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#### Options  Subscribe or Unsubscribe   Log In   Get Password Subject: Re: QUERY: gamma distribution SUMMARY

From:  Date: Thu, 14 Oct 1999 15:20:37 +0100

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 ```I asked for an algebraic explanation of my observation that the geometric mean of a gamma distributed sample with parameter alpha is close to alpha-0.5. Several people pointed out that the geometric mean can be expressed as the ratio of two gamma functions, and Quentin Burrell managed to justify the approximation alpha-0.5 by expanding the digamma function. Separately there was some confusion about the parametrisation of the gamma distribution, with one or two parameters. Many thanks to Julian Besag, Quentin Burrell, Mike Franklin, Anthony Davison, Phil Brain, Rob Crawford and Roger Newson. Their replies follow. Tim Cole ----- Original Message ----- I have convinced myself that the geometric mean of a sample from a gamma distribution with parameter alpha (i.e. arithmetic mean = variance = alpha) has expectation (alpha-0.5) for alpha5 or so, based on the exponential of the corresponding digamma function. Can anyone show this identity algebraically? I couldn't see it in Abramowitz and Stegun. --------------------------------------------------------- From: Julian Besag <[log in to unmask]> Tim,   You mean you want to show that    {gamma(alpha + 1/n) / gamma(alpha)} to power n is approx alpha-0.5?   Julian --------------------------------------------------------- From: "Quentin L. Burrell" <[log in to unmask]> Tim I haven't seen this result before so I tried to derive it along the following lines. Writing G = geometric mean of n observations and psi for the digamma function we get E[G] = [ gam(alpha + 1/n) / gam(alpha) ] ^ n which, using Taylor's series expansion' gives E[G] ~ [ 1 + psi(alpha) / n + o(1/n^2) ] ^ n          ~ exp[psi(alpha)] which I guess reproduces what you had. If you now use the asymptotic formula 6.3.18 from Abramowitz and Stegun, (with sufficient arm-waving approximation!) :- psi(z) ~ ln z - 1/(2z) - o(1/z^3) to get E[G] ~ (alpha) exp(-1/2alpha)          ~ (alpha)[ 1 - (1/2alpha) + ... ]          ~ alpha - 1/2 Does this fulfil your requirements? --------------------------------------------------------- From: Michael Franklin <[log in to unmask]> Tim,     Not the answer but it might help. In my ancient Kendall and Stuart Vol 1, p177, Ex 6.13. 1963 there is an exercise on the moments of the gamma distribution for which you are asked to show the expectation of log y is (d/da)log gamma(a) where a is your alpha. Given Stirling's approximations to log gamma your result might follow. Mike Franklin --------------------------------------------------------- From: Anthony Davison <[log in to unmask]> YOu get a ratio of two gamma functions. I think that if you replace each using Stirling's formula (6.1.37) and simplify it will come out. But what's wrong with simple expansion of the top gamma function; for reasonable \$n\$ it should be OK. --------------------------------------------------------- From: "phil.brain" <[log in to unmask]> Dear Tim, I'm not sure if I've understood the question properly, so may be answering the wrong one... The geometric mean is E((X1 x X2 x ... Xn)**(1/n)). Assuming that X1, X2, etc are independent, this is the same as (E(X**(1/n)))**n so the problem is to evaluate E(X**(1/n)). I'm assuming that the gamma p.d.f. (f(x)) is given by f(x) = a(ax)**(b-1) exp(-ax) / Gamma(b) Then E(X**(1/n)) = Integral from 0 to infinity of (1/a**(1/n)* a*(ax)**(b+1/n-1) * exp(-ax) / Gamma(b) with respect to x = (Gamma(b+1/n)/Gamma(b)) / a**(1/n) Thus the geometric mean is ((Gamma(b+1/n)/Gamma(b))**n)/a I may have made a mistake, so do check my logic/algebra out! All the best, Phil Brain --------------------------------------------------------- From: Rob Crawford <[log in to unmask]> you must be using a different definition of gamma dist from me. the tattered old text book I keep on returning to (Freud and Walpole mathematical stats 3rd edition Prentice hall 1980) has for gamma distibution f(x) = 1/beta^alpha * 1/gamma(alpha) * x^(alpha-1) * exp( -x/beta) where gamma(.) is the usual gamma function ( ie gamma(n) = (n-1)!) they show that E(x) = alpha*beta and var(x) = alpha*beta^2 --------------------------------------------------------- From: Roger Newson <[log in to unmask]> The gamma distribution, as generally understood, has parameters r and lambda, such that the density function f(.) is equal to Gamma(r) * lambda**(r) * x**{r-1} * exp(-lambda*x) (for x>0), where Gamma(r)=(r-1)! for r an integer. Its arithmetic mean and variance are mean = r/lambda variance = r/(lambda**2) so that r is the inverse squared coefficient of variation. It follows that the variance of a gamma distribution is not always equal to the mean. (However, in the special case where r=1, then the gamma distribution is an exponential distribution, and the mean is equal to the standard deviation, as the coefficient of variation is one.) The geometric mean is smaller than the arithmetic mean by an amount depending on the coefficient of variation 1/r**2. I do not know where you got the idea that a gamma distribution has a variance equal to the mean, although you may be confusing the gamma distribution with the Poisson distribution. However, you may be interested in estimating geometric means and their ratios for y-variates with a gamma distribution, or, more generally, y-variates whose coefficient of variation is fairly constant between groups. To do this, you can use the Stata command -glm- with options -family(gamma) link(log) eform-. Or, alternatively, use the command -rglm- (in STB-50), which is like -glm-, but calculates standard errors so as to be robust to deviations from the assumption that the coefficient of variation is constant between groups. I hope this is useful. Regards Roger [log in to unmask] Phone +44(0)20 7905 2666 Fax +44(0)20 7242 2723 Epidemiology & Public Health, Institute of Child Health, London WC1N 1EH, UK %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% ```