My study group and I are stuck on this Bayes' estimator problem. The question is:

Let X~Pois($\lambda$) Find the Bayes estimator for $\lambda$ with respect to:

(i) The prior distribution: $\lambda$ ~ exp(1)

(ii) The squared error loss function

So for our posterior distribution we ended up with:

$g(\lambda|x_1...x_n)= \dfrac{\frac{\lambda^{\Sigma x_i}}{\Pi x_i!}e^{-\lambda(n+1)}}{\int_0^\infty \frac{\lambda^{\Sigma x_i}}{\Pi x_i!}e^{-\lambda(n+1)}, d\lambda} $

which is a [Gamma distribution][1] with parameters $\Sigma x_i + 1$ and $n+1$.

Then our Bayes' estimator with respect to the Prior and the Squared Error Loss Function is the expected value which we get:

$\frac{\Sigma x_i + 1}{n+1} $

Which we are just not sure if this is right or not.

Could someone please let us know if our thinking is correct on this one?



1 Answer 1


This is a fairly straightforward example to check for one simple reason: Your prior is the conjugate prior for Poisson data. Your prior $\lambda \sim Exp(1)$, can be written as a Gamma distribution, because $$\lambda \sim Exp(1) \Rightarrow \lambda \sim \Gamma(1,1).$$ The posterior distribution should then also be a Gamma. I won't go through the math on it, but you can check Wikipedia's Table of Conjugate Priors to verify the distribution. You are correct that the posterior distribution is a $\,\Gamma(\sum x +1, n+1)$.

A Bayes Estimator under squared error loss is just the posterior mean, which yields $$E (\lambda | \mathbf{x}) = \frac{\sum{x_i} + 1}{n+1},$$ just as you have written.

  • $\begingroup$ Thanks so much for the sanity check! We were really banging our heads over this one! $\endgroup$
    – Perdue
    Commented Apr 27, 2013 at 22:59
  • 1
    $\begingroup$ What would be the Bayes estimator for a function of λ? $\endgroup$
    – Harry
    Commented Oct 28, 2016 at 14:23

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