Posterior of exponential likelihood and gaussian prior

Question

Likelihood function: $$L(\zeta) = \zeta^{\alpha} \exp\left[\zeta\sum_{i=1}^{\alpha}(x_i - x)\right]$$

Prior function: $$p(\zeta) = \frac{1}{(\sqrt{2π}σ_\zeta)} \exp\left[-\frac{(\zeta-\zeta_0)^2}{(2σ_\zeta^2)}\right]$$

I am trying to find the log posterior of $\zeta$

Bayes Theorem: $$\log(posterior) \stackrel{\log}{\propto} \log(L(\zeta))+\log(p(\zeta)$$

Solving this I get:

$$\log(posterior) \stackrel{\log}{\propto}\log[\zeta^{\alpha}] +\zeta\sum_{i=1}^{\alpha}(x_i - x)-\frac{(\zeta-\zeta_0)^2}{2\sigma_{\zeta}}-\log(\sigma_{\zeta})$$

but I don't not know how to to proceed.

Edit:

Here

• $x_{i} ≤ x$ where $x$ is a limiting constant
• We have ${x_1,…x_{\alpha}} $independently and identically distributed random sample.

Answer 1

Calling Wolfram integrator, one gets \begin{align*}\int_0^\infty x^a \exp(-1/2 (x - b)^2) \text dx =& 2^{(a - 1)/2} \left\{\sqrt{2} b Γ(a/2 + 1) _1F_1((1 - a)/2, 3/2, -b^2/2)\\ + Γ((a + 1)/2) _1F_1(-a/2, 1/2, -b^2/2)\right\}\\ \end{align*} for $a>-1$, where $_1F_1$ denotes the hypergeometric function. Since the posterior can be written as the above integrand, its normalisation constant can thus be found this way, but the end result is not much more informative than keeping an unspecified constant.