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I did this recently, so im posting even though this is old.
Assuming X and Y and gaussian (normal/bell curve) Random Variables.
G = Y - X
Therefore G is a gaussian random variable.
the probability that G is less than 0 equals the probability that X > Y
Mean for G:
E[G] = E[Y] - E[X]
Var[X] = E[(X-E[X])^2] = stdev[X]^2
Var[Y] = E[(Y-E[Y])^2] = stdev[Y]^2
Var[G] = E[(G-E[G])^2]
Var[G] = E[(Y-X-(E[Y]-E[X]))^2]
Var[G] = E[((Y-E[Y])-(X-E[X]))^2]
Var[G] = E[((Y-E[Y])^2+(X-E[X])^2-2*(Y-E[Y])(X-E[X]))]
Var[G] = Var[Y] + Var[X] - 2*COV[X,Y]
If X and Y are independant then the covariance equals zero.
You now have the stdev and mean of G, and can use the characteristic function of a gaussian random variable to find the probability.
For the numbers you provided,
E[G] = mean of G = -5389.25
Var[G] = variance of G = 26045501
Stdev[G] = sqrt(var[G]) = 5103.479
Using E[G]/Stdev[G] = -1.056 and a table for characteristic function values, the probability that X > Y = 0.8531
In other words, 85%.
Correct me if you can.
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