Suppose I have three continuous random variables $X_1$, $X_2$, and $Y$, where $Y = X_1+X_2$, and $X_1$ and $X_2$ are dependent.
If I know the probability distributions separately for $X_1$, $X_2$, and $Y$, is there a general method or some methods to try for determining the joint distribution of $X_1$ and $X_2$ or equivalently the conditional distribution of $X_1$ given $X_2$? $X_1$ and $X_2$ are dependent, so I can't use the normal formula – since that needs $X_1+X_2$ and $X_1$ to be independent and I can't assume that. Also $X_1$, $X_2$, and $Y$ are not necessarily normal random variables.
I've been thinking about this for a while and I'm not sure it's possible in general without having even more information. I've been using guess and check, and that's not exactly optimal, but I'm not sure if a more general method is even possible.
I was hoping there might be some way to figure things out from combining these three equations:
Let $Z = X_2$, then we have $X_1 = Y-Z$ and $X_2 = Z$. Further we know
$$
f_Y (y) = \int_{-\infty}^\infty f_{X_1,X_2} (y-z,z) dz
$$
and
$$
f_{X_1} (x_1) = \int_{-\infty}^\infty f_{X_1,X_2} (x_1,x_2) dx_2
$$
and
$$
f_{X_2} (x_2) = \int_{-\infty}^\infty f_{X_1,X_2} (x_1,x_2) dx_1
$$
Since I know the left side of all of those, is there any way to figure out the joint distribution, keeping in mind there's dependence so $$f_{x1,x2}(x1,x2) \neq f_{X_1} (x_1) f_{X_2} (x_2)$$
Thanks for any insight or help, even if it's just to tell me it's impossible.
Best Answer
In the discrete case, a simple dimensional analysis shows that the answer is "No", in general.
Assume that $(X_1,X_2)$ takes values in $\{0,1,\ldots,n\}\times\{0,1,\ldots,m\}$ then $X_1+X_2$ takes values in $\{0,1,\ldots,n+m\}$. The distributions of $X_1$, $X_2$, $X_1+X_2$ and $(X_1,X_2)$ depend on $n$, $m$, $n+m$ and $nm+n+m$ parameters respectively. Hence the three first distributions cannot determine the last distribution when $nm+n+m\gt n+m+(n+m)$, that is, when $(n-1)(m-1)\geqslant2$, for example if $\{n,m\}=\{2,3\}$.
A fortiori the answer is also "No" when the distributions of $X_1$, $X_2$ and $X_1+X_2$ are absolutely continuous. An example is when $X_1$ and $X_2$ are Cauchy with parameter $a$ and $X_1+X_2$ is Cauchy with parameter $2a$. Then $(X_1,X_2)$ can be independent, or one can have $X_1=X_2$ with full probability.