# Probability theory.

1. Dec 12, 2005

Probability theory.......

So we were given some practise problems for the exam....
in it, we got a question that we have NEVER seen in class nor can I find it in the text book.

The question is:

Find P(min(X, Y ) > x) and hence give the probability density function of U =
min(X, Y ).

Okay.. so i know how to answer it.. i think.. he gave us a solution...

But just what on earth does P(min(X, Y ) > x) mean? could their be P(max(X, Y ) > x) and if so, what does that mean?

Somebody care to take a moment and explain please?

Thanks

2. Dec 12, 2005

### Galileo

It's the probability that the minimum of X and Y is larger than x. What is it you not understand? min(X,Y) is clearly larger than x if both X and Y are.

3. Dec 12, 2005

okay... i see...

what about P(max(X,Y)>x) can that exist

How about P(min(X,Y)<x)... thats just P(X<x)P(Y<x) right??

4. Dec 12, 2005

### Staff: Mentor

If you have the joint pdf of X and Y then you can get P(min(X, Y ) > x) by integrating the joint pdf over the proper region.

$$\int _{x}^{\infty}\int _{x}^{\infty} \Muserfunction{pdf}(X,Y)\,dY\,dX$$

The other ones that you mention are the same idea, just integrating over different regions of the (X,Y) plane. For example, P(max(X,Y)>x) would be

$$1 - \int _{{-\infty}}^{x} \int _{{-\infty}}^{x} \Muserfunction{pdf}(X,Y)\,dY\,dX$$

I am sure you can get the rest of these similarly.

-Dale

5. Dec 12, 2005