# Bayes formula and picking variance from distribution

1. Sep 29, 2011

This should be rather simple bayesian problem, but I can't figure it out for myself.

If i pick numbers from normal distributionS, where the variance of the distribution at each pick v1, is in turn picked out of a normal distribution with variance v2.

What is then the distribution of the random number? i tried this on my calculator a bit, and it looks as if it is normal itself, but what is the variance of x?

This is not homework, but something I would like to understand how to calculate.

//Cal

2. Sep 29, 2011

### mathman

You need to clarify this.

3. Sep 29, 2011

### Mute

If you have two continuous random variables (for example), X and Y, with joint pdf $\rho_{X,Y}(x,y)$, this can be written in terms of the conditional distribution $\rho_{X|Y}(x|y)$ or $\rho_{Y|X}(y|x)$ as:

$$\rho_{X,Y}(x,y) = \rho_{X|Y}(x|y)\rho_Y(y) = \rho_{Y|X}(y|x)\rho_X(x),$$

where the individual distributions are

$$\rho_X(x) = \int_{-\infty}^\infty dy~\rho_{X,Y}(x,y),$$
and similarly for y. So, if you knew the distribution for y and you knew the condition distribution for x given y, you can calculate the distribution of x as

$$\rho_X(x) = \int_{-\infty}^\infty dy~\rho_{X|Y}(x|y)\rho_Y(y).$$

Amusingly enough, I was dealing with this concept myself this week, although in the context of a much more intractable problem.

This said, I think you need to tweak your suggested problem. The variance of a distribution is positive, so it can't be normally distributed. If you use the standard deviation, I don't think you'll get a result, because you'll have a factor of $1/\sigma$ in your integration, why of course blows up at the origin, but the exponential will also be even in sigma, so the principal value might be zero.

Last edited: Sep 29, 2011
4. Sep 30, 2011