If i have a random variable with distribution $X \sim \Gamma(\alpha,\beta)$ then what would be the distribution of $Y = \lambda X^2$ (with $\lambda$ a scaling factor)? Can I say that $Y$ will follow a gamma distribution as well?

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    $\begingroup$ You can say that $Y=\lambda X^2$ will follow a gamma distribution if you like, but it would not be a true statement. $\endgroup$ – Dilip Sarwate May 26 '15 at 21:47
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    $\begingroup$ Because all aspects of this kind of question have been addressed in other threads, it likely can be answered by a suitable site search. To focus it, please tell us what approaches you have taken to solving this problem and where you got stuck. $\endgroup$ – whuber May 26 '15 at 21:53
  • $\begingroup$ @DilipSarwate It would be helpful if you would be willing to share your wisdom as why it would not follow a gamma distribution $\endgroup$ – michael May 27 '15 at 13:46
  • $\begingroup$ @whuber I did an extended Google search. And also a search on stackexchange from which I could not find this question answered previously (wrt square gamma random variable, the constant factor was just to be more specific to my particular problem). $\endgroup$ – michael May 27 '15 at 13:47

By definition, the probability element of $X$ has the form

$$f_X(x)dx \propto x^\alpha \exp(-x) \frac{dx}{x}$$

for $x\gt 0$ (and equal to $0$ otherwise).

Let $y = \lambda x^2$ (with $\lambda \gt 0$). Because $x$ is positive, this is a one-to-one transformation, entailing both

$$x = \left(\frac{y}{\lambda}\right)^{1/2}$$


$$dy = 2\lambda x dx,$$


$$\frac{dx}{x} = \frac{dy}{2\lambda x^2} = \frac{1}{2} \frac{dy}{y}.$$

Consequently the probability element for $Y$ is

$$f_Y(y) dy = \frac{1}{2}\left(\frac{y}{\lambda}\right)^{\alpha/2} \exp\left(-\left(\frac{y}{\lambda}\right)^{1/2}\right)\frac{dy}{y}.$$

By rescaling it we can recognize the PDF as having the basic form $y^{\alpha/2-1} \exp(-\sqrt{y})$. It is a Generalized Gamma distribution. Such distributions, by definition, are positive powers of Gamma distributions.

If $Y$ had a Gamma distribution, say with parameter $\beta$ and scale $\sigma$, then the ratio of the two (unnormalized) expressions for their PDFs would have to be some nonzero positive constant $C$. Equivalently, for all $y\gt 0$ it must be the case that

$$y^{\alpha/2 - \beta} = C\exp\left(\sqrt{y/\lambda}-y/\sigma\right).$$

The right hand side decreases exponentially as $y\to\infty$ whereas the left hand side is only a power law: the two cannot be the same.

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    $\begingroup$ Now that I have remembered this is a Generalized Gamma, it was easy to find a duplicate by searching the site for "Generalized Gamma" PDF. The search turned up some additional posts of interest, too: take a look. $\endgroup$ – whuber May 27 '15 at 19:00

Assuming you're using the parametrization $f(x;\alpha, \beta) = \frac{\beta^\alpha}{\Gamma(\alpha)}x^{\alpha - 1}e^{-\beta x}$, a result commonly seen in mathematical statistics classes is that

$$X \sim \Gamma(\alpha, \beta) \Rightarrow cX \sim \Gamma\left(\alpha, \frac{\beta}{c}\right).$$

After a ton of Googling it doesn't seem like there's a name for the distribution of a squared Gamma random variable. However you can find its distribution using standard transformation of a random variable.

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    $\begingroup$ Thank you for taking the time to write the answer wrt the constant factor. The most important part was indeed with respect to the square random variable. If I knew how to use the standard transformation of a random variable I would not have asked this question... $\endgroup$ – michael May 27 '15 at 13:54

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