0
$\begingroup$

Suppose the prior for $μ$ is $Gamma(k,λ)$, so $fμ(m)=Cm^{k−1}exp(−λm)I(m≥0)$.

Further, suppose the data $X_1,…,X_n$ given $μ$ is iid Poisson with parameter $μ$, so for each $[Xj∣μ=m]∼[X∣μ=m]$, $f_X∣μ=m(i)=exp(−m)m^i/i!I(i∈{0,1,…})$,

How can I show that the posterior for μ is also Gamma? Would the parameters of the Gamma be a function of $k$ and $\lambda$?

$\endgroup$
0
$\begingroup$

The prior is

$$\pi(\mu)\propto \mu^{k-1}e^{-\lambda\mu}$$

The likelihood is

$$p(\mathbf{x}|\mu)\propto e^{-n\mu}\mu^{\Sigma_i x_i} $$

Thus the posterior is

$$\pi(\mu|\mathbf{x})\propto \pi(\mu)\cdot p(\mathbf{x}|\mu) =\mu^{(k+\Sigma_i x_i)-1}e^{-(\lambda+n)\mu}$$

Thus the posterior is still gamma with parameters

$$Gamma(k+\Sigma_i x_i;\lambda+n)$$

Hint: to simplify the calculations it is suggested to waste any quantity that doesn't depend on $\mu$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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