Gamma distribution and pdf Let $X \sim \mathsf{Gamma}(2,3)$ and $Y = 2X.$
Find the pdf of $Y$ at $y=13.5.$
Attempt: $f_X(x)= 2*[1/9*\Gamma(2)]*x*e^{-x/3}.$
Do I have to integrate now?

$X\sim \text{Gamma}(2,3)\implies f_X(x)=\frac{1}{9\Gamma(2)}xe^{-x/3}$ for $x>0$. The method of CDF yields (for $y>0$):
$$\begin{align}F_Y(y)&=P(Y\leq y)\\&=P\left(X\leq\frac{y}{2}\right)\\&=\int_0^{y/2}\frac{1}{9}xe^{-x/3}dx\\&=\frac{-1}{3}\left[(x+3)e^{-x/3}\right]^{y/2}_0\\&=1-\frac{y+6}{6}e^{-y/6}\end{align}$$
Using the fact that $f_Y(y)=\frac{d}{dy}F_Y(y)$, we see that:
$$f_Y(y)=\begin{cases}\frac{1}{36}ye^{-y/6}\qquad x>0\\0\qquad\qquad\text{otherwise}\end{cases}$$
Thus, $Y\sim\text{Gamma}(2,6)$. Hope this helps.
