# Moment generating function of a distribution

I want to find the moment generating function (mfg) and mean deviation of this distribution:

$$f(x,\epsilon,k,\theta) = k\theta^{(1+1/k+\epsilon/k)}x^{(k+\epsilon)}\exp{(-\theta x^k )}/(\Gamma(1+(1+\epsilon))/k)$$

where $$\epsilon, k, \theta$$ are the three parameters of this distribution. $$x$$ ranges from $$0$$ to infinity.

• Hi zahida and welcome to the site. I changed your formula to proper LaTeX code. Please check carefully if they are still correct. Jun 21 '14 at 17:24
• This is not a well-defined distribution until you specify the range of $x$.
– whuber
Jun 21 '14 at 17:31
• And specify any constraints on the parameters/ Jun 21 '14 at 17:33
• If we assume the range is $0\le x \lt \infty$ and reasonable values for the parameters, then when $k\gt 0$ is rational the mgf can be expressed as a finite algebraic combination of generalized hypergeometric functions.
– whuber
Jun 21 '14 at 17:37
• It looks a bit like a Stacy distribution. Jun 21 '14 at 17:38

This is a Generalized Gamma distribution: up to a scale factor, $$x^k$$ has a Gamma distribution.

E. W. Stacy studied the Generalized Gamma distribution in a 1962 paper available on Project Euclid. His parameterization is more natural for statistical applications and leads to simpler formulas. The density function is determined by the density for a unit scale factor $$a=1$$ by

$$f(x;1,d,p)\mathrm{d}x =\frac{1}{C(p,d)} x^d \exp(-x^p) \frac{\mathrm{d}x}{x}$$

for $$0 \lt x \lt \infty$$ (see Stacy equation (1)). The constant of proportionality is readily found via the substitution $$x^p=u,$$ yielding

$$C(p,d) = \int_0^\infty x^d \exp(-x^p)\frac{\mathrm{d}x}{x} = \frac{\Gamma(d/p)}{p}.$$

To find the $$r^\text{th}$$ moment, integrate $$x^r$$ against $$f\mathrm{d}x.$$ No additional calculations are needed, because the preceding formula already gives the result upon replacing $$d$$ by $$r+d$$:

$$\mu_r(p,d) = \frac{1}{C(p,d)} \int_0^\infty x^{r+d} \exp(-x^p) \frac{\mathrm{d}x}{x} = \frac{C(p, r+d)}{C(p, d)} = \frac{\Gamma((r+d)/p)}{\Gamma(d/p)}.$$

By definition, these give the terms in the expansion of the moment generating function (mgf),

$$M_{p,d}(t) = \sum_{r=0}^\infty \frac{\mu_r(p,d)}{r!} t^r = \frac{1}{\Gamma(d/p)} \sum_{r=0}^\infty \frac{\Gamma((r+d)/p)}{r!} t^r.$$

(See Stacy equation (3).) As usual, the scale factor $$a \gt 0$$ is accommodated by replacing $$t$$ by $$a t$$ in the mgf.

The parameterization given in the question identifies $$k$$ with $$p,$$ $$1+k+\epsilon$$ with $$d$$, and $$1/\theta^{1/k}$$ with $$a.$$

I presume the "mean deviation" refers to some function of these moments, but since I'm unsure which one, I will leave the algebra to the reader.