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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.

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  • $\begingroup$ To begin with, you must specify the distribution of $X$ somehow. $\endgroup$ – whuber Oct 29 '19 at 14:37
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    $\begingroup$ I think maybe the question is to find the mgf of $X$ (assuming that it exists) given these moments. $\endgroup$ – Jarle Tufto Oct 29 '19 at 15:09
  • $\begingroup$ If this is for some course, do add the self-study tag and read the tag wiki. $\endgroup$ – StubbornAtom Oct 29 '19 at 15:12
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    $\begingroup$ @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? $\endgroup$ – whuber Oct 29 '19 at 15:26
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    $\begingroup$ I think using the infinite series representation of the exponential solves the problem $E[e^{tX}] = E[\sum_{k=1}^\infty (tX)^k/k!] = \sum_{k=1}^\infty (1/k!) E[(tX)^k]$ $\endgroup$ – AdamO Oct 29 '19 at 16:08
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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.

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  • $\begingroup$ 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 $x=t/2.$ $\endgroup$ – whuber Oct 30 '19 at 14:24
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    $\begingroup$ @whuber I would have used the derivative trick in the last step too, so I would not call this complicating things. But yes, one less step would have been enough. $\endgroup$ – StubbornAtom Oct 30 '19 at 14:28

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