Probability distribution of the expectation value of a poisson variable given an observed value.

Question

what is probability density $P(\lambda|Poisson(\lambda) = N)$?

In other words, if I have a poisson variable $X(\lambda)$, where $\lambda$ is unknown, and I observe $X=N$, what is the probability density function of $\lambda$? (I don't know $P(N)$ or $P(\lambda)$.)

Answer 1

The probability of any given value of a continuous random variable is 0. If you have one observation only, then you can draw no inference at all. If you have some observations then you can use these to create a confidence interval, the more you have the tighter the interval will be.

Answer 2

The answer is direct from the most basic elements of Bayesian theory...

One assumes that the distribution of $X$ conditionally on some parameter $\Lambda$ is Poisson $\Lambda$ and that the distribution of $\Lambda$ has density $g$.

Then the density $h_n$ of the conditional distribution of $\Lambda$ conditionally on $X=n$ is such that $$ h_n(\lambda)=\frac{g(\lambda)p_\lambda(n)}{G(n)},\qquad G(n)=\int_0^\infty g(\lambda)p_\lambda(n)\mathrm d\lambda, $$ where $p_\lambda$ is the Poisson distribution of parameter $\lambda$, that is, $p_\lambda(n)=\mathrm e^{-\lambda}\lambda^n/n!$ for every integer $n\geqslant0$. Alternatively, $$ h_n(\lambda)=\frac{g(\lambda)\mathrm e^{-\lambda}\lambda^n}{\int_0^\infty g(t)\mathrm e^{-t}t^n\mathrm dt}. $$ Of course, the result depends very much on the density $g$ of the prior distribution of $\Lambda$.