What is the Expectation of (aX-bY)?

  • Context: Graduate 
  • Thread starter Thread starter nikozm
  • Start date Start date
  • Tags Tags
    Expectation
Click For Summary

Discussion Overview

The discussion revolves around the expectation of the expression E[a/X - b/Y], where X and Y are independent nonnegative Gamma distributed random variables, and a and b are nonnegative constants. Participants explore how to express this expectation in integral form and discuss the challenges involved in evaluating it.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant seeks assistance in expressing E[a/X - b/Y] in integral form.
  • Another participant clarifies the expression, asking if the interest is in aX - bY or a/X - b/Y, noting that the first is straightforward.
  • A participant confirms the focus on a/X - b/Y and requests specifics on the addition formula and integration bounds.
  • One participant suggests that splitting the expectation by linearity is a reasonable first step, but emphasizes the difficulty in finding E(1/X) for a given distribution.
  • Another participant mentions a known result for reciprocal PDFs and expresses a desire to apply the original Gamma PDFs directly.
  • A participant provides a formula for E(1/X) involving the density function and discusses the use of Stieljes integrals when no density is available.

Areas of Agreement / Disagreement

Participants generally agree on the complexity of evaluating E(1/X) for Gamma distributions and the need for specific integration techniques. However, there is no consensus on the best approach to express or evaluate E[a/X - b/Y].

Contextual Notes

The discussion highlights the challenges in deriving expectations for reciprocal random variables, particularly in relation to the Gamma distribution, and the potential need for additional mathematical tools or methods.

Who May Find This Useful

Readers interested in probability theory, particularly those studying expectations of functions of random variables, may find this discussion relevant.

nikozm
Messages
51
Reaction score
0
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.

Thanks in advance
 
Physics news on Phys.org
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.
 
Thanks for the reply.

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 separately (i.e, E[c/X]-E[b/Y])?
 
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
 
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..
 
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)
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K