I have an idea of Erlang PDF by intuition but I want to get the answer by analytical derivation of its CDF i.e.
\begin{equation} F_{Y_k}(y)=1-\sum_{n=0}^{k-1}\dfrac{(\lambda y)^n e^{-\lambda y}}{n!} \end{equation}
The PDF of $Y_k$ can be obtained by differentiating the above expression i.e.
\begin{equation} f_{Y_k}(y)=\dfrac{d}{dy}F_{Y_k}(y)= \dfrac{\lambda^k y^{k-1}e^{-\lambda y}}{(k-1)!} \end{equation}
Kindly if someone has done it in analytical manner (without any intuition), just with differentiating steps, please share. Thanks in advance.
I think nobody has tried it, so I was able to do it. I am answering in aspiration of helping someone for future.
\begin{align*} f_{Y_k}(y)=\dfrac{dF_{Y_k}(y)}{dy}&=-\sum\limits_{n=0}^{k-1}\dfrac{1}{n!}\dfrac{d[(\lambda y)^ne^{-\lambda y}]}{dy}\\ &=-\sum\limits_{n=0}^{k-1}\bigg[\dfrac{n\lambda^ny^{n-1}e^{-\lambda y}}{n!}-\dfrac{\lambda^{n+1}y^ne^{-\lambda y}}{n!}\bigg]\\ &=-e^{-\lambda y}\bigg[\sum\limits_{n=0}^{k-1}\dfrac{n\lambda^ny^{n-1}}{n!}-\sum\limits_{n=0}^{k-1}\dfrac{\lambda^{n+1}y^n}{n!}\bigg]\\ &=-e^{-\lambda y}\bigg[\sum\limits_{n=1}^{k-1}\dfrac{\lambda^ny^{n-1}}{(n-1)!}-\sum\limits_{n=0}^{k-1}\dfrac{\lambda^{n+1}y^n}{n!}\bigg]\\ &=e^{-\lambda y}\bigg[\sum\limits_{n=0}^{k-1}\dfrac{\lambda^{n+1}y^n}{n!}-\sum\limits_{n=1}^{k-1}\dfrac{\lambda^ny^{n-1}}{(n-1)!}\bigg]\\ &=e^{-\lambda y}\bigg[\dfrac{\lambda^{k}y^{k-1}}{(k-1)!}+\sum\limits_{n=0}^{k-2}\dfrac{\lambda^{n+1}y^n}{n!}-\sum\limits_{n=0}^{k-2}\dfrac{\lambda^{n+1}y^n}{n!}\bigg]\\ &=\dfrac{\lambda^{k}y^{k-1}e^{-\lambda y}}{(k-1)!} \end{align*}