Gamma-distributed variable to the power $3$

Question

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 $3$?

Answer 1

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.