Re: Expectation of positive random variable
- From: "Stephen J. Herschkorn" <sjherschko@xxxxxxxxxxxx>
- Date: Sun, 27 May 2007 16:40:54 -0400
Ji-Woo Kim wrote:
I have problem regarding the title.
Could you explain why following statement is true?
If X is a random variable with P(X >= 0) = 1, then E{X} = int_0^{inf}
P(X > x) dx
I have posted this at least twice before over the past several years, but it is easier to retype than find.
Let I(u) be the indicator variable for {X > u}. Then X = integral(u=0..infty, I(u)). Thus,
EX = E integral(u=0..infty, I(u)) = integral(u=0..infty, EI(u)), where the interchange of integral and expectation is justified since I is nonnegative (Tonelli's theorem).
--
Stephen J. Herschkorn sjherschko@xxxxxxxxxxxx
Math Tutor on the Internet and in Central New Jersey and Manhattan
.
- References:
- Expectation of positive random variable
- From: Ji-Woo Kim
- Expectation of positive random variable
- Prev by Date: Hilbert cube has the fixed point property
- Next by Date: Integral of even functions
- Previous by thread: Expectation of positive random variable
- Next by thread: [Graph theory]Sum of chromatic number
- Index(es):
Relevant Pages
|
