Bayes formula and picking variance from distribution

1. Sep 29, 2011

freshmanaskin

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

freshmanaskin

Thank you mute, all you said seems very right and true to me. with large enough mu compared to sigma it will become pretty normal. I might have to use some other distribution to pick the variance if i use this for something. But this was more of a though experiment.

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