The probability mass function for the Poisson distribution with parameter $\theta$ is

$$ \mathbb P(N=n;\theta)=\frac{e^{-\theta}\theta^n}{n!} $$

Since this is a probability mass function, we have:

$$ \sum_{n=0}^\infty\mathbb P(N=n;\theta)=\sum_{n=0}^\infty \frac{e^{-\theta}\theta^n}{n!}=1 $$

This is true for any distribution. However, we also have:

$$ \int_0^\infty \mathbb P(N=n;\theta)d\theta=\frac1{n!}\int_0^\infty e^{-\theta}\theta^n d\theta=1 $$

Since we are now integrating over $\theta$, rather than summing over $n$, the expression $\frac{e^{-\theta}\theta^n}{n!}$ is more properly interpreted as the likelihood function for the distribution. Likelihood functions do not, in general, integrate over their domain to give $1$. What is special about the Poisson distribution that causes its likelihood function to have this property?

  • $\begingroup$ My previous answer indicated that the last integral is just the Gamma function. Obviously, the property exists because the Poisson pmf becomes the integrand of the Gamma function. What more are you looking for? Are you wondering if there is some reason that the distribution has an pmf that becomes the Gamma function? What is the source of the specialness you're looking for? $\endgroup$ – Emily Jul 30 '13 at 16:24
  • $\begingroup$ My main motivation for asking this question is to find a nice probabilistic interpretation for the gamma function integral. So I want to find an answer that will tell me why the integral of the likelihood function is $1$ without actually having to evaluate the integral itself. My guess is that the likelihood on its own is not going to be enough: indeed, if we make the substitution $\theta\mapsto\frac\theta 2$ then we still have a valid Poisson distribution, but the likelihood function now integrates to give you $\frac12$, not $1$. $\endgroup$ – John Gowers Jul 30 '13 at 19:33
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    $\begingroup$ Are you sure about that? $\endgroup$ – Emily Jul 30 '13 at 19:36
  • $\begingroup$ No, that can't be right, because it is the pdf of the Gamma distribution. $\endgroup$ – John Gowers Jul 30 '13 at 19:38
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    $\begingroup$ I have an incompletely developed idea about why things happen the way they do in your problem, and if it pans out, I will write an answer, but consider that if $X \sim N(\mu,\sigma^2)$ where $\mu$ is unknown, then upon observing that $X$ had value $x$ on an experiment, the likelihood function is $$L(\mu; x) = f_X(x) = \frac{1}{\sigma\sqrt{2\pi}}\exp(-(x-\mu)^2/2\sigma^2) = \frac{1}{\sigma\sqrt{2\pi}}\exp(-(\mu-x)^2/2\sigma^2)$$ which, regarded as a function of $\mu$ with $x$, the data, being a constant, is also a density function, in fact that of a $N(x,\sigma^2)$ random variable. $\endgroup$ – Dilip Sarwate Aug 1 '13 at 19:07

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