Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $P(\lambda)$.)

share|improve this question
Got something from an answer below? –  Did Jul 19 '13 at 0:56
add comment

2 Answers 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 $\Lambda$.

share|improve this answer
add comment

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.

share|improve this answer
Oh of course, I meant to ask for the probability density. –  Anthony Barker Jul 11 '13 at 12:46
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.