Understanding Probability Theory: P(min(X, Y ) > x) Explained

rad0786
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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 textbook.

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
 
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rad0786 said:
But just what on Earth does P(min(X, Y ) > x) mean?
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.
 
okay... i see...

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

How about P(min(X,Y)<x)... that's just P(X<x)P(Y<x) right??
 
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}<br /> \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}<br /> \int _{{-\infty}}^{x}<br /> \Muserfunction{pdf}(X,Y)\,dY\,dX

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

-Dale
 
yes thanks this helps!
 
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