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.

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    $\begingroup$ Hi zahida and welcome to the site. I changed your formula to proper LaTeX code. Please check carefully if they are still correct. $\endgroup$ Jun 21 '14 at 17:24
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    $\begingroup$ This is not a well-defined distribution until you specify the range of $x$. $\endgroup$
    – whuber
    Jun 21 '14 at 17:31
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    $\begingroup$ And specify any constraints on the parameters/ $\endgroup$
    – wolfies
    Jun 21 '14 at 17:33
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    $\begingroup$ 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. $\endgroup$
    – whuber
    Jun 21 '14 at 17:37
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    $\begingroup$ It looks a bit like a Stacy distribution. $\endgroup$
    – wolfies
    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.


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