Need Some Mathematical Guidance Regarding Random Variables

[joshthekid]

This is not a homework question but I project I am working on and need someone with more mathematical prowess than myself. I am using a computer program to draw random numbers from two independent distributions, x1 and x2, for two different cases and I want to establish a theoretical mathematical relationship for the probability that x1 will be less then x2. The first case is two different exponential distributions, i.e. exp(-αx), and the second from a power law, x^(-β), over a limited range a to b. I have been working on this for about a week so any help or guidance would be much appreciated. Thanks

Josh

 

[eigenperson]

What you need to do is write down the joint probability distribution function. This is a function $p_{X,Y}$ in variables x and y such that for any region A, $\iint_A p_{X,Y} = P((X, Y) \in A)$.

Because your random variables are independent, $p_{X,Y}(x,y) = p_X(x)p_Y(y)$, where $p_X$ and $p_Y$ are the pdfs of your individual random variables.

Then you just integrate over the region A that is the set of points where X < Y, and you have P(X < Y).

I'll do the exponential distribution as an example. $p_X(x) = \alpha e^{-\alpha x}$, and $p_Y(y) = \beta e^{-\beta y}$. So the joint pdf is $p_{X,Y}(x,y) = \alpha \beta e^{-\alpha x - \beta y}$. Now you integrate:

$P(X < Y) = \int_0^\infty \int_0^y \alpha \beta e^{-\alpha x - \beta y}\,dx\,dy = \alpha \beta \int_0^\infty e^{-\beta y} {1 - e^{-\alpha y}\over \alpha}\,dy = 1 - {\beta \over \alpha + \beta} = {\alpha \over \alpha + \beta}.$

So that's the answer.

 

[joshthekid]

thanks Eigenperson!

Josh