gamma distribution to chi squared transform

Question

I thought If X~gamma($\alpha$, $\beta$) then $\frac{2X}{\alpha}$ ~ $\chi^2_n$ where n=2$\beta$. but when I solve exercise in Mathematical statistics with application (ex9.85e) If Y~Gamma$(\alpha$, $\theta$) then $\frac{2Y}{\alpha}$ should follow $\chi^2_{2\theta}$ ? but, my professor said $\frac{2Y}{\alpha}$ follow $\chi^2_{4n}$. why is that?! Am i missing something right?

Answer 1

The textbook mentioned was confirmed (via comments, deleted afterwards, back in Dec. 2016) to be by Dennis Wackerly et al

There's probably a mix up of the parameters when trying to recall the functional form by memory. The transformation of dividing by a parameter should be dividing the scale parameter $\beta$. Both statements as posted here by the professor and by you are "off" in that respect.

Parameters Clarification

There are two common parameterizations for Gamma distribution so just to be clear, here $X \sim \text{Gamma}(\alpha, \beta)$ means

$$f_X(u) = \frac1{\beta} \frac1{\Gamma(\alpha)}\left( \frac{u}{\beta} \right)^{\alpha - 1}e^{ -\frac{u}{\beta}}$$

written in such form to emphasize the role played by the scaling parameter $\beta$, where $\alpha$ is the shape parameter.

(the other common parametrization is the use of $\lambda = 1/\beta$ as the rate of a Poisson process)

One can view a Chi-square distribution as defined by taking $\alpha = \frac{n}2$ and $\beta = 2$, where $n \in \mathbb{N}$ is the degree of freedom. That is, Chi-sq is a special case of Gamma. This is what Dennis Wackerly's book does in Sec.4.6 on p.187 (or the corresponding place where Gamma distribution is introduced in other editions).

Chi-sq distribution can of course be defined with different motivations. Regardless of how it is constructed, the density for a $S \sim \chi^2_n$ takes the form

$$f_S(u) = \frac12 \frac1{\Gamma(\frac{n}2)}\left( \frac{u}2 \right)^{ \frac{n}2 - 1}e^{ -\frac{u}2}$$

Answer to the Stated Question

Now, back to examining our $X \sim \text{Gamma}(\alpha, \beta)$. Define a scaled version $W\equiv \frac{2X}{\beta}$ such that the inverse transform is $x = h(w) = \frac{\beta}2 w$

$$ \begin{align} f_W(u) &= \bigg| \frac{ \text{d} h(u) }{ \text{d} u} \bigg|\cdot f_X\left( h(u) \right) \\ &= \frac{\beta}2 \cdot \frac1{\beta} \frac1{\Gamma(\alpha)}\left( \frac{ \beta u}{2\beta} \right)^{\alpha - 1}e^{ -\frac{\beta u}{2\beta}} \\ &= \frac12 \frac1{\Gamma(\alpha)}\left( \frac{u}2 \right)^{\alpha - 1}e^{ -\frac{u}2} \end{align}$$

which gives the Chi-sq distribution when $\alpha$ is a half-integer (which includes integers), that is, a subset of such $W$ gives $W \sim \chi_n^2$ when $\alpha = \frac{n}2$.

Note that one can have a set of (collection of) Chi-sq with arbirary jumps in degree of freedom. In particular,

• when $\alpha$ is any even-integers, $\alpha = 2m$, this subset of $W = \frac{2X}{\beta}$ gives $W \sim \chi^2_m$ where $m$ is any natural number.

• when $\alpha$ is any multiple of four, $\alpha = 4m$ for integer $m$, the subset of $W = \frac{2X}{\beta}$ gives us $W \sim \chi^2_{2m}$ only the even degrees of freedom.

One can say that you were right and the professor was wrong (besides the parameters mix-up).

It can also be seen that any attempt to change the degree of freedom in the transformation will fail: the shape parameter $\alpha$ appears in the exponent in the density so, if the goal is to replace $\alpha$ by something else, first one must make $\alpha$ disappear via incorporating some transformation that involves power, like $W \equiv X^{\alpha-1}$. However, besides the Jacobian term bringing back the $\alpha$, that power will ruin the $e^{-u}$ term and will not give you a Chi-sq.