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

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- Jul 18th 2008, 06:54 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, 04:53 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 $\displaystyle \Pr(X > u) \geq \Pr(Y > u)$:

1. $\displaystyle \Pr(X > Y) = \int_{y=-\infty}^{\infty} g(y) \int_{x = y}^{\infty} f(x) \, dx \, dy = \int_{u=-\infty}^{\infty} g(u) \Pr(X > u) \, du$.

2. $\displaystyle \Pr(X < Y) = \Pr(Y > X) = \int_{x=-\infty}^{\infty} f(x) \int_{y = x}^{\infty} g(y) \, dy \, dx = \int_{u=-\infty}^{\infty} f(u) \Pr(Y > u) \, du$.

3. $\displaystyle \Pr(X > Y) + \Pr(X < Y) = 1$.

Now use 1., 2., 3. and $\displaystyle \Pr(X > u) \geq \Pr(Y > u)$ to show that $\displaystyle \Pr(X > Y) + \Pr(X > Y) \geq 1 \Rightarrow \Pr(X > Y) \geq \frac{1}{2}$. - Jul 19th 2008, 06:27 PMvioletsf
Thank you very much!

how about #3, I mean where is P(X=Y) - Jul 19th 2008, 07:32 PMmr fantastic
- Jul 20th 2008, 11:04 AMvioletsf
Another question: f(u) and g(u) are not the same, right?

then how can you compare #1 and #2 ?

Thanks! - Jul 20th 2008, 09:44 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).