X and Y are independent exponential random variables

Question

Suppose that X and Y are independent exponential random variables each with mean 1. What is P(Y > X²)?

Answer 1

I[0, ∞) I[x²,∞) e-^xe-^y dy dx = I[0, ∞) e^{-x-x^2} dx p

This may seem impossible to integrate, but if we complete the square on the exponent by adding and substracting 1/4 and recognize that this is a normal distribution with variance 1/2 centered at -1/2 if we have 1/sqrt(pi)) in the denominator, we have

e^1/4 * sqrt(pi) * I[0, inf) 1/sqrt(pi) e - (x+1/2)^2

We can now get the integral from the distribution of the normal, as the standard deviation is 1/sqrt(2), and the integral begins 1/2 to the right of the mean equals 1/2/(1/sqrt(2)) = 1/sqrt(2) standard deviations to the right, so the integral is e^1/4 * sqrt(pi) * normsdist(-1/sqrt(2)) (an integral to the right is 1 - the integral to the left, but we start 1/sqrt(2)) standard deviations to the right)

e^1/4 * sqrt(pi) * normsdist(-1/sqrt(2)) = 0.545641360765047