A variable x is said to have an exponential distribution wit

A variable x is said to have an exponential distribution with parameter lambda > 0 if the density function for x is [Note: this distribution is the same as in lecture, where lambda is a parameter and lambda = 1/mu.] Find the mean of the exponential distribution in terms of lambda. [i.e. Find ] The variance of a continuous distribution can be found by using Using the mean found in part(a), find the variance of the exponential distribution.

Solution

a) = xf(x) dx

= _-⁰0*dx + ₀ x*e^-xdx

= ₀ x*e^-xdx

now using integration by parts

x*e^-xdx = x*e^-xdx- dx/dx*(e^-xdx)dx

= -(x/)e^-x - (-1/)*e^-xdx

= -(x/)e^-x - 1/² *e^-x

So

₀ x*e^-xdx = *|-(x/)e^-x - 1/² *e^-x|₀

= -*(-1/²) = 1/

i.e. = 1/

b) ² = (x-)²f(x)dx

= _-⁰0*dx + ₀ (x-)²*e^-xdx

= ₀ (x)²*e^-xdx + ²**₀ e^-xdx - 2*₀ x*e^-xdx

now by integration by parts

(x)²*e^-xdx = x²*-1/*e^-x - 2x*-1/*e^-x

= -x²*1/*e^-x + 2/*x*e^-xdx

So

₀ (x)²*e^-xdx + ²**₀ e^-xdx - 2*₀ x*e^-xdx

= *|-x²*1/*e^-x + 2/*x*e^-xdx|₀ - ^2*(0-1) - 2*1/

= -*(2/*-1/²) + 1/² - 2/²

= 1/²

So

  ² = 1/² = ²