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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.
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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
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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
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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
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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
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