Conjugate prior distribution 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 $\xi^x e^{-\xi}$?
 A: 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.)
