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 HW Helper P: 1,391 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.