What is the variance of a probability distribution function of the form: f(x)=ke^(-2x)f(x)=ke2x?

1 Answer
Oct 27, 2015

The distribution is an exponential distribution. k = 2 and E (x) = 1/2 ,
E ( x^2x2 )= 1/2 => V (x) = E ( x^2x2) - {E (x)}^2{E(x)}2 - 1/2 - (1/2) ^2(12)2 = 1/2 - 1/4 = 1/4.

Explanation:

The limit of the distribution is (0, oo ) To find k,
int_0^BB0 k e^- (2x)e(2x) dx = k Gamma (1)/ 2 = 1 => k/2 = 1 => k = 2.
E (x) = #int_0^Bx