Conjugate prior distribution

Question

Suppose data consists of a single observation $x$ on Poisson random variable $X$,where $X\mid\xi\sim\mathcal{P}(\xi)$.How do I show that the likelihood function for $\xi$ is $f(x\mid\xi)$ proportional to $\xi^x e^{-\xi}$?

Answer 1

Poisson likelihood. If the observation is $x \sim Pois(\lambda),$ then the relevant density function is $p(x|\lambda) = e^{-\lambda}\lambda^x/x!.$ The likelihood function is viewed as a function of $\lambda$ and $x$ is a known observed value (now viewed as a constant). Often, the likelihood function is expressed 'up to a constant multiple' by writing $p(x|\lambda) \propto e^{-\lambda}\lambda^x,$ where the righthand side is sometimes called the 'kernel' of the likelihood. (The symbol $\propto$, read "proportional to" and rendered in TeX as \propto, indicates absence of an unneeded constant factor.)

Conjugate gamma prior. In the title of your question, you mention a conjugate prior distribution. A natural conjugate prior for this likelihood would be a gamma distribution $p(\lambda) \propto \lambda^{\alpha-1} e^{-\kappa \lambda},$ so that the information in the prior is expressed by selecting appropriate values of $\alpha$ and $\kappa.$

Gamma posterior. With this prior and likelihood, the posterior distribution is $p(\lambda|x) \propto \lambda^{\alpha-1} e^{-\kappa \lambda} \times e^{-\lambda}\lambda^x,$ which, upon collecting terms in exponents, is easily seen to be another member of the gamma family.

When prior and likelihood are conjugate (mathematically compatible) in this fashion, it is possible to recognize the posterior distribution just by looking at the kernel. (Otherwise, there is an integral in the denominator of the righthand side of Bayes' Theorem that needs to be evaluated.)