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I'm unable to understand the proof behind determining the Moment Generating Function of a Poisson which is given below: $N \sim \mathrm{Poiss}(\lambda)$ $$ E[e^{\theta N}] = \sum\limits_{k=0}^\infty e^{\theta k} \frac{e^{-\lambda}\lambda^k }{k!} = e^ {\lambda[e^{\theta}-1]} $$

Edit: Q.1 I don't understand how we go from $E[e^{\theta N}] = \sum\limits_{k=0}^\infty e^{\theta k} \frac{e^{-\lambda}\lambda^k }{k!}$. Q.2 Also, I didn't know how it goes from $\sum\limits_{k=0}^\infty \frac{(\mathrm e^{\theta}\lambda)^k }{k!} = \mathrm e^ {-\lambda}\,\mathrm e^{\mathrm e^\theta \lambda}$

Thank You

user131983
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3 Answers3

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To derive the MGF of poisson:

Poisson: $ \frac{\lambda^x e^{- \lambda}}{x!}$

$MGF = E[e^{t x}] = \sum_\limits{x=0}^{\infty} e^{tx} \frac{\lambda^x e^{- \lambda}}{x!}$

We don't care about anything not related to X so factor out $e^{-\lambda}$, we'll also group the two values with common powers i.e $e^{tx}$ and $\lambda^x$ are both to the power of x.

$E[e^{\lambda t}] = e^{-\lambda} \sum_\limits{x=0}^{\infty} \frac{(\lambda e^t)^x }{x!}$

Now the summation looks very similar to the exponential function from:

https://en.wikipedia.org/wiki/List_of_mathematical_series

i.e

$\sum_\limits{x=0}^{\infty} \frac{a^x}{x!} = e^a$

Thus sub the result of the exponential function in.

$E[e^{\lambda t}] = e^{-\lambda}{e^{\lambda e^t}}$

And simplify:

$E[e^{\lambda t}] = e^{\lambda(e^t -1)}$

Maximilian Janisch
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Kevinj22
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$$\sum\limits_{k=0}^\infty\mathrm e^{\theta k} \frac{\mathrm e^{-\lambda}\lambda^k }{k!} = \mathrm e^{-\lambda}\sum\limits_{k=0}^\infty \frac{(\mathrm e^{\theta}\lambda)^k }{k!} = \mathrm e^ {-\lambda}\,\mathrm e^{\mathrm e^\theta \lambda}=\mathrm e^{(\mathrm e^\theta-1) \lambda}$$

Did
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1

For a discrete random variable $X$ with support on some set $S$, the expected value of $X$ is given by the sum $$\mathrm{E}[X] = \sum_{x \in S} x \Pr[X = x].$$ And the expected value of some function $g$ of $X$ is then $$\mathrm{E}[g(X)] = \sum_{x \in S} g(x) \Pr[X = x].$$ In the case of a Poisson random variable, the support is $S = \{0, 1, 2, \ldots, \}$, the set of nonnegative integers. To calculate the MGF, the function $g$ in this case is $g(X) = e^{\theta X}$ (here I have used $X$ instead of $N$, but the math is the same). Hence $$\mathrm{E}[e^{\theta N}] = \sum_{k = 0}^\infty e^{\theta k} \Pr[N = k],$$ where the PMF of a Poisson distribution with parameter $\lambda$ is $$\Pr[N = k] = e^{-\lambda} \frac{\lambda^k}{k!}, \quad k = 0, 1, 2, \ldots.$$

heropup
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