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


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?

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

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.

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