X and Y are independent random variables. If P(X>u)>=P(Y>U) for all u, then P(X>Y)>=1/2

Thanks a lot

Printable View

- Jul 18th 2008, 07:54 AMvioletsfprobability question
X and Y are independent random variables. If P(X>u)>=P(Y>U) for all u, then P(X>Y)>=1/2

Thanks a lot - Jul 19th 2008, 05:53 AMmr fantastic
Assume X and Y are continuous random variables with pdf f(x) and g(y) (the argument is easily modified for the discrete case). Then the joint pdf is f(x) g(y) due to independence of X and Y.

What my argument boils down to is that in the XY-plane, the 'weighted area' below the line Y = X is greater that the 'weighted area' above the line Y = X since :

1. .

2. .

3. .

Now use 1., 2., 3. and to show that . - Jul 19th 2008, 07:27 PMvioletsf
Thank you very much!

how about #3, I mean where is P(X=Y) - Jul 19th 2008, 08:32 PMmr fantastic
- Jul 20th 2008, 12:04 PMvioletsf
Another question: f(u) and g(u) are not the same, right?

then how can you compare #1 and #2 ?

Thanks! - Jul 20th 2008, 10:44 PMmr fantastic
Right.

I'll need to think about this! It seems less simple now than when I posted.

I'm confident the approach is correct, but it might not be the most fruitful approach (I don't see what other approach could be taken at this moment).