
#1
May1710, 04:04 PM

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! 



#2
May1710, 04:12 PM

Emeritus
Sci Advisor
PF Gold
P: 16,101

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 



#3
May1710, 04:25 PM

P: 41

The formula for general RV X is [tex]EX = \int_{\infty}^{0} F(x) dx + \int_{0}^{\infty} (1  F(x)) dx[/tex]. 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.




#4
May1810, 02:33 AM

P: 1,270

Finding E(X) from distribution functionThanks! 



#5
May1810, 04:47 AM

P: 41

Try to prove it by manipulating the general formula I posted, it's not hard.



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