Find the unique distribution of a random variable knowing the moments of the random variable

Question

This problem comes from Allan Gut's 'An Intermediate Course in Probability', but I cannot solve the problem.

The random variable $X$ has the property that $$EX^{n}=\frac{2^{n}}{n+1}, \quad n = 1,2, ...$$

Find some (in fact, the unique) distribution of X having these moments.

I know that the moment-generating function (MGF) can be expressed as $$\psi_{X}(t)= E\, e^{tX} = 1 + \sum_{k=1}^{\infty}\frac{t^{k}}{k!}EX^{k}.$$

I tried to use this expression to find an expression of the MGF that I can use to identify the distribution:

$$\psi_{X}(t) = 1 + \sum_{k = 1}^{\infty}\frac{t^{k}}{k!}\frac{2^{k}}{k+1} = 1 + \sum_{k = 1}^{\infty}\frac{(2t)^{k}}{k!(k+1)}$$ $$ = 1 + \sum_{k = 1}^{\infty}\frac{(2t)^{k+1}}{2t(k+1)!} = 1 + \frac{1}{2t}\sum_{k = 0}^{\infty} \frac{(2t)^{k}}{k!} = 1+\frac{1}{2t}e^{2t}$$

But this is not an MGF I recognize, so I'm not sure how to proceed.

Answer 1

Hint: let $Y=X/2$; then $\mathbb E\left[Y^n\right]=1/(n+1)$ which is the integral of $y\mapsto y^n$ over an interval I let you find.

Answer 2

You had a mistake in your calculation of the moment generating function, when reindexing the sum. In fact, $$ 1+\sum_{k=1}^{\infty}\frac{(2t)^{k+1}}{2t(k+1)!}=1+\sum_{k=2}^{\infty}\frac{(2t)^{k}}{2tk!}=1+\frac{e^{2t}-2t-1}{2t}, $$ which is different what you wrote (you had the sum starting at $0$ and not at $2$). Simplifying this expression yields $$ \mathbb Ee^{tX}=\frac{e^{2t}-1}{2t}, $$ which you may recognize as the moment generating function of the uniform distribution on the interval $[0,2]$.