0
$\begingroup$

If you have a random variable U(X,Y) that is a function of two other random variables X and Y such that

$U(X,Y)=X+Y$

and you know the PDFs of X and Y are defined to be exponential such that

$f(t) = \lambda e^{-\lambda t}u(t) $

then you know $f_X(x) = f_Y(y)$ (i.e. X and Y's PDFs are equal).

You can use this information to compute the PDF of U:

$f_U(u) = f_X(x) * f_Y(y)$

where * means the convolution.

I computed $f_U(u)=\lambda^2ue^{-\lambda u}$ using the definition of convoution.

However, I can't use convolution to compute

$f_V(v)$ when V = $\frac{X}{X+Y}$

Convolution only works for sums of random variables (like U=X+Y), but not when you're dividing random variables.

How do I find the PDF of V?

$\endgroup$
  • 1
    $\begingroup$ You might want to apply the usual method. $\endgroup$ – Did Oct 22 '14 at 20:27
1
$\begingroup$

To find the PDF in more complicated cases, the general method is to make changes of variable in integrals with unknown functions.

Let $g$ be a measurable positive function.

$$ Eg\left(\frac{X}{X+Y}, X+Y\right) = \dots = \int f(u,v) g(u,v) du dv $$

The goal here is to arrive to the RHS. Then $f$ is the joint PDF of $(U,V)$, which is more informative. In bonus, the PDF splits in a product $f_1(u)f_2(v)$ iff $U,V$ are independent.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.