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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? * * For searches and help try: * http://www.stata.com/help.cgi?search * http://www.stata.com/support/statalist/faq * http://www.ats.ucla.edu/stat/stata/

**References**:**st: mean of a distribution***From:*carol white <wht_crl@yahoo.com>

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