1
$\begingroup$

If $X_1, X_2$ are $2$ random variables such that $(X_1, X_2)$ and $(-X_1, X_2)$ have the same joint distributions then show that $X_1$ and $X_2$ are uncorrelated.

I know that to be uncorrelated the $Cov(X_1, X_2) = E(X_1X_2)-E(X_1)E(X_2) = 0$

which implies $E(X_1X_2)=E(X_1)E(X_2)$

But how do I proceed from here?

$\endgroup$
2
$\begingroup$

If $(X_1,X_2)$ and $(-X_1,X_2)$ have the same joint distribution, and $f(y,z):\mathbb{R}^2\to\mathbb{R}$, then you'd expect $f(X_1,X_2)$ and $f(-X_1,X_2)$ to have the same distribution as well.

This gives you a few nice pieces of information:

  1. $X_1$ and $-X_1$ have the same distribution.
  2. $X_1X_2$ and $-X_1X_2$ have the same distribution.

Now, if two random variables have the same distribution, then they must have the same expectation. (Why?) Can you see where to go from here?

$\endgroup$
  • $\begingroup$ If two random variables have the same distribution, then they must have the same expectation because for a given distribution, we integrate to get the expected value. If you give the same distribution twice, you get the same result of the integral. $\endgroup$ – Note Apr 24 '18 at 20:16
  • $\begingroup$ so then $E[X_1X_2] = -E[X_1X_2]$ same for $E[X_1] = -E[X_1]$ so then $E[X_1X_2] -E[X_1]E[X_2]= -E[X_1X_2]+E[X_1]E[X_2] \Rightarrow 2E[X_1X_2] = 2 E[X_1]E[X_2] $ $ \Rightarrow E[X_1X_2] = E[X_1]E[X_2]$ $\endgroup$ – Note Apr 24 '18 at 20:32
  • $\begingroup$ Yep, that's the idea! $\endgroup$ – Nick Peterson Apr 24 '18 at 20:34
1
$\begingroup$

Hint: Show that $E(X_1 \cdot X_2) = - E(X_1 \cdot X_2)= 0$ and $E(X_1)=-E(X_1)=0$.

$\endgroup$

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.