Let's say we have two normally distributed populations A and B, which have different means and standard deviations. We then pick one item from population A and one from population B. How could we analytically find out the probability that the item from population B is larger in value than the item from population A?
I know that this can be solved fairly easily with some Monte-Carlo simulation, but I was curious to see how this problem could be solved analytically.
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Think of it as a three-dimensional anthole. If A and B are independent then this is the same as the Probability (A-B) > 0 coming from a distribution of (A-B) which has mean $\mu_A - \mu_B$ and $ var(A-B) = var(A)+var(B)$ Then you just have to work out the Z-score of 0 in that distribution.
For instance, Let's say A is given by $N(3,5)$ and B is given by $N(2,12)$ (I'm writing these in mean, standard deviation form) then A-B is given by $N((3-2),\sqrt{25+144}) = N(1,13)$ A-B=0 occurs with a Z-score of $\frac{-1}{13}$ which is 0.2206 or 22%. 22% of the time B will be greater than A and 78% of the time A will be greater than B.
Lets say the joint pdf of A and B is $f(a,b)$. The analytical solution is:
$\int_{b\geq a} f(a,b) dadb$, this will just be the integral above the line $a=b$ on the domain of $f(a,b)$