# Mean and variance of loggamma distribution

1. Nov 2, 2014

### Bipolarity

The loggamma distribution is defined by
$$g(x) = \frac{1}{ \Gamma ( α) θ^{ α} } \frac{(ln( x))^{ α - 1}}{x^{1+\frac{1}{θ}}}$$, for $$1 < x < ∞$$

where α is a positive integer.
I've been trying to find the mean and variance of this distribution. It's been somewhat frustrating because the integral is rather difficult to compute, so I was thinking of using the moment-generating function to do this computation. However, the resulting expression seems very messy so I'm not sure it's the way to proceed.

Thanks.
BiP

2. Nov 4, 2014

### chiro

Hey Bipolarity.

If you use a substitution you should get u = ln(x) and du/dx = 1/x. This means integrating gives

= C*u^(alpha-1)*(1/(e^u)^(1/s))*1/x (C is a constant independent of the integration variable)
= C*u^(alpha-1)*e^(-u/s)*du

which has the same form as the normal gamma distribution PDF.

For expectations you just have to multiply by x before doing the substitution and this will give you (for E[X]) the following:

x*g(x) with substitution u = ln(x) gives a final integral of
= C*u^(alpha-1)*(1/(e^u)^(1/s - 1))*1/x
= C*u^(alpha-1)*e^(-u/(s-1))*du

Notice that the only thing that changes is the parameter within the exponential and this is of the gamma family of distributions.

If you do the same thing for E[X^2] you get

C*u^(alpha-1)*e^(-u/(s-2))*du since multiplying by x^2 will change the denominator of x^(1/s+1) to x^(1/s+1-2) = x^(1/s-1).

Using this you should be able to compute means and variances pretty easily using a trick where you pull a constant out of your integral and make sure that everything inside the integrand is just a density function for a particular set of values of a Gamma distribution. You evaluate this to one and whatever is left outside the integral (a constant not involving the integrand) is your mean or second moment.

It's very common to do this trick for these kinds of distributions (i.e. Gamma).