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st: RE: mean of a distribution


From   "Nick Cox" <n.j.cox@durham.ac.uk>
To   <statalist@hsphsun2.harvard.edu>
Subject   st: RE: mean of a distribution
Date   Sun, 18 Oct 2009 19:12:56 +0100

Not sure what you're asking here. 

One way to think about this parameterisation, or any other, is by using
dimensional analysis. The density of a univariate distribution for x
must have units that are the reciprocal of the units of x. It follows
that lambda has such units, and the mean, having the units of x, must be
proportional to the reciprocal of lambda. That doesn't give you the
proportionality constant of 1, which follows from the rest of the
definition, but it makes the reciprocation intuitive. 

David Finney wrote a splendid article about dimensional analysis and
statistics:

D. J. Finney. 1977.  
Dimensions of statistics. 
Journal of the Royal Statistical Society. Series C (Applied Statistics),
26: 285-289. 

This is on JSTOR, if you have access to that. 

Nick 
n.j.cox@durham.ac.uk 

carol white

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?

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