# Proof of Expectation for Continuous Uniform Distribution

• Aug 2nd 2012, 06:45 PM
mh03
Proof of Expectation for Continuous Uniform Distribution
Hi everyone, I have been looking at some proofs recently and came across this one for the expectation of a Continuous Uniform Distribution. Not sure how it was done, and was wondering if anybody could guide me through the steps, and if there are any extra workings so that I can understand it better?

If X ~ U(a,b), then:

E(X) = http://www.mathsrevision.net/alevel/...xpectation.GIF

Any help would be very much appreciated
• Aug 3rd 2012, 01:14 AM
emakarov
Re: Proof of Expectation for Continuous Uniform Distribution
Which of the four equalities don't you understand?
• Aug 3rd 2012, 01:31 AM
mh03
Re: Proof of Expectation for Continuous Uniform Distribution
the last three really. I am rusty with my calculus, and was hoping somebody could explain the process of how they moved from one to the next.
• Aug 3rd 2012, 01:38 AM
emakarov
Re: Proof of Expectation for Continuous Uniform Distribution
The second inequality uses the Fundamental Theorem of Calculus because $\displaystyle (x^2/2)'=x$. The notation $\displaystyle [g(x)]_a^b$ meand $\displaystyle g(b) - g(a)$. The last inequality holds because $\displaystyle b^2-a^2=(b-a)(b+a)$.
• Aug 3rd 2012, 05:02 AM
HallsofIvy
Re: Proof of Expectation for Continuous Uniform Distribution
That is, $\displaystyle \left[\frac{x^2}{2}\right]_a^b$ means $\displaystyle \frac{b^2}{2}- \frac{a^2}{2}= \frac{b^2- a^2}{2}= \frac{(b-a)(b+a)}{2}$