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

Question

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.

Answer 1

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 $$f_Y(y)=\frac{f_X(g^{-1}(y))}{|g'(g^{-1}(y)|}$$