Let X be a random variable with $E(X^m) = (m+1)!2^m$,m = 1,2,3,..... Find the density of X. I assumed it to be exponential and it turned out to be true. But i couldn't prove it.
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To prove it, once you have made the correct conjecture, use the given information to compute the moment generating function of $X$, and compare with the known mgf of a suitable exponentially distributed random variable. – André Nicolas Feb 22 '16 at 18:19
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how to compute the the moment generating function of x? – maths lover Feb 22 '16 at 18:20
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The mgf is $\sum_0^\infty \frac{E(X^m)}{m!}t^m$, which is $\sum_0^\infty (m+1)2^m t^m$. This sum can be found by differentiating $\sum w^{m+1}$. – André Nicolas Feb 22 '16 at 18:26
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We know that the moment-generating function of a random variable $X$ obeys the property $$M_X(t) = \operatorname{E}[e^{tX}] = \sum_{k=0}^\infty \frac{\operatorname{E}[X^k]}{k!} t^k = \sum_{k=0}^\infty \frac{M_X^{(k)}(0)}{k!} t^k,$$ so that $M_X^{(k)}(0) = \operatorname{E}[X^k]$. Now if we substitute, we find $$M_X(t) = \sum_{k=0}^\infty \frac{(k+1)!2^k}{k!} t^k = \sum_{k=0}^\infty (k+1) (2t)^k.$$ But observe that this is the formal derivative of the power series $$\frac{1}{2} \sum_{k=0}^\infty (2t)^{k+1} = \frac{1}{2} \frac{2t}{1-2t} = \frac{t}{1-2t},$$ so $$M_X(t) = \frac{d}{dt}\left[\frac{t}{1-2t}\right] = \frac{1}{(1-2t)^2}.$$ I leave the rest as a simple exercise.
heropup
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The MGF is $$ E(\exp(sX)) = \sum_{m=0}^\infty E(X^m) \frac{s^m}{m!}$$
However, your formula does not match the MGF of an exponential distribution. What it does match is a Gamma distribution.
Robert Israel
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