shape and rate of the square of a variable having a gamma distribution

from this answer (Expectation of a squared Gamma) I would like to know the shape and rate parameters of a squared gamma. I struggle a bit here.

Gamma(shape, rate)^2 -> Gamma(?, ?)


Thanks.

• How do you know that the square of a Gamma even has a shape and rate parameter analogous to those for the Gamma. May 12 '17 at 2:26
• Your title and your body seem to ask different things. The title asks about a variable whose square is gamma while the body seems to ask about a variable which is the square of a gamma (which are different things). Both seem to rely on the mistaken idea that the result will itself be gamma. It is not so. Please clarify what you are asking. May 12 '17 at 3:03

If $X \sim \Gamma(\alpha, \lambda)$, then $X^2 \sim \Gamma(\mathrm{?}, \mathrm{?})$.
If this were the case, then unfortunately, $X^2$ does not follow a Gamma distribution. To see this, assume $X \sim \Gamma(\alpha, \lambda)$, let us write $Y = X^2$. Then clearly, both $X$ and $Y$ have $\mathbb{R}^+$ as their supports. Now, by transformation of random variable, we may find the p.d.f. of $Y$.
For $y>0$, we have:
\begin{align} f_Y(y) &= \frac{\mathrm{d}}{\mathrm{d}y} \mathrm{Pr}(Y\le y) \\\\ &= \frac{\mathrm{d}}{\mathrm{d}y}\mathrm{Pr}(X\le \sqrt{y}) \\\\ &= \frac{\mathrm{d}\sqrt{y}}{\mathrm{d}y}\frac{\mathrm{d}}{\mathrm{d}\sqrt{y}}\mathrm{Pr}(X\le \sqrt{y}) \\\\ &= \frac{1}{2\sqrt{y}}f_X(\sqrt{y}) \\\\ &=\frac{1}{2\sqrt{y}}\frac{\lambda^\alpha}{\Gamma{(\alpha)}}(\sqrt{y})^{\alpha - 1}\mathrm{e}^{-\lambda\sqrt{y}} \\\\ &= \frac{\lambda^\alpha}{2\Gamma{(\alpha)}}(\sqrt{y})^{\alpha - 2}\mathrm{e}^{-\lambda\sqrt{y}} \\\\ \end{align} and $f_Y(y) = 0$ otherwise.
From the form of the p.d.f., we may note that $Y$ does not follow a Gamma distribution, at least not a "usual" one. Nevertheless, for $y>0$, we may write: \begin{align} f_Y(y) &= \frac{\lambda^\alpha}{2\Gamma{(\alpha)}}(\sqrt{y})^{\alpha - 2}\mathrm{e}^{-\lambda\sqrt{y}} \\\\ &= \frac{(\lambda^2)^{\frac{\alpha}{2}}}{2\Gamma{(\alpha)}}(y)^{\frac{\alpha}{2} - 1}\mathrm{e}^{-(\lambda^2 y)^{\frac{1}{2}}} \\\\ \end{align} Then it can be viewed as a Generalized Gamma Distribution with parameters $p=\frac{1}{2}$, $d=\frac{\alpha}{2}$ and $a=\lambda^{-2}$. Noted other parametrizations are also possible.