# Finding E(X) from distribution function

by kingwinner
Tags: distribution, function
 P: 1,270 Theorem: Let F(x) be the distribution function of X. If X is any r.v. (discrete, continuous, or mixed) defined on the interval [a,∞) (or some subset of it), then E(X)= ∞ ∫ [1 - F(x)]dx + a a 1) Is this formula true for any real number a? In particular, is it true for a<0? 2) When is this formula ever useful (computationally)? Why don't just get the density function then integrate to find E(X)? Thanks for clarifying!
 Emeritus Sci Advisor PF Gold P: 16,098 1) If there was a restriction on a, the statement of the theorem should have said something. If the statement is right for positive a, then it's surely right for negative a as well. 2) Why would you go through all the trouble of looking for the density distribution*, multiplying by x, and integrating that when you could just use that formula? *: The hypotheses cover the situation where the density cannot be written as a function
 P: 41 The formula for general RV X is $$EX = -\int_{-\infty}^{0} F(x) dx + \int_{0}^{\infty} (1 - F(x)) dx$$. This formula works in a much more general setting than you might expect. Some distributions don't have densities (singular distributions), for example http://en.wikipedia.org/wiki/Cantor_distribution but the formula still applies.
P: 1,270
Finding E(X) from distribution function

 Quote by Mandark The formula for general RV X is $$EX = -\int_{-\infty}^{0} F(x) dx + \int_{0}^{\infty} (1 - F(x)) dx$$. This formula works in a much more general setting than you might expect. Some distributions don't have densities (singular distributions), for example http://en.wikipedia.org/wiki/Cantor_distribution but the formula still applies.
I've seen this general formula. But does it imply that the "theorem" above is true for a<0 (e.g. a=-2, or a=-2.4) as well? I've done some manipulations and I think the theorem above is true for ANY a, but I would like someone to confirm this.

Thanks!
 P: 41 Try to prove it by manipulating the general formula I posted, it's not hard.

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