Bayes formula and picking variance from distribution

[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

 

[mathman]

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.

You need to clarify this.

 

[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.

 

[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.

 

[CellCoree]

culated response//

Bayes formula is a mathematical tool used in Bayesian statistics to update the probability of a hypothesis as new evidence becomes available. In this scenario, we can use Bayes formula to calculate the distribution of the random number picked from a normal distribution with variance v1, which is in turn picked from a normal distribution with variance v2.

The formula for this problem would be P(v1|v2) = P(v2|v1) * P(v1) / P(v2), where P(v1|v2) is the probability of picking a number with variance v1 given that the variance of the distribution it was picked from is v2, P(v2|v1) is the probability of picking a variance v2 given that the number picked has a variance v1, P(v1) is the prior probability of picking a number with variance v1, and P(v2) is the prior probability of picking a variance v2.

By using this formula, we can calculate the probability distribution of the random number picked from the normal distribution. The resulting distribution will also be normal with a mean of 0 and a variance of v1 + v2. This means that the variance of the random number will be the sum of the variances from the two normal distributions.

It is important to note that this is a simplified scenario and in real-world problems, there could be other factors to consider. However, Bayes formula can still be a useful tool in understanding and solving Bayesian problems involving picking numbers from distributions.