2
$\begingroup$

This is one of my homework question, which the answer sheet has already been given out. However, I still don't understand it.

Exercise 1.1. It is well known that for two normal random variables, zero covariance implies independence. Why does this not apply to the following situation:

$X ∼ N(0, 1),\, \text{Cov}(X, X^2) = E[X^3] − E[X]\times E[X^2] = 0 − 0 = 0 $

but obviously $X^2$ is totally dependent on X?

So, I assume that I have to show :

$f (x_1, x_2) = f_{X_1} (x_1) * f_{X_2} (x_2) $

for the random variables $X$ and $X^2$ to be not independent. The problem is that I do not know how to compute the joint density function $f (x_1, x_2) $ for the two random variables $X$ and $X^2$.

First of all, is this correct?

$f_{X_1} (x) =$ $1\over{\sqrt{2\pi\sigma^2}}$ $e^{- x^2 \over 2\sigma^2}$

$f_{X_2} (x) =$ [$1\over{\sqrt{2\pi\sigma^2}}$ $e^{- x^2 \over 2\sigma^2}]^2$.

Then, how do I compute the joint density function $f (x_1, x_2) $ ?

I am so confused. Please help!!!!!

$\endgroup$
0
$\begingroup$

Since $X^2$ isn't a normal random variable, there's no contradiction.

$\endgroup$
0
$\begingroup$

And, even if you did have normal variates, it is well known that for jointly normal random variables, zero covariance implies independence. Can you see why from the joint pdf?

$\endgroup$

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.