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Mute
#3
Sep29-11, 03:23 PM
HW Helper
P: 1,391
If you have two continuous random variables (for example), X and Y, with joint pdf [itex]\rho_{X,Y}(x,y)[/itex], this can be written in terms of the conditional distribution [itex]\rho_{X|Y}(x|y)[/itex] or [itex]\rho_{Y|X}(y|x)[/itex] as:

[tex]\rho_{X,Y}(x,y) = \rho_{X|Y}(x|y)\rho_Y(y) = \rho_{Y|X}(y|x)\rho_X(x),[/tex]

where the individual distributions are

[tex]\rho_X(x) = \int_{-\infty}^\infty dy~\rho_{X,Y}(x,y),[/tex]
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

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

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 [itex]1/\sigma[/itex] 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.