I have a random variable $X$ and I define an additional random variable $Z=X+Y$. Now $X$ and $Z$ are dependent.
I know the distributions of $X$, $Y$ and I know that $X$ and $Y$ are independent. In particular, for my case $X\sim Gamma(N,\lambda)$ and $Y\sim Exp(\lambda)$.
How do I find the Joint CDF $F_{X,Z}$ ?
First, we observe that $F_{X,X+Y}(x',z')=\mathbb{P}(\{X\leq x'\}\cap\{Z\leq z'\})$, what is equal to: $$ \int_{-\infty}^{x'}\int_{-\infty}^{z'}f_{X,Z}(x,z)dzdx\quad, $$ where an integrand is a joint density of a random vector $[X,Z]^{T}$. So we need to find this density. To do this, we can exploit a formula for a joint density of transformed random vector: $$ f_{X,Z}(x_0,z_0)=f_{X,Y}(h^{-1}(x_0,z_0))\times |det(\frac{dh^{-1}}{d[x,z]}(x_0,z_0))|\quad, $$ where $det(\frac{dh^{-1}}{d[x,z]}(x_0,z_0))$ is the Jacobian of an inverse transformation of our random vector, evaluated at point $[x_0,z_0]$. In our case mappings look like: $$ \begin{bmatrix}x\\y\end{bmatrix}\mapsto\begin{bmatrix}h_1(x,y)\\h_2(x,y)\end{bmatrix}= \begin{bmatrix}x\\z\end{bmatrix}=\begin{bmatrix}x\\x+y\end{bmatrix}\\\begin{bmatrix}x\\z\end{bmatrix}\mapsto\begin{bmatrix}h_1^{-1}(x,z)\\h_2^{-1}(x,z)\end{bmatrix}=\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}x\\z-x\end{bmatrix}\quad. $$ The Jacobian of $h^{-1}(\cdot)$ is: $$ det(\frac{dh^{-1}}{d[x,z]}(x_0,z_0))=det(\begin{bmatrix}\frac{dh_1^{-1}}{dx}&\frac{dh_1^{-1}}{dz}\\\frac{dh_2^{-1}}{dx}&\frac{dh_2^{-1}}{dz}\end{bmatrix}\mid_{(x_0,z_0)})=det(\begin{bmatrix}1&0\\-1&1\end{bmatrix})=1\quad. $$ So we can write: $$ f_{X,Z}(x,z)=f_{X,Y}(x,z-x)\mathbb{1}_{[0,\infty)}(x)\mathbb{1}_{\{z-x\geq 0\}}(x,z)=f_X(x)f_Y(z-x)\mathbb{1}_{\{0\leq x\leq z\}}(x,z)\quad=\frac{\lambda^{N}x^{N-1}}{\Gamma(N)}e^{-\lambda x}\lambda e^{-\lambda (z-x)}\mathbb{1}_{\{0\leq x\leq z\}}(x,z)=\frac{\lambda^{N}x^{N-1}}{\Gamma(N)}\lambda e^{-\lambda z}\mathbb{1}_{\{0\leq x\leq z\}}(x,z)\quad. $$ Now we can back to distribution function: $$ F_{X,Z}(x',z')=\int_{-\infty}^{x'}\int_{-\infty}^{z'}\frac{\lambda^{N}x^{N-1}}{\Gamma(N)}\lambda e^{-\lambda z}\mathbb{1}_{\{0\leq x\leq z\}}(x,z)dzdx=\int_{0}^{x'}\frac{\lambda^{N}x^{N-1}}{\Gamma(N)}\int_{x}^{z'}\lambda e^{-\lambda z}dxdz=\frac{\lambda^{N}}{\Gamma(N)}\int_{0}^{x'}x^{N-1}(e^{-\lambda x}-e^{-\lambda z'})dx=\frac{\lambda^{N}}{\Gamma(N)}\int_{0}^{x'}x^{N-1}e^{-\lambda x}dx-e^{-\lambda z'}\frac{(\lambda x')^N }{\Gamma(N+1)}=F_X(x')-e^{-\lambda z'}\frac{(\lambda x')^N }{\Gamma(N+1)} $$
