Question

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

Answer 1

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

Explanation:

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