0
$\begingroup$

My intuition on this says, no, since the $\chi^2$ random variable $U=\sum_{i=1}^k Z_i^2$ where each $Z_i$ is an independent Standard Normal random variable.

That said, I was told by a professor some time ago that the sample mean is independent from the sample variance of a Normal distribution, so does it follow that U is independent from Z?

$\endgroup$
1
  • 2
    $\begingroup$ First think about this: In $U=Z^2$ (i.e. $k=1$), is $U$ independent of $Z$? $\endgroup$
    – gunes
    Aug 4, 2019 at 14:41

1 Answer 1

1
$\begingroup$

Your professor's comment is about Basu's theorem, see the Example section also. As for your question, which is also counter-intuitive, think about the case where $k=1$, i.e. $U=Z^2$. A simple proof of dependence would be to show a non-zero covariance value, however these variables have zero covariance. Another one is to think of conditional distribution $f_{U|Z}(u|z)$. Independence forces us to have $$f_{U|Z}(u|z)=f_U(u)$$ which is not correct since $f_{U|Z}(u|z)$ have a degenerate PDF which is equal to $\delta(u-z^2)\neq f_U(u)$, not a Chi-Squared distribution. Intuitively, if we know $Z=z$, then $U$ will certainly be $u=z^2$ without any uncertainty left.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.