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RE: st: RE: mean of a distribution

From   "Nick Cox" <>
To   <>
Subject   RE: st: RE: mean of a distribution
Date   Sun, 18 Oct 2009 19:39:37 +0100

The word `calculate' is a little ambiguous. For distributions like the exponential, the formula you quote gives the answer through calculus. 


carol white

Thanks for all responses. 

I just wanted to find out how could one have calculated the mean for example of the exponential distribution if the mean were not known (inverse of lambda)? Or as Martin called it shortcut, if the shortcut were not known?

--- On Sun, 10/18/09, Nick Cox <> wrote:

> From: Nick Cox <>
> Subject: st: RE: mean of a distribution
> To:
> Date: Sunday, October 18, 2009, 11:12 AM
> 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 
> 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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