I am trying to take consecutive powers of a Gamma Distribution. For example, if

$X \sim \textrm{Gamma}(k, \theta)$,

I would like to find $X^2$, $X^3$, and in general $X^m$ for $m>0$.

The pdf using the shape-scale parametrization is

$f(x;k,\theta) = \frac{x^{k-1}e^{-\frac{x}{\theta}}}{\theta^k\Gamma(k)} \quad \text{ for } x > 0 \text{ and } k, \theta > 0$.

I have started to do this by working by the following:

$P(X^m \leq x) = P(-\sqrt[m]{x} \leq X \leq \sqrt[m]{x}) = \int_{-\sqrt[m]{x}}^{\sqrt[m]{x}} \frac{t^{k-1}e^{-\frac{t}{\theta}}}{\theta^k\Gamma(k)} \,dt$.

Then, $\quad f_{X^m}(x) = \frac{d}{dx}\int_{-\sqrt[m]{x}}^{\sqrt[m]{x}} \frac{t^{k-1}e^{-\frac{t}{\theta}}}{\theta^k\Gamma(k)} \,dt$ $\quad$

which by the Fundamental Theorem of Calculus equates to:


After this, I am stuck because there are many different cases to consider, such as when $m$ is an integer, etc.

Could anyone tell me if my strategy is correct? Thanks!

  • 4
    $\begingroup$ "I would like to find $X^2$, $X^3$," -- you can't mean what you say. Do you mean you want to find the distribution of $X^2$ (and so on)? Do you mean you want to find the expectation of $X^2$ etc? Or something else? Please clarify via an edit of your question. $\endgroup$
    – Glen_b
    Jan 16, 2015 at 10:29
  • 2
    $\begingroup$ You could use directly the Jacobian formula and by all means forgo the use of the $\sqrt[n]{x}$ notation. $\endgroup$
    – Xi'an
    Jan 16, 2015 at 12:25
  • $\begingroup$ I am trying to show that $T^m$ for $m>0$ is a gamma distribution as well, that is, the gamma distribution is closed under taking powers. $\endgroup$
    – user123276
    Jan 16, 2015 at 15:53

1 Answer 1


Your strategy will fail. Powers of $X$ follow Generalized Gamma distributions. Their densities are given by

$$f(x; k, \gamma, 1) = \frac{1}{\Gamma (k)} x^{\gamma k} e^{-x^{\gamma }} \left(\gamma \frac{dx}{x} \right).$$

(The third parameter, here set to a unit value $\sigma=1$, is a scale parameter.) For the power $m$ of $X \sim \Gamma(k)$, the shape parameter is $\gamma = 1/m$. This all becomes obvious when you think of $x^\gamma$ as being another variable, say $u$, and note that

$$\frac{du}{u} = \frac{d(x^\gamma)}{x^\gamma} = \gamma \frac{d x}{x}.$$

The remaining factors, $u^k e^{-u}/\Gamma(k)$, give the PDF of a Gamma distribution (with respect to the measure $du/u$).

The issue before us is whether for any given $k$ it is possible to find other parameters $k^\prime$ and $\sigma^\prime$ such that $f(x,k, \gamma, \sigma)=f(x;k^\prime, 1, \sigma^\prime)$, where $\gamma = 1/m$ for some integral $m \gt 1$: that is what it would mean for $X^m$ to have an (ordinary) Gamma distribution.

We can settle this by looking at moments. Integration shows that the (non-central) moments of the generalized Gamma distribution are

$$\mu_i = \sigma^i\frac{\Gamma(k+m i)}{\Gamma(k)}.$$

For integral values of $m$ their successive ratios are

$$\frac{\mu_{i+1}}{\mu_i} = \frac{\sigma^{i+1}\Gamma(k + mi + m)}{\sigma^i\Gamma(k + m i)} = \sigma(k + mi)(k+mi+1)\cdots (k+mi+m-1).$$

Trying to solve equations (or rather, prove they have no solutions) involving these formulas looks messy. Instead, consider the limiting values

$$\lim_{i\to \infty}\frac{\mu_{i+1}}{i\mu_i}.$$

For $m=1$ (a true Gamma distribution) this is the limiting value of $\sigma(k+i)/i$, obviously equal to $\sigma$, whereas for $m\gt 1$ this is the limiting value of $\sigma(k+mi)\cdots(k+mi+m-1)/i$, which (being of order $i^{m-1}$) diverges. Therefore the distribution of $X^m$ for $m\gt 1$ (and integral) cannot have the same moments as any possible Gamma function, proving it is not a Gamma function.

It should be clear that the generalized Gamma distribution family is indeed closed under taking (positive) powers.

  • $\begingroup$ thank you for your explanation. you mentioned that the generalized gamma distribution family is indeed closed by taking powers. to do this would I just use my integration method? $\endgroup$
    – user123276
    Jan 21, 2015 at 5:45
  • $\begingroup$ It's even simpler than that. Suppose $X$ has a generalized Gamma distribution. That means $X$ has the same distribution as a (positive) power $\gamma$ of some gamma-distributed variable $Y$. Therefore for any positive power $\alpha$, the distribution of $X^\alpha$ is that of $(Y^\gamma)^\alpha=Y^{\gamma\alpha}$, showing that $X^\alpha$ also has a generalized Gamma distribution. Consequently, generalized gammas are closed under taking (positive) powers. $\endgroup$
    – whuber
    Jan 21, 2015 at 15:31

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