I want to know if my CDF calculation is correct.
Let $X,Z$ be independent random variables with exponential distribution and parameter $\gamma_x,\gamma_z$. Compute $P(Z+X<t)$
\begin{align*}P(Z+X<t)&=P(Z<t-x,X=x)\\ &=\int\limits_{0}^\infty \int\limits_{0}^{t-x} \gamma_xe^{-\gamma_x t}\gamma_ze^{-\gamma_zm}\mathrm{d}m\mathrm{d}x=\int\limits_{0}^\infty \gamma_z\gamma_xe^{-\gamma_x t}\int\limits_{0}^{t-x} e^{-\gamma_zm}\mathrm{d}m\mathrm{d}x\\ &=\int\limits_{0}^\infty \gamma_z\gamma_xe^{-\gamma_x t} \left[\frac{-e^{-\gamma_zm}}{\gamma_z}\right]_{0}^{t-x}\mathrm{d}x\\ &=\int\limits_{0}^\infty \gamma_xe^{-\gamma_x t} (1-e^{-\gamma_z(t-x)})\mathrm{d}x\\ &=\gamma_x\int\limits_{0}^\infty e^{-\gamma_x t}-e^{-\gamma_zx-\gamma_zt-\gamma_x t}\mathrm{d}x=\gamma_x\left[\frac{-e^{-\gamma_x t}}{\gamma_x}\right]_0^\infty+\gamma_xe^{-\gamma_zt-\gamma_x t}\left[\frac{e^{-\gamma_zx}}{\gamma_x}\right]_0^\infty \mathrm{d}x\\ &=1-e^{-t(\gamma_z+\gamma_x )}\qquad t\geq 0\\ \end{align*}
I don't know how to check such things. What is a good way?
The pdf for $x$ and $z$ are:
$\frac{1}{\lambda_x}e^{\frac{-x}{\lambda_x}}$ and $\frac{1}{\lambda_z}e^{\frac{-z}{\lambda_z}}$
Let $Y=X+Z$
In order for $y=x+z$ we need $X=x$ and $Z=y-x$
The probability that $X=x$ and $Z=y-x$ is the product of the pdfs at these two values, i.e. $P(Y=X+Z) = \frac{1}{\lambda_x}e^{\frac{-x}{\lambda_x}} \cdot \frac{1}{\lambda_z}e^{\frac{-(y-x)}{\lambda_z}}$
Now integrate over all possible values of $x$. Note that $x$ cannot be greater than $y$ since the sum of $x$ and $z$ is $y$
$P(Y=y) = \int_{0}^{y} \frac{1}{\lambda_x}e^{\frac{-x}{\lambda_x}} \cdot \frac{1}{\lambda_z}e^{\frac{-(y-x)}{\lambda_z}}dx$
$P(Y=y) = \frac{e^{\frac{-y}{\lambda_x}}-e^{\frac{-y}{\lambda_z}}}{\lambda_x-\lambda_z}$
My notation is a bit sloppy. $P(Y=y)$ is the pdf of $Y$. Now integrate the pdf to get the cdf
$\text{CDF}(y)=\int_0^t \frac{e^{\frac{-y}{\lambda_x}}-e^{\frac{-y}{\lambda_z}}}{\lambda_x-\lambda_z}dy = \frac{\frac{1}{\lambda_x}(1-e^{\frac{t}{\lambda_x}})-\frac{1}{\lambda_z}(1-e^{\frac{t}{\lambda_z}})}{\frac{1}{\lambda_x}-\frac{1}{\lambda_z}}$