Finding the pdf of a transformation of these independent random variables

Question

Hi, I am looking at the question above. We know that $f_X(x) = \lambda^nx^{n-1}e^{-\lambda x}/\Gamma(n)$ and $f_Y(y) = \lambda e^{-\lambda y}$.

We also know that the range for each of these pdfs are $x>0$ and $y>0$.

Now, after computing we get $f_{UV}(u,v) = \lambda^n(u-v)^{n-1}e^{-\lambda(u-v)}/\Gamma(n)\hspace{3mm} \cdot \hspace{3mm} \lambda e^{-\lambda v} $

Now I have to find the pdf of U. I know I have to integrate that equation between the limits of $v=u$ and $v=0$ to find said pdf of U. But I am stuck on the integral. Any help? Thanks!

Answer 1

\begin{align} \int_0^u f_{UV}(u, v) \mathop{dv} &= \lambda^{n+1} e^{-\lambda u} \frac{1}{\Gamma(n)} \int_0^u (u-v)^{n-1} \mathop{dv} \\ &= \lambda^{n+1} e^{-\lambda u} \frac{1}{\Gamma(n)} \left[- \frac{1}{n}(u-v)^n\right]_{v=0}^u \\ &= \lambda^{n+1} e^{-\lambda u} \frac{1}{\Gamma(n)}\cdot \frac{1}{n} u^n. \end{align} Can you take it from here? Note $\Gamma(n) \cdot n = \Gamma(n+1)$.