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I would like to ask this question that I received from my lecturer. The question was to prove the given mgf is actually a poisson distribution.

Here's the mgf; $M_x(t)=8^{(e^t)-1}$

The pmf of the poisson distribution; $P(X=x)={(\frac{3}{2})}^x {\frac{(\ln 2)^x}{x!{2^{(3-x)}}}}$

What I know is: the mean and variance of the Poisson are the same (= $\lambda$), but I can't figure out how to do it.

Thank you.

StubbornAtom
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RejaR
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  • Show, that mgf of the distribution you mentioned above is $8^{e^t-1}$ and then use uniqueness: https://math.stackexchange.com/questions/458680/how-to-prove-moment-generating-function-uniqueness-theorem – Botnakov N. Nov 11 '21 at 04:09
  • The pmf of a Poission distribution, with rate parameter $\lambda$, is $\mathsf P(X,{=},x)~=~\dfrac{\lambda^x \mathrm e^{-\lambda}}{x!};\mathbf 1_{x\in\Bbb N}$ . – Graham Kemp Nov 11 '21 at 04:12
  • @GrahamKemp: the OP's given pmf is actually a Poisson with parameter $\log8$. Just a matter of rewriting it. – tommik Nov 11 '21 at 09:42
  • Thank you for the answer – RejaR Nov 12 '21 at 05:40

1 Answers1

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The MGF of a poisson is

$$M_X(t)=e^{\theta(e^t-1)}$$

You can rewrite your expression in the following way:

$$8^{e^t-1}=e^{\log8^{(e^t-1)}}=e^{\log 8(e^t-1)}$$

Which is the MGF of a poisson $\text{Poi}(\log8)$ that is

$$\mathbb{P}[X=x]=\frac{\left(\frac{3}{2} \right)^x(\log 2)^x}{x!\cdot2^{3-x}}=\frac{e^{-\log8}(\log 8)^x}{x!}\cdot\mathbb{1}_{\{0;1;2;3;\dots\}}(x)$$

tommik
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