# Thread: How to find P(x1<x2)

1. ## How to find P(x1<x2)

X2 and x1 being 2 independent exponential variables with mean 5 and 2.
I found f(x1) and f(x2) and defined y as x2 - x1 but I'm stuck there. How do I calculate f(y)? or is there another way of finding P(x1<x2).
Thank you!

2. ## Re: How to find P(x1<x2)

Just doing the integration directly shouldn't be too bad for this problem.

3. ## Re: How to find P(x1<x2)

Thanks but if I knew what to integrate I wouldn't have posted the question!

4. ## Re: How to find P(x1<x2)

Do you know the joint distribution? Can you draw out the region the defines where x1 < x2?

5. ## Re: How to find P(x1<x2)

@Dason. No I don't know the joint distribution

6. ## Re: How to find P(x1<x2)

The joint distribution of independent random variables is just the product of their corresponding pdfs.

7. ## Re: How to find P(x1<x2)

I did not do it with the joint function. I integrated f(x1) from 0 to x2 and subtracted the result from 1.
I was wondering if I should have used y=x1-x2 instead and solve P(y>0)

8. ## Re: How to find P(x1<x2)

Originally Posted by statgoon13
I did not do it with the joint function. I integrated f(x1) from 0 to x2 and subtracted the result from 1.
How does that get you the answer? Wouldn't that be a function of x2 still?
I was wondering if I should have used y=x1-x2 instead and solve P(y>0)
That is one way to do it but it isn't nearly as easy as the method I mentioned earlier.

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