pdf of sum of two dependent random variables What is the pdf of sum of two dependent random variables given we know their joint
pdf and individual pdfs. I have seen already some posts but none of them answered when they are dependent. Every one solved for only the independent case but i need for dependent case in terms of the joint pdf and individual pdfs in an explicit form. 
 A: If $f(x,y)$ denotes the PDF of $(X,Y)$ and $Z=X+Y$ then:
$$f_Z(z)=\int^{\infty}_{-\infty}f(x,z-x)dx$$
Substituting $y=u-x$ we find:$$F_Z(z)=\int\int_{x+y\leq z}f(x,y)dxdy=\int^{\infty}_{-\infty}\left[\int^{z-x}_{-\infty}f(x,y)dy\right]dx=$$$$\int^{\infty}_{-\infty}\left[\int^{z}_{-\infty}f(x,u-x)du\right]dx=\int^{z}_{-\infty}\left[\int^{\infty}_{-\infty}f(x,u-x)dx\right]du$$
Taking the derivative of $F_Z$ we come to the mentioned result.
A: You can use Jacobian determinant and temporary variables. Consider :
$$Z=aX+bY$$
$$W = cX+dY$$
Therefore, we get:
$$X=AZ+BW\ \ \ \ \ , \ \ \ \ Y=CZ+DW$$ 
And for a given pair of  $X=x$ and $Y=y$, we have one solution as:
$$x=Az+Bw\ \ \ \ \ , \ \ \ \ y=Cz+Dw$$ 
Thus Jacobian determinat would be:
$$J(x,y)=ad-bc$$
which translates to:
$$f_{W,Z}(w,z)=\frac{1}{|ad-bc|} f_{x,y}(Az+Bw, Cz+Dw)$$
Here after, you can calculate the marginal PDF of W or Z by a simple integration. In other words:
$$f_{W}(w)=\int_{0}^{\infty}f_{W,Z}(w,z)dz$$
$$f_{Z}(z)=\int_{0}^{\infty}f_{W,Z}(w,z)dw$$

Note:$a,b,c,d$ are defined in the way that:
 $$ad-bc\neq 0$$
A: Let $X,Y$ be dependant variables with PDF $f(x,y)$.
Then sum PDF is: $$f_{X+Y}(k)= \int f(x,k-x) dx$$
