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Let $X$ be a random variable and it is known that the mgf of $X$ exists.

If the $k$th moment is given by $m_k=\mathbb E[X^k]=\frac{(2k+1)!}{k!2^k}$ for $k=0, 1, ...$

Problem: Find the mgf of $X$.

My attempt: The mgf of $X$ is $M_X (t)= \sum_{k=0}^\infty \frac{m_k}{k!}t^k=\sum_{k=0}^\infty \frac{(2k+1)!}{(k!)^22^k}t^k$.

However, I have no idea how to proceed further. What should I do?

bellcircle
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  • Sadly this is not showing any afford, show something more please. – MAN-MADE Jul 07 '17 at 14:17
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    @MANMAID, I disagree. The OP clearly knows the definition of the MGF, but has no idea how to simplify the expression. S/he has nicely formatted the question and his/her initial work, showing clear effort on his/her part. The question is really in simplifying the already found expression of the MGF... – Bob Krueger Jul 07 '17 at 14:26
  • I wish I could help though. I don't know how to simplify the given expression of the MGF without perhaps knowing any more about where $X$ came from. – Bob Krueger Jul 07 '17 at 14:30
  • @Bob1123 There is no other information about the random variable $X$. – bellcircle Jul 07 '17 at 15:33
  • Applying the ratio test to those coefficients yields $\frac{4k^2+10k +6}{2(k+1)}\to \infty$ as $k\to \infty$. Are you sure those are the right moments? – Nap D. Lover Jul 07 '17 at 16:00
  • @LoveTooNap29 The ratio test gives $\frac{4k^2+10k+6}{2(k+1)^2} \to 2$ as $k \to \infty$. Try it again. – bellcircle Jul 07 '17 at 16:03
  • @bellcircle you are right. I lost the square on the $k!$ in the denominator. Thanks for the correction. – Nap D. Lover Jul 07 '17 at 16:05

1 Answers1

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$$\sum_{k=0}^{\infty}\binom{2k}{k}x^k = (1-4x)^{-1/2}$$

Put $x = \frac{t}{2}$,

$$\sum_{k=0}^{\infty}\binom{2k}{k}\left(\frac{t}{2}\right)^k = (1-2t)^{-1/2}$$

Take derivative wrt $t$,

$$\sum_{k=0}^{\infty}\binom{2k}{k}\frac{k}{2}\left(\frac{t}{2}\right)^{k-1} = (1-2t)^{-3/2}$$

multiply with $t$,

$$\sum_{k=0}^{\infty}\binom{2k}{k}k\left(\frac{t}{2}\right)^{k} = t(1-2t)^{-3/2}$$

Finally,

$$\sum_{k=0}^{\infty}\binom{2k}{k}(2k+1)\left(\frac{t}{2}\right)^{k} = 2t(1-2t)^{-3/2} + (1-2t)^{-1/2} = (1-2t)^{-3/2}$$

So, the distribution is,

$$gamma\left(\frac{3}{2}, 2\right)$$

Check this question for the proof of first equation.

Dhruv Kohli
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