# st: RE: mean of a distribution

 From "Nick Cox" To Subject st: RE: mean of a distribution Date Sun, 18 Oct 2009 19:12:56 +0100

```Not sure what you're asking here.

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.

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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