Finding the probability density function of $X+Y$ Question: Let $X$ and $Y$ be two independent and identically distributed exponential random variables with parameter $\lambda>0$. Compute the probability density function of $X+Y$. 
My Answer: I have found the joint probability density function of $X$ and $Y$ to be $f_{X,Y}(x,y)=\lambda^2e^{-\lambda x - \lambda y}$. I then let $Z=X+Y$ and calculated $F_Z(z)=\lambda^2e^{-\lambda z}$.
I know I need to integrate $F_Z(z)$ to calculate the density function but am unsure on what the limits should be or if I found $F_Z(z)$ correctly. I was also wondering if there is a quicker way to tackle such questions or if this method is okay. 
 A: You can use convolution to solve the problem.
Let $z \geq 0$, 
\begin{align}f_Z(z) &= \int_0^z f_X(x)f_Y(z-x) dx  \\
&= \int_0^z \lambda \exp(-\lambda x) \lambda \exp(-\lambda (z-x)) dx  \\
&= \int_0^z \lambda^2 \exp(-\lambda z) dx  \\
&=  \lambda^2 z\exp(-\lambda z) 
\end{align}
As for the problem of how to find the limit of integration, we want $0 \leq x < \infty$ and $0  \leq z-x<\infty $. Solving for $x$ in the inequalities give you $0 \leq x \leq z$. 
Remark $1$: Sum of independent identical exponential distributions is known as Erlang Distribution, which is a special case of gamma distribution.
Remark $2$: To find pdf from CDF, we differentiate rather than integrate.
Remark $3$: Now, suppose we  insist to find CDF from joint distribution and then find the pdf by differentiating it.
Let $z > 0$, 
\begin{align}
F_Z(z) &= Pr(Z \leq z)\\
&= \int_0^z \int_0^{z-x} \lambda^2 \exp(-\lambda x) \exp( - \lambda y)dy dx \\
&=  \int_0^z  \lambda \exp(-\lambda x) \left( 1-\exp(-\lambda(z-x))\right) dx \\
&=  \int_0^z  \lambda \left( \exp(-\lambda x)-\exp(-\lambda z)\right) dx \\
&= 1- \exp(-\lambda z) - \lambda z \exp(-\lambda z) \\
&= 1 - (1+\lambda z) \exp(-\lambda z)
\end{align}
Differentiating:
\begin{align}
f_Z(z) &=-\lambda \exp(-\lambda z) -(1+\lambda z)(-\lambda)\exp(-\lambda z)\\
&= \lambda^2 z \exp(-\lambda z)
\end{align}
