PDF for sum of dependent random variables When the variables $X, Y$ are independent, then the PDF of $Z = X + Y$ can be computed using convolutions:
$$
f_Z(z) = \int_{-\infty}^{\infty} f_X(x)f_Y(z - x) dx
$$
When the variables are dependent, apparently you can use
$$
f_Z(z) = \int_{-\infty}^{\infty} f_{XY}(x, z - x) dx
$$
I am wondering where the expression came from for the dependent case? It looks very similar to the independent case except you can't separate the joint distribution into marginals.
 A: From the cumulative distribution function: $$F_Z(z)= P(Z\le z ) = P(X+Y \le z) = P(Y \le z -X) = \int_{-\infty}^{\infty} 
\int_{-\infty}^{z-x} f_{XY}(x,y)\, dy\, dx  $$
Now, $$f_Z(z) = \frac{d F_Z(z) }{ d z} = 
\int_{-\infty}^{\infty} \frac{d }{ d z} \left(
\int_{-\infty}^{z-x} f_{XY}(x,y) dy\, \right) dx = 
\int_{-\infty}^{\infty} f_{XY}(x,z-x)  dx  $$
If $X$ and $Y$ are independent we can further write
$$f_Z(z) = \int_{-\infty}^{\infty} f_X(x) f_Y(z-x)  dx  $$
A: The characteristic function of $Z=X+Y$ is
$$E[e^{iuZ}]=E[e^{iu(X+Y)}]=\int_{\mathbb{R}^2} e^{iu(x+y)}p_{X,Y}(x,y)dxdy$$
The pdf is found by inverse Fourier Transform. By exchanging integrals
$$f_{X+Y}(z)=\frac{1}{2\pi}\int_{\mathbb{R}^2}\bigg(\int_{\mathbb{R}} e^{iu(x+y)}e^{-iuz}du\bigg)p_{X,Y}(x,y)dxdy=$$
$$=\int_\mathbb{R^2}\delta(z-(x+y))p_{X,Y}(x,y)dxdy$$
By symmetry of the Dirac delta $\delta(w)=\delta(-w)$
$$\int_\mathbb{R^2}\delta((x+y)-z)p_{X,Y}(x,y)dxdy=$$
$$(x+y)-z=0 \implies y=z-x $$
$$=\int_\mathbb{R}p_{X,Y}(x,z-x)dx$$

$$P(Z\leq z)=P(X+Y\leq z)=\int P(X+y\leq z|Y=y)p_Y(y)dy=$$
$$=\int_{\infty}^\infty\int^{z-y}_{-\infty}p_{X|Y}(x,y)p_y(y)dxdy=\int_{\infty}^\infty\int^{z-y}_{-\infty}p_{X,Y}(x,y)dxdy$$
$$f_Z(z)=\frac{d}{dz}P(Z\leq z)=\int_{-\infty}^\infty p_{X,Y}(z-y,y)dy$$
By using Leibniz integral rule.
A: There is a property for linear combinations that says if X, Y have joint pdf $f(x, y)$ and $Z=aX+bY+c$, then Z has pdf $$g(z)=\int_{-\infty}^\infty f\left(x, \frac{z-c-ax}b\right)\frac1{|b|}dx$$
For $Z=X+Y$, $$\begin{split}G(z)&=\int_{-\infty}^\infty\int_{-\infty}^{z-x}f(x,y)dydx\end{split}$$
Change of variables $w=x+y$
$$\begin{split}G(z)&=\int_{-\infty}^\infty\int_{-\infty}^{z}f(x,w-x)dwdx\\
&=\int_{-\infty}^z\int_{-\infty}^{\infty}f(x,w-x)dxdw\end{split}$$
Derivative with respect to $z$
$$\begin{split}g(z)&=\left[\frac{d}{dz}(z)\right]\int_{-\infty}^\infty f(x, z-x)dx-0\\
&=1\cdot\int_{-\infty}^\infty f(x, z-x)dx=\int_{-\infty}^\infty f(x, z-x)dx\end{split}$$
A: A pdf for a random vector $X:\Omega\to\Bbb R^n$ is any function $g:\Bbb R^n\to \Bbb R$ such that $E[h(X)]=\int_{\Bbb R^n}h(x)g(x)\,dx$ for, say, all bounded summable Borel functions $h:\Bbb R^n\to\Bbb R$.
So, if your random vector is $V=(X,Y)$, then, without too many ceremonies, you can use the change of variables $(t,s)= (x,y+x)$ and Fubini-Tonelli in the integral $$E[h(X+Y)]=\int_{\Bbb R^2} h(x+y)f_{X,Y}(x,y)\,dxdy=\int_{\Bbb R^2}h(s)f_{X,Y}(t,s-t)\,dsdt=\\=\int_{-\infty}^\infty f(s)\left(\int_{-\infty}^\infty f_{X,Y}(t,s-t)\,dt\right)\,ds$$
And there you have it.
