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The moment generating function of a random variable $X$ is given by:

$$M(t) = (1/3^{2k})(7+2e^t)^k, \quad \forall t$$

a) Determine $P(X = 3)$

b) Derive the $r^{th}$ factorial moment of $X$

I assume you must deduce what distribution this is from the mgf, I just have no idea which one it is.

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Let $X_i\stackrel{\mathrm{i.i.d.}}\sim\mathsf{Ber}(p)$ random variables for $i=1,2,\ldots$, then the moment generating function of $X_1$ is $$ \mathbb E[e^{tX_1}] = 1-p + pe^t. $$ Set $S_k = \sum_{i=1}^k X_i$, $k=1,2,\ldots$. Then the moment generating function of $S_k$ is given by \begin{align} \mathbb E[e^{tS_k}] &= \mathbb E\left[e^{t\sum_{i=1}^k X_i} \right]\\ &= \prod_{i=1}^k \mathbb E[e^{tX_i}]\\ &= \prod_{i=1}^k \mathbb E[e^{tX_1}]\\ &= \mathbb E[e^{tX_1}]^k \\ &= (1-p + pe^t)^k. \end{align} Set $p=\frac29$ and we have your function $M(t)$. Then $$ \mathbb P(X=3) =\binom k 3 \left(\frac29\right)^3\left(\frac79\right)^{k-3}= \frac8{343}\binom k 3 \left(\frac79\right)^k. $$ To determine the factorial moments of $X$, we instead use the probability generating function $P(s) = \mathbb E[s^X]$. Here $P(s) = \left(\frac79 + \frac29s\right)^k$, and the $r^{\mathrm{th}}$ factorial moment is given by \begin{align} \lim_{s\to1^-} \frac{\mathsf d^r}{\mathsf ds^r} P(s) &= \lim_{s\to1^-} \frac{\mathsf d^r}{\mathsf ds^r} \left(\frac79 + \frac29s\right)^k\\ &= \left(\frac29\right)^r\left(\frac79 + \frac29s\right)^{k-r}\frac{k!}{(k-r)!}, \ r\leqslant k, \end{align} and zero for $r>k$.

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