Uniform prior and poisson likelihood, what posterior distribution will be produced?

If i have a uniform distribution over a fixed specified and a finite range, and a Poisson likelihood distribution, what posterior will be produced?

The likelihood has this form $$P(\pmb{X}| \pmb{\Lambda}, \pmb{Y}) \propto \prod_{a=1}^{c} \frac{\lambda_{a}^{X_a}}{X_a!~}$$

The full posterior is $$P(\pmb{X},\pmb{\Lambda}|\pmb{Y})$$

and he states that:

the $$\lambda_a$$ are assigned independent priors uniform over a fixed, specified, and finite range, so that the posteriors are simply truncated gamma distributions that for $$\lambda_a$$ have shape pa-rameter $$X_i$$+1 and scale parameter 1;

Explicitly for the Poisson distribution:

With likelihood

$$f(x|\lambda) = \frac{\lambda^x e^{-\lambda}}{x!}$$

and prior

$$f(\lambda) = \frac{1}{b-a} \textbf{1}_{a \leq x \leq b}$$

where $$\textbf{1}_A$$ is the indicator function,

, you get posterior

$$f(\lambda|x) \propto \frac{\lambda^x e^{-\lambda}}{x!} \cdot \frac{1}{b-a} \textbf{1}_{a \leq x \leq b} \propto \lambda^x e^{-\lambda}\cdot \textbf{1}_{a \leq x \leq b}$$

This expression is of the form of a truncated gamma distribution.

More generally, if you truncate the prior, then your posterior will truncate in the same way. As a consequence, a truncated conjugate distribution is itself a conjugate distribution.

The Gamma distribution is a conjugate prior for the Poisson distribution where the range is unlimited. So a truncated Gamma a conjugate prior when the range is limited.

In the computations here, I am ignoring a constant and express Bayes theorem as $$f(\theta|x) \propto f(x|\theta)f(\theta)$$ instead of the possibly more familiar $$f(\theta|x) = \frac{f(x|\theta)f(\theta)}{f(x)}$$ Ignoring the constant makes the computations more easy and it doesn't matter for the final result anyway (the constant of proportionality doesn't change the shape of the function which it is about). In this question you find another example Obtain Bayes estimator with conjugate prior .

• @StephanKolassa I formulated the answer on purpose indirectly, because I believe that the OP is required to do some algebra themselves to practice. --- From my answer it follows that, since the Gamma is a conjugate prior for the Poisson distribution when the range is unlimited. So a truncated Gamma a conjugate prior when the range is limited. --- It can be more explicitly derived when the expressions like $f(x|\theta)$ are filled into the general expressions of this answer. Commented Jul 24 at 10:10
• @StephanKolassa you could see the uniform distribution as the limit of a gamma with infinite variance. --- The nature of the question, not homework but clarifying a statement from an article, was not initially clear to me. Commented Jul 24 at 10:24
• +1 I could see as soon as I read the question this was simply a truncated gamma. If you start considering a purely flat (improper) prior, the likelihood provides the kernel of a gamma and then the actual uniform prior simply truncates that. I think much more than the guidance that is here may be risking overdoing it. To me it does outline the steps in the calculations pretty clearly as is. Commented Jul 24 at 18:44