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Suppose the moment generating function for a given Poisson distribution is given by F(t).

If I have another weird random variable which I analyze and find that the moment generating function is F(t)+C (where C is just a constant term), is this also a Poisson distribution Random Variable? It has same expectation and variance as the first one.

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A moment generating function is $1$ at $t=0$. So if $C\ne 0$ and $F(t)$ is an mgf, then $F(t)+C$ cannot be an mgf.

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  • $\begingroup$ Waow. I did not know that. There is probably something wrong with my calculation then. Thank you!! $\endgroup$ – Mathew Sep 21 '15 at 1:15
  • $\begingroup$ You are welcome. Note that if $t=0$, then $E(e^{tX})=E(1)=1$. $\endgroup$ – André Nicolas Sep 21 '15 at 1:16
  • $\begingroup$ I've been trying to find the error for the last 40 min but no luck. The f(t) term I have is that of a Poisson distribution with parameter lambdap and the constant term is -exp^(-lambda). To make C zero, lambda would tend to infinity. But the mgf would also say that the mean and variance are lambdap. Does that make any sense? I would have posted a pic of my working but I need 4 more reputation points to do that $\endgroup$ – Mathew Sep 21 '15 at 1:58
  • $\begingroup$ I do not know what the other "weird" random variable is, and how you decided on its mgf. Perhaps it would be best if you asked a separate question, describing the random variable in detail, and asking about its distribution, or whatever you need to know. $\endgroup$ – André Nicolas Sep 21 '15 at 2:16

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