# Bayesian prior

• Feb 22nd 2010, 03:42 AM
straussvogel
Bayesian prior
Dear all,

I have a small understanding problem with the concept of Bayesian priors.
I use a computer program which samples a posterior distribution $\displaystyle P(\theta | D)$ using Markov-Chain Monte-Carlo given some Data $\displaystyle D$ and prior information on a parameter $\displaystyle \theta$.
$\displaystyle P(\theta | D) = P(D | \theta) * \frac{P(\theta)}{P(D)}$

Lets assume the the simple case that my data D is missing, so my posterior distribution is determined only by the prior and not the likelihood function:
$\displaystyle P(\theta | D) = P(\theta)$

The prior information about theta is that it should have a value of $\displaystyle 152\pm 4$.

The computer program incorporates priors the following way:

$\displaystyle P(\theta)=\frac{1}{2}\left( \frac{log\theta - log\theta^*}{\sigma_{log\theta}} \right)^2$

where $\displaystyle \theta^*$ is my prior knowledge: 152.
The logarithm of the parameters is taken to not get the parameter values get negative.

My question is now: How do I put my prior information $\displaystyle \theta^*=152\pm 4$ into the prior function or in other words: what should be the value of
$\displaystyle \sigma_{log\theta}$?

The manual of the software says, if, for example I want to constrain $\displaystyle \theta$ with 95% probability between $\displaystyle \theta / 100$ and $\displaystyle \theta * 100$, $\displaystyle \sigma_{log\theta}$ must be log(10), because I think in the normal distribution, a value lies with 95% probablities within 2 standard deviations of the mean, so:
$\displaystyle \theta / 100 + 4 * \sigma = \theta * 100$, solving for $\displaystyle \sigma$ yields
$\displaystyle \sigma = log(10)$

However, I tried to do the same for my 152+-4 prior as follows:

To constrain $\displaystyle \theta$ between $\displaystyle \frac{\theta}{\sqrt{156/148}}$ and $\displaystyle \theta * \sqrt{156/148}$, I calculated $\displaystyle \sigma=\frac{1}{4} * log(156/148)$.

But here comes the problem: When I put this value as $\displaystyle \sigma_{log\theta}$ in the program, the sampled distribution of $\displaystyle \theta$ does really have a mean of 152, like I would expect. But the standard deviation of $\displaystyle \theta$ is always about 2 and not 4, as I would expect.

Does anybody see, what I got wrong here? I'd appreciate any help, I went over it a lot of times and checked by calculations, but this does not make any sense to me!
By the way, sorry for the long post, but I hope my problem is clear...

Greetings,
straussvogel