Expectation prob. Proof

1. Apr 16, 2007

theperthvan

1. The problem statement, all variables and given/known data
Prove that
E(X) > a.P(X>a)

2. Relevant equations
E(X) is expectation, a is a positive constant and X is the random variable.
(Note, > should be 'greater than or equal to' but I'm not too sure how to do it)

3. The attempt at a solution
Well I can show it easy enough for X~Exp(h), but of course this is not a general proof.
And I played around a bit with P(X>a)=1-F(a) where F is the cdf, but yeah. That's all I could really do. I just don't really get how the a factor comes in.

2. Apr 16, 2007

Dick

Take P(x) to be a normal distribution centered at 0. The E(x)=0. Yet a.P(x>a) is clearly positive for a>0. There's something wrong with the problem statement. '>=' is fine for greater than or equal to.

3. Apr 16, 2007

theperthvan

Sorry, X is non-negative.

4. Apr 17, 2007

Dick

Ok then. Define E(X) as an integral and split the integral into the ranges 0-a and a-infinity. Drop the first integral and think about approximating the second by something smaller.