Obtain the moment generating function

Question

How do I come up with the mgf given this: $E(X^r)=\frac{(r+1)!}{2^r} , r = 1, 2, 3, ...$ ?

The exercise does not require to prove that the distribution exists. It only asks to obtain the mgf and its pdf. I tried getting the moments by substituting the r values, and then I get a series. I have no idea what's next. I am quite confused on how to get the final mgf, that would make it easier to determine the pdf. Thank you.

Answer 1

Note that the given expression for $E\,[X^j]$ holds for all $j\in\{0,1,2,\ldots\}$.

Existence of moments does not imply existence of MGF in general; it's the converse which is true.

So assuming the MGF $E\,[e^{tX}]$ is finite when $t$ is contained in some open interval containing zero, it can be obtained from the moments as follows:

\begin{align} E\,[e^{tX}]&=E\small\left[\sum_{j=0}^\infty \frac{(tX)^j}{j!}\right] \\&=\sum_{j=0}^\infty\frac{t^j}{j!}\color{green}{E\,[X^j]}\qquad\quad\small(\text{by dominated convergence theorem}) \\&=\sum_{j=0}^\infty \frac{t^j(j+1)!}{2^jj!} \\&=\sum_{j=0}^\infty \left(\frac{t}{2}\right)^j(j+1) \\&=\sum_{j=0}^\infty j\left(\frac{t}{2}\right)^j+\sum_{j=0}^\infty \left(\frac{t}{2}\right)^j \end{align}

You should be able to complete the answer from here, keeping in mind the radius of convergence of the power series.