If $E(X^r)=[(r+1)!](2^r), \ \ r=1,2,3,\ldots ,$ find the moment generating function of $X$.
I approached this problem by using the Maclaurin series for the moment generating function, but ended up getting stuck:
First, by definition, $$E(X^r)=M^{(r)}(t)\bigg|_{t=0}$$
where $M^{(r)}(t)\bigg|_{t=0}$ represents the $r$th derivative of the moment generating function of $X$, $M(t)$, evaluated at $t=0$.
Then, by the Maclaurin series,
\begin{align}
M(t)&=\frac{M(0)}{0!}t^0+\frac{M^{(1)}(0)}{1!}t^1+\frac{M^{(2)}(0)}{2!}t^2+\cdots+\frac{M^{(r)}(0)}{r!}t^r+\cdots \\&=1+\frac{2!*2^1}{1!}t^1+\frac{3!*2^2}{2!}t^2+\cdots+\frac{(r+1)!*2^r}{r!}t^r+\cdots \\ &= 1+2(2t)+3(2t)^3+\cdots+r(2t)^r+\cdots
\end{align}
So this is where I got stuck. I think I have to find a certain function that also has the above Maclaurin series, but is specifically a m.g.f. Do I have to do some type of trial and error here, or is there some technique I'm unaware of?
Best Answer
You have a couple of errors in the last line. \begin{align} M(t) &= 1+2(2t)+3(2t)^2+\dots+(r+1)(2t)^r+\dots \\ &= \frac{d}{dt}\left(t+\frac{(2t)^2}{2}+\frac{(2t)^3}{2}+\dots+\frac{(2t)^{r+1}}{2}+\dots\right) \\ &= \frac{1}{2}\cdot\frac{d}{dt}\left(2t+(2t)^2+(2t)^3+\dots+(2t)^{r+1}+\dots\right) \\ &= \frac{1}{2}\cdot\frac{d}{dt}\frac{2t}{1-2t} \\ &= \frac{1}{2}\cdot\frac{(1-2t)2-2t(-2)}{(1-2t)^2} \\ &= \frac{1}{(1-2t)^2} \end{align}