Suppose that $Y$ follows a gamma distribution with parameters $a$ and $b$. That is,
$f(y)$= $y^{a-1}e^{-y/b}/\gamma(a)b^a$; $y>0$
For $a>1$, show that the mode of $Y$ is $(a-1)b$.
My working:
I know that the mode is the value of $y$ where its density is maximized; hence for given $f(y)$ I have:
$f^{'}(y) = 0$ and I solve for values of $y$. The trouble is when I put the value of $\gamma(a)$ in the given $f(y)$ and take its derivative by applying quotient rule the numerator becomes complicated for me to handle it since there is improper integral involved. Can any one help me. I will appreciate it
\begin{align*} f(y)&= \frac{1}{\Gamma(a)b^a}y^{a -1}e^{-\frac yb}\\ f'(y)&= \frac{1}{\Gamma(a)b^a}\frac d{dy}y^{a-1}e^{-\frac yb}=\frac{1}{\Gamma(a)b^a}\left[(a -1)y^{a-2}e^{-\frac yb}-\frac 1by^{a-1}e^{-\frac yb}\right]\\ &=\frac{e^{-\frac yb}y^{a-2}}{\Gamma(a)b^a}\left((a-1)-\frac1by\right)=0\\ (a-1)-\frac 1by&= 0\longrightarrow -\frac1by=-(a-1)\longrightarrow y=(a-1)b \end{align*} Note that $\Gamma(a)b^a$ is just a normalizing constant, and does not depend on $y$. The definition of the gamma function does not have to be used.