# Showing if $\frac{X}{c}\sim\text{Gamma}(a,b)$ then $X\sim\text{Gamma}(a,cb)$

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?

• $m_Z(u)=m_X(u/c)$. – StubbornAtom Jul 6 '18 at 5:55
• Typo fixed. Thanks for spotting that! My result still follows though – user557493 Jul 6 '18 at 5:56
• Which gamma pdf are you using? – StubbornAtom Jul 6 '18 at 5:59
• 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? – user557493 Jul 6 '18 at 6:03
• If $X\sim\text{Gamma}(a,b)$, then $$f(x)=\frac{1}{\Gamma(a)b^a}e^{-\frac{x}{b}}x^{a-1} \ \ \ x>0$$ – user557493 Jul 6 '18 at 6:36

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 $$X$$.

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 $Z$.

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