# Obtain the moment generating function

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.

• To begin with, you must specify the distribution of $X$ somehow. – whuber Oct 29 '19 at 14:37
• I think maybe the question is to find the mgf of $X$ (assuming that it exists) given these moments. – Jarle Tufto Oct 29 '19 at 15:09
• If this is for some course, do add the self-study tag and read the tag wiki. – StubbornAtom Oct 29 '19 at 15:12
• @Jarle You're right--thank you for pointing that out. I will vote to re-open. But since the answer requires nothing besides quoting the definition of the mgf, I still wonder whether the question has been phrased as intended. Maybe the intention behind this exercise includes proving that such a distribution exists? – whuber Oct 29 '19 at 15:26
• I think using the infinite series representation of the exponential solves the problem $E[e^{tX}] = E[\sum_{k=1}^\infty (tX)^k/k!] = \sum_{k=1}^\infty (1/k!) E[(tX)^k]$ – AdamO Oct 29 '19 at 16:08

Note that the given expression for $$E\,[X^j]$$ holds for all $$j\in\{0,1,2,\ldots\}$$.
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}
• Your last algebraic step appears to complicate the evaluation rather than simplify it. The penultimate expression is immediately recognizable as the derivative of $1/(1-x)=\sum_j x^j$ evaluated at $x=t/2.$ – whuber Oct 30 '19 at 14:24