0
$\begingroup$

$X$ and $Y$ are i.i.d and follow standard normal distribution. What is the conditional expectation $E(X^2\mid X+Y=1)$?

Also, I have seen methods of calculating conditional expectation scattered in literatures. Some use a fraction of integral of probability density functions; while others use "hacky" properties of random variables' distributions. What is a good starting point of calculating a conditional expectation?

$\endgroup$
  • $\begingroup$ Do you wish to assume that $X,Y$ are both $N(0,1)$ or $N(\mu,\sigma)$ ? $\endgroup$ – N8tron May 27 '18 at 2:52
  • $\begingroup$ @N8tron Edited. Standard normal distribution. $\endgroup$ – Yi Bao May 27 '18 at 2:54
  • 2
    $\begingroup$ There is no 'master formula' for computing conditional expectation in general setting. So a 'hacky' way is really helpful when it is available. Otherwise, you need to look at the joint distribution... $\endgroup$ – Sangchul Lee May 27 '18 at 3:24
  • $\begingroup$ Since $(X,X+Y)$ is jointly normal, $X\mid X+Y$ is univariate normal, from which the expectation follows. $\endgroup$ – StubbornAtom Dec 16 '18 at 18:15
6
$\begingroup$

I'm afraid you'll find this "hacky" but...

Consider the transformed variables $S=X+Y$ , $R=X-Y$. It's easy to see that these variables (which correspond to a scaled 45 degrees rotation) are iid, $N(0,2)$.

Now $X=(S+R)/2$. Then we want

$$\begin{align} E[X^2 \mid S] &= E\left[\left(\frac{S+R}{2}\right)^2 \mid S\right]\\ &=\frac{1}{4}\left(E[S^2\mid S] + 2 E[S R \mid S] + E[R^2\mid S] \right)\\ &=\frac{1}{4}\left(S^2 + 2 S E[R ] + E[R^2] \right)\\ &=\frac{1}{4}(S^2+2) \end{align} $$

Or

$$ E[X^2 \mid X+Y=1] = \frac{1}{4}(1^2+2)=\frac{3}{4}$$

Quick sanity check: recall that we must have $E[E[X^2 \mid S]]= E[X^2]=1$. And, indeed $E[\frac{1}{4}(S^2+2)]=\frac{1}{4}(2+2)=1$.

$\endgroup$

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.