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(?, ?)


  • 2
    $\begingroup$ How do you know that the square of a Gamma even has a shape and rate parameter analogous to those for the Gamma. $\endgroup$ Commented May 12, 2017 at 2:26
  • $\begingroup$ 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. $\endgroup$
    – Glen_b
    Commented May 12, 2017 at 3:03

1 Answer 1


It seems that by squared gamma, you mean:

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.


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