# Uncorrelated Random Variables

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?

## 2 Answers

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?

• 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. – Note Apr 24 '18 at 20:16
• so then $E[X_1X_2] = -E[X_1X_2]$ same for $E[X_1] = -E[X_1]$ so then $E[X_1X_2] -E[X_1]E[X_2]= -E[X_1X_2]+E[X_1]E[X_2] \Rightarrow 2E[X_1X_2] = 2 E[X_1]E[X_2]$ $\Rightarrow E[X_1X_2] = E[X_1]E[X_2]$ – Note Apr 24 '18 at 20:32
• Yep, that's the idea! – Nick Peterson Apr 24 '18 at 20:34

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