# Gamma-distributed variable to the power $3$

If $X^\frac 13\sim\operatorname{GAM}(\theta,3)$ (so $\kappa=3$). Then what is the distribution of $X$? Is there any way to do this? I have tried by making use of the original MGF and doing this to the power $3{:}$ $\frac{1}{(1-\theta t)^\kappa}=\frac{1}{(1-\theta t)^3}$ for $X^\frac 13$ and $\left(\frac{1}{(1-\theta t)^3}\right)^3=\frac{1}{(1-\theta t)^9}$ giving $X\sim\operatorname{GAM}(\theta,9)$ but am not sure whether this is the right way, are you allowed to 'just' do the original MGF to the power $3$?

Maybe it depends on what you mean by finding the distribution. For example, to say something is Gamma-distributed doesn't just identify what it's probability density is, but locates it within a family of distributions that are known to have certain properties. But if you only want to know the density function, then your problem is tractable. \begin{align} f_X(x) = {} & \frac d {dx} \Pr(X\le x) = \frac d {dx} \Pr( X^{1/3} \le x^{1/3}) \\[10pt] = {} & \frac d {dx} \int_0^{x^{1/3}} \frac 1 {\Gamma(3)} \left( \frac u \theta \right)^{3-1} e^{-u/\theta} \, \left(\frac{du} \theta\right) \\[10pt] = {} & \frac 1 {\Gamma(3)} \left( \frac {x^{1/3}} \theta \right)^{3-1} e^{-x^{1/3}/\theta} \cdot \frac 1 \theta \cdot \frac d {dx} x^{1/3} \\[10pt] = {} & \frac 1 {6\theta^3} e^{-x^{1/3}/\theta}. \\[10pt] & \text{All of the above is for $x\ge0$.} \end{align}

POSTSCRIPT

The comment below this answer indicates a desire to find $\operatorname E(X).$ You don't need to know the density of $X$ to do that; in fact $99\%$ of that problem has already been done for you, as follows. You have $X=Y^3$ and $Y$ has a Gamma distribution, so \begin{align} & \operatorname E(X) = \operatorname E(Y^3) = \int_0^\infty y^3 f_Y(y) \, dy \\[10pt] = {} & \int_0^\infty y^3\cdot \frac 1 {\Gamma(3)} \left(\frac y \theta\right)^{3-1} e^{-y/\theta} \, \left( \frac{dy} \theta\right) \\[10pt] = {} & \frac {\theta^3} {\Gamma(3)} \int_0^\infty \left( \frac y \theta \right)^{6-1} e^{-y/\theta} \, \left( \frac{dy} \theta \right) \\[10pt] = {} & \frac{\theta^3}{\Gamma(3)} \int_0^\infty w^{6-1} e^{-w} \, dw \tag 1 \\[10pt] = {} & \frac{\theta^3}{\Gamma(3)} \cdot \Gamma(6). \end{align} You see that the integral in line $(1)$ is one that you've already done before; its value is $\Gamma(6).$ That's what I meant by saying $99\%$ is already done, i.e. what is above is a trivial reduction to an integral that you already know.

Finally $$\theta^3 \frac{\Gamma(6)}{\Gamma(3)} = \theta^3 \cdot \frac{5 \cdot 4 \cdot 3 \cdot \Gamma(3)}{\Gamma(3)} = 60\theta^3.$$

If this was an examination question and the problem was just to find $\operatorname{E}(X),$ then almost certainly the method given in this postscript is what was intended, for two reasons: (1) Working with the density of $X$ itself seems too cumbersome for an examination unless it's just by reducing it to the same integral that we see on line $(1)$ above, and (2) I would think you be expected to know how to do it that way. Probably this item (2) is why the question was there.

• Thank you for the explanation but I had already been given the pdf, however, in the test question we got the distribution of X^1/3 but we had to know E(X), so I thought we had to find X's distribution and use this distribution's characteristics as to find the mean. Wondered if there was any other way to 'easily' do this aside from using a generalized gamma. – Walter Nap May 30 '18 at 19:19
• @WalterNap : I've added a postscript to my answer in response to your comment above. You were doing it the hard way. – Michael Hardy May 30 '18 at 19:53
• @WalterNap : If this was a question on an examination, then I have little doubt that the intended method is what I wrote in the postscript, rather than by finding the density for $X$ itself. $\qquad$ – Michael Hardy May 30 '18 at 19:55