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I am trying to show that if $$\frac{X}{c}\sim\text{Gamma}(a,b) \ \ \ \ \ \ \text{then} \ \ \ \ \ \ \ X\sim\text{Gamma}(a,cb)$$

My first approach was to use a PDF transformation. I let $$Z=\frac{X}{c}\Rightarrow X=Zc$$ Then $$f_X(x)=f_Z\Big(\frac{x}{c}\Big)\Big|\frac{dz}{dx}\Big|=\frac{1}{\Gamma(a)b^a}e^{-\frac{x}{cb}}\Big(\frac{x}{c}\Big)^{a-1}\frac{1}{c}=\frac{1}{\Gamma(a)(cb)^a}e^{-\frac{x}{cb}}x^{a-1}$$ Which is clearly the density function of $\text{Gamma}(a,cb)$. Hence $X\sim\text{Gamma}(a,cb)$.

But, I know tried to confirm this using an MGF approach.

$$m_z(u)=\mathbb{E}\Big(e^{Zu}\Big)=\mathbb{E}\Big(e^{\frac{X}{c}u}\Big)=m_X\Big(\frac{u}{c}\Big)=\Big(1-b\frac{u}{c}\Big)^{-a}$$ which is the MGF for the $\text{Gamma}\Big(a,\frac{b}{c}\Big)$ distribution.

Where have I made my mistake in the MGF approach?

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    $\begingroup$ $m_Z(u)=m_X(u/c)$. $\endgroup$ – StubbornAtom Jul 6 '18 at 5:55
  • $\begingroup$ Typo fixed. Thanks for spotting that! My result still follows though $\endgroup$ – user557493 Jul 6 '18 at 5:56
  • $\begingroup$ Which gamma pdf are you using? $\endgroup$ – StubbornAtom Jul 6 '18 at 5:59
  • $\begingroup$ Oh I see, I assumed that $X\sim\text{Gamma}(a,b)$ which is incorrect. What's the best way to show $X\sim\text{Gamma}(a,cb)$ using MGFs? $\endgroup$ – user557493 Jul 6 '18 at 6:03
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    $\begingroup$ If $X\sim\text{Gamma}(a,b)$, then $$f(x)=\frac{1}{\Gamma(a)b^a}e^{-\frac{x}{b}}x^{a-1} \ \ \ x>0$$ $\endgroup$ – user557493 Jul 6 '18 at 6:36
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Let $Z = \dfrac{X}{c} \sim \text{Gamma}(a, b)$.

Then $$M_{Z}(u) = (1-bu)^{-a} = \mathbb{E}[e^{uZ}]\tag{*}$$ and thus, observing that $X = cZ$,

$$M_{X}(u)=\mathbb{E}[e^{uX}] = \mathbb{E}[e^{ucZ}]=\mathbb{E}[e^{(uc)Z}]=M_{Z}(uc) = [1-b(uc)]^{-a}=[1-(bc)u]^{-a}$$ hence $X \sim \text{Gamma}(a, bc)$.


Where did you go wrong in your work? First of all,

$$m_z(u) \neq \Big(1-b\frac{u}{c}\Big)^{-a}$$ We can see that since $Z \sim \text{Gamma}(a, b)$ that its MGF should be given by (*) above.

Second of all, $$m_X\Big(\frac{u}{c}\Big) \neq \Big(1-b\frac{u}{c}\Big)^{-a}$$ What you did here - assuming you were doing the work from left to right - was you assumed that $X \sim \text{Gamma}(a, b)$ to begin with (how else would you know what $m_{X}(u)$ is equal to that?), but we actually don't have that assumption available to us. That is, you shouldn't make any assumptions about what $M_{X}$ is. You know what $M_{Z}$ is because you have an assumption of what the distribution of $Z$ is available to you, but this is not the case for $X$.

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As usual, the calculations depend on whether the distribution is parametrized by shape and rate; i.e., $$f_X(x) = \frac{b^a x^{a-1} e^{-bx}}{\Gamma(a)}, \quad M_X(t) = (1 - t/b)^{-a},$$ or shape and scale: $$f_X(x) = \frac{x^{a-1} e^{-x/\theta}}{\theta^a \Gamma(a)}, \quad M_X(t) = (1 - \theta t)^{-a}.$$ If you use the shape-rate parametrization, then $Y = X/c$ has density $$f_Y(y) = c f_X(cy),$$ which implies $Y$ has rate $cb$. This is corroborated by the MGF approach: $$M_Y(t) = \operatorname{E}[e^{tY}] = \operatorname{E}[e^{(t/c)X}] = M_X(t/c) = (1 - t/(cb))^{-a}.$$ If you use the shape-scale parametrization, you get $$f_Y(y) = cf_X(cy)$$ as before, but this means $Y$ has scale $\theta/c$. The MGF is $$M_Y(t) = M_X(t/c) = (1 - (\theta/c) t)^{-a},$$ which is again consistent.


It is worth noting that I have done the reverse transformation; that is to say, you assumed $X/c \sim \operatorname{Gamma}(a,b)$ and I have taken $X \sim \operatorname{Gamma}(a,b)$. My $Y$ is your $X$, and my $X$ is your $Z $.

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  • $\begingroup$ So, are you saying that if $$\frac{Y^2}{\sigma^2}\sim\text{Gamma}(a,b)\Rightarrow Y^2\sim\text{Gamma}\Big(a,b\sigma^2\Big)??$$ $\endgroup$ – user557493 Jul 6 '18 at 9:01

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