# Expectation of (aX-bY) ?

1. Dec 17, 2013

### nikozm

Hello,

I 'm trying to express the following in integral form:

E[a/X-b/Y], where E[.] stands for the expectation operator.

Let a,b be some nonnegative constants and X,Y are independent nonnegative Gamma distributed random variables.

Any help would be useful.

2. Dec 17, 2013

### mathman

Do you want aX - bY or a/X - b/Y? The first is easy.
For the second you need to first get the distribution functions for the reciprocals. Then use the addition formula.

3. Dec 17, 2013

### nikozm

i 'm interested in evaluating the second on (i.e., a/X - b/Y). Can you be more specific ? the addition formula with what integration bounds exactly?
what if i would use the expectation of X and Y seperately (i.e, E[c/X]-E[b/Y])?

4. Dec 17, 2013

### Office_Shredder

Staff Emeritus
Splitting the expectation by linearity is the obvious first step. The hard question is given a distribution X, find E(1/X). This is non-trivial in general, but luckily for you for a gamma distribution this has already been done

http://en.wikipedia.org/wiki/Inverse-gamma_distribution

5. Dec 17, 2013

### nikozm

This a well-known result for resiprocal PDFs. In fact, this is a generic rule for every distribution function (i.e. taking the inverse variable within the initial PDF and multiplying it with the derivative of the inverse variable gives the corresponding reciprocal PDF).

I was hoping for a solution by applying the original Gamma PDFs directly..

6. Dec 18, 2013

### mathman

Let X be a random variable with density function f(x), then E(1/X) = ∫(f(x)/x)dx.
If there is no density and the distribution function is F(x), you need a Stieljes integral, E(1/X) = ∫(1/x)dF(x)