Sum of two independent exponential distributions

Let $Y_1\sim \exp(\lambda_1)$ and $Y_2\sim \exp(\lambda_2)$ be two independent r.v.'s.

Show that the pdf $p_V(x)$ for their sum $V=Y_1+Y_2$ has the following form $$p_V(x)=\frac{e^\frac{-v}{\lambda_1}-e^\frac{-v}{\lambda_2}}{\lambda_1-\lambda_2};\quad v\ge0$$

My attempt:

distribution of $Y_i$, $$p_{Y_i}(y_i)=\frac{1}{\lambda_i}e^\frac{-{y_i}}{\lambda_i};\quad x\ge0;\quad i=1,2$$

joint distribution of $Y_1$ and $Y_2$, $$p_{Y_1,Y_2}(y_1,y_2)=\frac{1}{\lambda_1\lambda_2}e^{\frac{-{y_1}}{\lambda_1}-\frac{{y_2}}{\lambda_2}};\quad {y_1},{y_2}\ge0$$

Given, $V=Y_1+Y_2$

let $U=Y_2$

So the Jacobian of Transformation is $1$

and hence joint distribution of $V$ and $U$, $$p_{V,U}(v,u)=\frac{1}{\lambda_1\lambda_2}e^{\frac{-{v-u}}{\lambda_1}-\frac{{u}}{\lambda_2}};\quad {v},{u}\ge0$$

so the distribution of $v$ $$p_{V}(v)=\int_0^\infty p_{V,U}(v,u)du=\frac{e^\frac{-v}{\lambda_1}}{\lambda_1-\lambda_2};\quad v\ge0$$

which doesn't match with the result.

Solution 1:

We could calculate the cumulative distribution function of $V$, and then differentiate. It is quicker to use the convolution $$\int_{-\infty}^\infty f_1(z-t)f_2(t)\,dt,$$ where $f_1$ and $f_2$ are the densities of our random variables.

The density function $f_V(v)$ of $V$ is $0$ for $v\lt 0$. So we look only at the case $v\ge 0$. In our case, since the exponentials have density $0$ to the left of $0$, the actual expression is $$f_V(v)=\int_{t=0}^v \lambda_1 \lambda_2 e^{-\lambda_1(v-t)}e^{-\lambda_2 t}\,dt.$$ We are basically integrating $e^{-(\lambda_2-\lambda_1)t}$, which is not difficult.