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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!

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  • $\begingroup$ Convolution is one way to do this. $\endgroup$ – Sean Roberson Dec 24 '18 at 17:03
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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.

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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.

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  • $\begingroup$ Then how to do $f_{Z|X}(z|x)$? $\endgroup$ – Dave Dec 25 '18 at 2:12
  • $\begingroup$ For discrete case, does pdf exist? $\endgroup$ – Danny Pak-Keung Chan Dec 25 '18 at 18:43
  • $\begingroup$ Sorry @Dave, I see what your question is really asking. So in general, if we know X, Y dependent variables, there are infinitely many options for $f_{XY}(x,y)$. So we cannot simply calculate it. See this question. $\endgroup$ – ptolemy0 Dec 25 '18 at 19:42
  • $\begingroup$ And only if we know the joint distribution explicity does the above work. If we did want to find it, such as in your case, wikipedia gives an answer, but this is honestly beyond my knowledge, so I can't help any further. Sorry for an annoying answer, though I've added this to my answer. $\endgroup$ – ptolemy0 Dec 25 '18 at 19:43
  • $\begingroup$ Actually, I reconsidered it, and amended my answer, which I think should clear things up! Hope that is better. $\endgroup$ – ptolemy0 Dec 25 '18 at 21:40

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