# Show for $\ c>0\$ that $\ cY\sim \ \text{Gamma}(\alpha,c\beta)$

Show for any constant $\ c>0\$ that $\ cY\sim \ \text{Gamma}(\alpha,c\beta)$ $$Y\sim\text{Gamma}(\alpha,\beta)$$ $$f_Y(y)=\frac{1}{\Gamma (\alpha)\beta^\alpha}e^{\frac{-y}{\beta}}y^{\alpha-1} \ \ \ \ \ y>0$$

Attempt: I had an intuition that computing the moment generating function of $Y$ may help. Hence I calculated that $$m_Y(u)=(1-\beta u)^{-\alpha} \ \ \ \ \ \ \text{for} \ \ u<\frac{1}{\beta}$$

Now I am unsure of how to proceed. Any advice would be very appreciated.

• Why not write $\Gamma(\alpha,c\beta)$? – gebruiker Jun 30 '18 at 9:23
• @gebruiker: It's more common to write $Gamma(\alpha ,c\beta )$ than $\Gamma(\alpha ,c\beta )$. The notation $\Gamma(a,b)$ is more used for the Incomplete gamma function. – Surb Jun 30 '18 at 9:25
• Did not know that. Thank you – gebruiker Jun 30 '18 at 9:25
• Why not compute moment generating function of $cY$? – AnyAD Jun 30 '18 at 9:28
• @AnyAD: Wouldn't this just be $c \ m_Y(u)$ (as $c$ is constant)? – user557493 Jun 30 '18 at 9:46

Hint

Method 1 : Using the fact that $$\mathbb P\{Y\leq y\}=\int_0^\infty \frac{x^{\alpha -1}e^{-\frac{x}{\beta }}}{\Gamma(k)\beta ^k}dx,$$ I think that $$\mathbb P\{cY\leq y\}$$ can be easily deduce.

Method 2 : If $g$ is a bijection, and if $Y=g(X)$, then $$f_Y(y)=\frac{f_X(g^{-1}(y))}{|g'(g^{-1}(y)|}$$

• I don't understand. Doesn't $$f_Y(y)=\mathbb P\{Y= y\}=\frac{y^{\alpha -1}e^{-\frac{y}{\beta }}}{\Gamma(\alpha)\beta ^\alpha}?$$ Where did $P\{Y\leq y\}$ come from? – user557493 Jun 30 '18 at 9:43
• @Bell The density evaluated ar $y$ does not equal probability at a point which is always zero for a continuous distribution. – StubbornAtom Jun 30 '18 at 11:34
• And if I find $\mathbb{P}(cY\leq y)$, how does this help? Sorry, but I'm still a bit unsure. – user557493 Jun 30 '18 at 22:38
• @Bell $P(Y\le y)$ is the distribution function of $Y$ which is to be differentiated to get the pdf $f_Y$ of $Y$. – StubbornAtom Jul 1 '18 at 9:39