0

Can someone please explain why the following two scenarios lead to different distributions?

1) $X\sim\chi^2(m)$ and $Y\sim\chi^2(n)$. Using moment-generating functions, show that $Y-X$ does not have a $\chi^2(n-m)$ distribution, for $n>m$ with $X$ and $Y$ independent.

Refer to the following for a derivation of this: Distribution of Difference of Chi-squared Variables

2) Show using MGF's that if $X\sim\chi^2(m)$ and $S = Y+X \sim\chi^2(n-m)$, then $S-X\sim\chi^2(n)$.

Here is my derivation: $$ E[e^{(S-X)t}]=E[e^{St}]E[e^{-Xt}]\\ =\frac{M_S(t)}{M_X(t)}\\ = (1-2t)^{-(m+n)/2}(1-2t)^{m/2}\\ = (1-2t)^{-n/2} $$ My question is: why can't we use this type of derivation (putting $M_X(t)$ in the denominator instead of using $M_X(-t)$) in (1)? Or alternatively, is there another way to show this using $M_X(-t)$ in the derivation?

Thanks!

Community
  • 1
Katelyn
  • 23
  • What are you asked to prove in (2), exactly? The hypotheses are unclear, you might want to restate. – Did Jan 30 '13 at 10:59

2 Answers2

1

Your mistake is that you want $E[1/Z]$ to be $1/E[Z]$ for a random variable $Z$ (in fact, when $Z > 0$ they are never equal unless $Z$ is almost surely constant).

Robert Israel
  • 448,999
  • Interesting... my professor definitely recommended this approach. Is there an alternative way to show (2)? – Katelyn Jan 30 '13 at 00:37
0

there is another mistake that X and S are not independent, so the first equation is wrong.(E(XY)~=E(X)E(Y))

user124316
  • 1