How can I detemine the joint p.d.f. of $(X,Y)$, i.e., $f_{X,Y}(x,y)$? Consider $Z=X+Y$, where $X,Y$ and $Z$ are random variables with p.d.f.s denoting $f_X(x)$,  $f_Y(y)$ and $f_Z(z)$, respectively. Then, how can I detemine the joint p.d.f. of $(X,Y)$, i.e., $f_{X,Y}(x,y)$?
In addition, is there possible to calculate $f_{X,Z}(x,z)$ and $f_{Y,Z}(y,z)$?
Appreciate!
 A: In general, even if random variables $X$ and $Y$ have pdf $f_{X}$
and $f_{Y}$, it may happen that the random vector $(X,Y)$ does not
have pdf $f_{XY}$.
Let us clarify some terminoloies: Let $(\Omega,\mathcal{F},P)$ be
a probability space. Given a random variable $X$, its distribution
$\mu_{X}$ is a Borel measure $\mu_{X}:\mathcal{B}(\mathbb{R})\rightarrow[0,1]$
defined by $\mu_{X}(B)=P\left(X^{-1}(B)\right),$ $B\in\mathcal{B}(\mathbb{R})$.
If there exists a Borel function $f_{X}:\mathbb{R}\rightarrow\mathbb{R}$
such that $\int_{B}f_{X}(x)dx=\mu_{X}(B)$ for any $B\in\mathcal{B}(\mathbb{R})$,
we say that $X$ has a pdf. Since $\mu_{X}\geq0$, we have that $f_{X}\geq0$
($m$-a.e., where $m$ is the Lebesgue measure on $\mathbb{R}$).
Moreover, $f_{X}$ is not unique but is only unique $m$-a.e. Moreover,
$X$ has pdf if and only if $\mu_{X}$ is absolutely continuous with
respect to the Lebesgue measure $m$ (in the sense: $m(B)=0\Rightarrow\mu_{X}(B)=0$).
This setting can be extened to multi-dimensional case. For example,
the (joint) distribution $\mu_{XY}$ of the random vector $(X,Y)$
is a Borel measure $\mu_{XY}:\mathcal{B}(\mathbb{R}^{2})\rightarrow[0,1]$
such that $\mu_{XY}(B)=P\left((X,Y)^{-1}(B)\right)$. Here $(X,Y)$
is regarded as a map: $(X,Y):\Omega\rightarrow\mathbb{R}^{2}$, $\omega\mapsto(X(\omega),Y(\omega))$.
Similarly, if there exists a Borel function $f_{XY}:\mathbb{R}^{2}\rightarrow\mathbb{R}$
such that $\mu_{XY}(B)=\int_{B}f(x,y)\,dm_{2}(x,y)$, where $m_{2}$
is the Legesbue measure on $\mathbb{R}^{2}$, then we say that $(X,Y)$
has a (joint) pdf. Again, $(X,Y)$ has a pdf if and only if $\mu_{XY}$
is absolutely continuous with respect to $m_{2}$. In this case, the
pdf $f_{XY}$ is unique up to $m_{2}$-a.e. and $f_{XY}\geq0$ $m_{2}$-a.e.
Counter-example that $X,$ $Y$ both have pdf but $(X,Y)$ does not
have pdf: Choose a probability space $(\Omega,\mathcal{F},P)$ such
that there exists a random variable $X:\Omega\rightarrow\mathbb{R}$
with $X\sim N(0,1)$. Define $Y=X$. Clearly, $X$, $Y$ both have
pdf, denoted by $f_{X}$ and $f_{Y}$ (in fact, $f_{X}=f_{Y}$). We
prove that $(X,Y)$ does not have a pdf. Let $L=\{(t,t)\mid t\in\mathbb{R}\}$.
Note that $L$ is a Borel set and $(X,Y)^{-1}(L)=\Omega$, so $\mu_{XY}(L)=P(\Omega)=1$.
On the other hand, $m_{2}(L)=0$. Hence $\mu_{XY}$ is not absolutely
continuous with respect to $m_{2}$ and hence $(X,Y)$ does not have
a pdf. 
A: Firstly, to find $f_{XY}(x,y)$. Your question phrases it like we have a particular senario, when $X, Y$ are independent. If this is the case, then:
$$f_{XY}(x,y) = f_X(x)f_Y(y)$$
So to find the joint distribution we simply multiply the marginal distributions.
Secondly, you ask how to find $f_{XZ}(x,z)$. In this case, we have X, Z, which are not independent (since $Z = X + Y$). Then we find it like so:
\begin{align} 
f_{XZ}(x,z) &= Pr(X=x \ \ and \ \ Z=z)     \\
            &= Pr(X=x \ \ and \ \ X+Y=z)   \\
            &= Pr(X=x \ \ and \ \ Y=z-x)   \\
            &= f_{XY}(x,z-x)               \\
            &= f_X(x)f_Y(z-x)              \\
\end{align}
With the last step following from independence. Of course this is not very general, and only works in this case (since $Z=X+Y$). So when our relationship is different, as long we know the conditional distribution, then we can use Bayes Theorem, extended to PDFs.
$$f_{XZ}(x,z) = f_{Z|X}(z|x)f_X(x)$$
Of course, we must know this conditional distribution.
Although even more generally, if we were to only know two dependent marginal distributions, and allow any general relationship. There will often be infinitely many joint distributions, so it will become a lot complex. See wikipedia.
