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Re: st: mean of a distribution


From   Jeph Herrin <junk@spandrel.net>
To   statalist@hsphsun2.harvard.edu
Subject   Re: st: mean of a distribution
Date   Sun, 18 Oct 2009 20:31:06 -0400


I think you are confusing the random variable X and the values
in its domain, x. It's true taht

E(X) = integral xf(x)

but you have to integrate over the real line - it's an integral,
not just an anti-derivative. In this case, you find the anti derivative
(I think using integration by parts) and plug in the end points
of the reals (plus & minus infinity), and the answer is 1/lambda.

hope this helps,
Jeph



carol white wrote:
Hi, How to calculate the mean of the distribution of a random
variable? Take the exponential distribution with the probability
density function f(x)=lambda.exp(-lambda.x) where lambda is a
constant and x is a random variable. The mean of this distribution is
the reciprocal of lambda. If the mean is the expected value of x,
which for a continuous random variable E(x) = Integral (x.f(x))dx,
how could E(x) be the reciprocal of lambda?

Regards,

Carol



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