Let Xk be independent exponentially distributed random variables with mean 2 . Find E[min{X1,X2,X3...,X100}] Solution If the mean is 2, = 1/2, and Fxi(x) = 1 - e-1/2x for all i, i = 1, 2, ..., 100 Then, P(min(x1,x2,x3,...x100) <= X) = 1 - P(all x1, x2, x3, ..., x100 >= X) = 1 - P(x1 >= X)P(x2 >= X) ...P(x100 >= X) = 1 - (1 -e-1/2x) (1 - (1 -e-1/2x)... (1 - (1 -e-1/2x)= 1 - (1 - (1 -e-1/2x) 100 = 1 - (e-1/2x)100 = 1 - e-50x This is the CDF for the negative exponential distribution with parameter 50 Thus, E(x) = 1/50.