align*} f_Z(z) = \int_{-\infty}^{\infty}f(x)f\left(\frac{z}{x}\right)\frac{1}{|x|}dx. \end{align*} Let \begin{align*} f_{X|Z}(x|z) = \frac{f_{X, Z}(x, z)}{f_Z(z)} \end{align*} be the conditional density function. Then \begin{align*} E(X\mid XY=z) &= \int_{-\infty}^{\infty} x f_{X|Z}(x|z) dx\\ &= \frac{\int_{-\infty}^{\infty} x f(x)f\left(\frac{z}{x}\right)\frac{1}{|x|} dx}{f_Z(z)}\\ &=0, \end{align*} since the function $x f(x)f\left(\frac{z}{x}\right)\frac{1}{|x|}$ is odd. Therefore, \begin{align*} E(X\mid XY) = 0. \end{align*}

Community
  • 1
Gordon
  • 4,541
0

Suppose Case 1: XY is positive. Then either:

1) X and Y both positive. 2) X and Y both negative.

We have zero information to suspect one over the other. XY may hint about the magnitude of X, however it does not tell us if it is positive or negative. That is to say, P(X=k | XY) = P(X=-k | XY). So E(X)=0 in Case 1, because equal chance for X=k or X=-k.

Suppose Case 2: XY is negative. Then either:

1) X positive, Y negative OR 2) X negative, Y positive

Again, XY hints at the magnitude of X, but not whether it is positive or negative. P(X=k | XY) = P(X=-k |XY). So expected value is still 0.

Either way E(X|XY)=0.

XSPX
  • 96