I was just wondering if someone could help me understand this derivation of the probability generating function for a Poisson distribution, (I understand it, until the last step):

$$\pi(s)=\sum^{\infty}_{i=0}e^{-\lambda}\frac{\lambda^i}{i!}s^i$$ $$\pi(s)=e^{-\lambda}\sum^{\infty}_{i=0}\frac{e^{\lambda s}}{e^{\lambda s}}\frac{(\lambda s)^i}{i!}$$ $$= e^{-\lambda}e^{\lambda s} $$

This is a re-production from some lecture notes, but I'm not sure how it jumps from the 2nd last step to the last step?

If someone can show me the intermediate steps I would be very grateful!!

  • $\begingroup$ Hint: do you know of alternative definitions for $e^{\lambda s}$. Maybe as a sum? $\endgroup$ – Drew75 Nov 2 '13 at 9:06
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    $\begingroup$ This is not a "derivation" in any significant sense, because it starts with the result. A meaningful derivation might begin with the construction of the Poisson as a limit of Binomial$(\lambda/n, n)$ distributions as $n$ grows large. Because the PGFs of these distributions are $\left(1 + \frac{\lambda}{n}(s-1)\right)^n$, their limit as $n\to\infty$ is $e^{\lambda(s-1)} = e^{-\lambda}e^{\lambda s},$ QED. (Use of characteristic functions makes this argument rigorous.) $\endgroup$ – whuber Nov 2 '13 at 13:58

The last step simply uses the fact that for each real number $t$, $$\exp(t)=\sum_{i=0}^\infty\frac{t^i}{i!}.$$ Here $t=\lambda s$. (the introduction of $\frac{e^{\lambda s}}{e^{\lambda s}}$ does not seem to be of use here)


It's a simple matter of playing 'spot the probability function':

$$\pi(s)=e^{-\lambda}\sum^{\infty}_{i=0}\frac{e^{\lambda s}}{e^{\lambda s}}\frac{(\lambda s)^i}{i!}=e^{-\lambda}e^{\lambda s}\cdot\sum^{\infty}_{i=0}e^{-\lambda s}\frac{(\lambda s)^i}{i!}=e^{-\lambda}e^{\lambda s}\cdot 1$$

since the term in the sum is just the sum over the probability function of a Poisson$(\lambda s)$


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