#### Poisson Likelihood [Equation of the Week]

Astrophysics, especially high-energy astrophysics, is all about counting photons. And this, it is said, naturally leads to all our data being generated by a Poisson process. True enough, but most astronomers don’t know exactly how it works out, so this derivation is for them.

Suppose *N* counts are randomly placed in an interval of duration *τ* without any preference for appearing in any particular portion of *τ*. i.e., the distribution is uniform. The counting rate *R = N/τ*. We can now ask, what is the probability of finding *k* counts in an infinitesimal interval *δt* within *τ*?

First, consider the probability that one count, placed randomly, will fall inside *δt*,

ρ = δt/τ ≡ Rδt/N ≡ ν/N

where *ν = R δt* represents the expected counts intensity in the interval *δt*. When *N* counts are scattered over *τ*, the probability that *k* of them will fall inside *δt* is described with a binomial distribution,

p(k|ρ,N) =^{N}C_{k}ρ^{k}(1-ρ)^{N-k}

as the product of the probability of finding *k* events inside *δt* and the probability of finding the remaining events outside, summed over all the possible distinct ways that *k* events can be chosen out of *N*. Expanding the expression and rearranging,

= N!/{(N-k)!k!} (R δt/N)^{k}(1-(R δt/N))^{N-k}

= N!/{(N-k)!k!} (ν^{k}/N^{k}) (1-(ν/N))^{N-k}

= N!/{(N-k)!N^{k}} (ν^{k}/k!) (1-(ν/N))^{N}(1-(ν/N))^{-k}

Note that as *N,τ —> ∞* (while keeping *R* fixed),

N!/{(N-k)!N^{k}} , (1-(ν/N))^{-k}—> 1

(1-(ν/N))^{N}—> e^{-ν}

and the expression reduces to

p(k|ν) = (ν^{k}/k!) e^{-ν}

which is the familiar (in a manner of speaking) expression for the Poisson likelihood.

## hlee:

I think Poisson probability mass function is statistician’s jargon matching your Poisson likelihood. A few times, I saw Poisson statistics from astro-ph preprints instead which I believe Poisson pmf. Thanks for the neat derivation!

07-08-2008, 4:30 pm