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.



  • $x_{i} ≤ x$ where $x$ is a limiting constant
  • We have ${x_1,…x_{\alpha}} $independently and identically distributed random sample.
  • 2
    $\begingroup$ You've got Bayes' Theorem wrong: it says $$\textrm{posterior}\propto L(\zeta)\times p(\zeta)$$ On the log scale that's $$\log\textrm{posterior}=C+\log L(\zeta)+\log p(\zeta)$$ where $e^C$ is the constant of proportionality in the first equation. The constant $C$ is defined by the fact that the posterior density is a density, and so integrates to 1. Also, I assume $x_i$ are the data, but what are $\alpha$ and $x$? And are you sure about the likelihood, because it doesn't look familiar. $\endgroup$ Mar 3 at 7:36
  • $\begingroup$ Also, the constant of proportionality usually has to be estimated numerically -- is there some reason you expect it to be explicitly computable in this example. $\endgroup$ Mar 3 at 7:39
  • $\begingroup$ What would the normalisation condition be for log(posterior)? Is it $Int(C+logL(ζ)+logp(ζ)) =0$ $\endgroup$
    – rexy qmech
    Mar 3 at 7:47
  • 1
    $\begingroup$ @ThomasLumley I replaced the $\propto$ with a $\stackrel{\log}{\propto}$ and corrected the likelihood function, assuming the data was a shifted exponential starting at $x$. The OP made a simultaneous edition to make it an Exponential stopping at $x$. $\endgroup$
    – Xi'an
    Mar 3 at 7:52

1 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.


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