I am working through past examination questions from the Royal Statistical Society and came across this one from 2009 in Module 5 (Question 2(i) and Solution):
The random variable $X$ has a $\chi^{2}_{k}$ distribution ($k=1,2,3,
…$) which has the moment generating function (mgf) $m(t) = (1 −
2t)^{−k/2}$ for $t < \frac{1}{2}$.Using the mgf, find the mean and variance of $X$.
I know the mean will be $k$ and variance $2k$ but I can't derive it. Here is my best attempt which is incorrect:
$$ E[X] = d/dx[M_X(0)] = -2 \times \frac{-k}{2}(1-2X)^{-k/2 -1} = k^{\frac{-k}{2} -1} $$
I'm pretty sure I should be doing an expansion at $0$ but I can't see how to do it.
Best Answer
Your differentiation is almost right, but you've put $X$ where you should have put $t$: $$\newcommand{\diff}{\mathrm{d}} \frac{\diff M_X(t)}{\diff t}=-2\left(\frac{-k}{2}\right)(1-2t)^{-\frac{k}{2}-1} $$ Now all you need to do is set $t = 0$ in that expression, recalling that $1$ to the power of anything is $1$.