0
$\begingroup$

i have the following problem:

given a variable $x \in \mathbb{R}$ drawn from a Gaussian distribution with known mean $\mu$ and unknown precision $\tau$ (the inverse of the variance). So: $$p(x \vert \tau) = \mathcal{N}(x; \mu, \tau^{-1})$$

Now we assume as our prior that $\tau$ is Gamma distributed: $$p(\tau) = \mathcal{G}(\tau; k, \theta) = \frac{\tau^{k-1}}{\Gamma(k)\theta^k} \exp\left(\frac{-\tau}{\theta}\right)$$

The goal now is to compute the posterior distribution : $p(\tau \vert x)$

What I have so far:

I want to calculate the posterior via the Bayes theorem: $$p(\tau \vert x) = \frac{p(x \vert \tau) \cdot p(\tau)}{\int p(x \vert \tau) \cdot p(\tau) d\tau}$$ After doing some calculations I came up with $$p(x \vert \tau) \cdot p(\tau) = \frac{\tau^{k-1}}{\sqrt{2 \pi \tau^{-1}} \cdot\Gamma(k)\theta^k} \exp\left(-\frac{(x - \mu)^2}{2\tau^{-1}} + \frac{-\tau}{\theta}\right)$$

So is there any other way than integrating this term and to some nasty calculations or is this the way I have to go? Or am I completely wrong here using the Bayesian approach?

Tanks for answers!

$\endgroup$
0
$\begingroup$

There is an error in you posterior formula, normalization should be wrt $\tau$. The correct formula is

\begin{equation} p(\tau|x)=\frac{p(x|\tau)p(\tau)}{\int d\tau p(x|\tau) p(\tau)} \end{equation}

You have that

\begin{equation} p(x|\tau)p(\tau)\propto \tau^{k-\frac{1}{2}}e^{-\tau\Bigg(\frac{(x-\mu)^2}{2}+\frac{1}{\theta}\Bigg)} \end{equation}

which is proportional to $\mathcal{G}\Bigg(\tau,k-\frac{1}{2},\Bigg(\frac{(x-\mu)^2}{2}+\frac{1}{\theta}\Bigg)^{-1}\Bigg)$ for the purposes of integration in $\tau$ and thus you can calculate the renormalization constant. I leave to you the details of the calculation and collecting all factors.

$\endgroup$
  • $\begingroup$ oh yes, you are right. Thank you! $\endgroup$ – pprime May 21 at 16:22

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.