# Thread: trying (and failing) to get mean of 2 parameter exponential distribution

1. ## trying (and failing) to get mean of 2 parameter exponential distribution

I have the following 2 parameter exponential density function:
a*e^(-a*[x-b]), where x>=b, a>0, neg infinity < b < pos infinity

I start by evaluating:
Mean = int[x*a*e^(-a*[x-b])]dx bounded by neg. and pos. infinity

I solve the integral with integration by parts to:
-x*e^(-a*[x-b]) - (1/a)*e^(-a*[x-b]) evaluated between neg. and pos. infinity.

The Mean comes to be pos. infinity. If I change the lower bound the integral is evaluated over I get that the Mean is (1/a), which, I think, is the right answer. I am stumped why I can't generate (1/a) with the bounds of integration as neg. and pos. infinity. Ideas?

2. let Y=X-b and this becomes the usual exponential distribution.
the mean and variance of Y are well known and easy to compute.
even if you wish to integrate this directly, you should use calc 1 subsitution, y=x-b.
AND the lower bound of x is b in your integral.
You need to show your work.