Explain p.d.f. of the sum of random variables

Click For Summary

Discussion Overview

The discussion revolves around finding the probability density function (pdf) of the sum of two random variables given their joint pdf. It explores various methods and considerations, including the implications of dependence between the variables and the validity of the resulting pdf.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant inquires about the method to find the pdf of the sum of two random variables with a joint pdf f(x,y).
  • Another participant notes that if the random variables are dependent, there is no straightforward solution.
  • A participant suggests that the pdf may diverge if an integral involves a constant, raising a question about the validity of the pdf in such cases.
  • One response asserts that the pdf should not diverge if it is valid, but acknowledges that it may not yield finite moments.
  • Another participant proposes using the joint characteristic function to find the pdf of the sum, detailing the process of inverse transforming.
  • A different approach is presented, where the probability that X+Y is less than a certain value z can be expressed as a double integral, with a method for evaluating it through a change of variables.

Areas of Agreement / Disagreement

Participants express differing views on the implications of dependence between the random variables and the conditions under which the pdf remains valid. The discussion includes multiple methods for finding the pdf, indicating that no consensus has been reached on a singular approach.

Contextual Notes

There are unresolved assumptions regarding the nature of the random variables and the conditions under which the pdf is considered valid. The discussion also highlights the complexity of evaluating integrals in this context.

exoCHA
Messages
2
Reaction score
0
Hi,
I need your help,
Say we have two random variables with some joint pdf f(x,y). How would I go about finding the pdf of their sum?
 
Physics news on Phys.org
If the two random variables are dependent there is no easy solution. (I tried Google!)
 
I think I found the answer: google docs... page 10

another question, if I end up with an integral like that and a constant inside, then does the pdf of the two variables diverges and there's no answer?
 
Hey exoCHA.

The PDF won't diverge if it's a valid PDF.

The PDF however may not give finite moments but that is a completely different story. Your PDF should be a valid PDF if you did everything right and it should contain a region that has finite realizations of Z if the domain of your X and Y in your joint PDF are also finite for something like Z = X + Y.

Remember that even a Normal distribution is valid for the entire real line just like the Z will also be valid across the entire real line.
 
Another way to find the pdf would be to calculate the joint characteristic function

$$\Gamma(\mu_x,\mu_y) = \langle e^{i\mu_x x + i\mu_y y} \rangle = \int_{-\infty}^\infty dx \int_{-\infty}^\infty dy~\rho(x,y) e^{i\mu_x x + i\mu_y y}.$$

One can recover the joint pdf by inverse transforming in the two different mu's, but you can also find the pdf of their sum by setting both ##\mu_x=\mu_y = \mu## and inverse Fourier transforming in ##\mu##:

$$\rho_Z(z) = \int_{-\infty}^\infty \frac{d\mu}{2\pi} \Gamma(\mu,\mu)e^{-i\mu z},$$
where z = x + y. (Whether or not this integral is easy to do is of course a separate issue).
 
If you have a density function f(x,y), then the probability that X+Y < z can be expressed as a double integral,
∫∫f(x,y)dxdy where the integral range is given by x+y < z. This can be evaluated by u=x+y replacing x, so du = dx. The double integral now has y range (-∞,∞) and u range (-∞,z) and g(u,y) = f(u-y,y) as the integrand.
 

Similar threads

  • · Replies 30 ·
2
Replies
30
Views
5K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
4
Views
3K