How to show mathematically whether the following conditional relationships hold? In the following Bayesian network, the variables $ x_{i} $ are mutually independent (let's assume that these are the positions of $N$ boats). The variables $ y_{i,j} $ are distance measurements between two parent nodes $ x_{i}, x_{j} $ (with that said, some measurement $ y_{i,j} $ depends only on these two parents). I do have a model that given a measurement $ y_{i,j} $ and a parent $ x_{j} $, it provides me with the pdf of the parent's $ i $ position.

The dots above show that there can be $ N $ boats in the system, and that for any $ (x_{i}, x_{j}) $ pair of parents we can have up to two measurements of their distance (one from $i$ to $j$ and one from $j$ to $i$). In the provided example, if dots are unclear, feel free to forget about them and just replace N with 4 so that you have some exact instance of the system (where for example, there is no measurement between Node 1 and Node N=4). So these measurements act like connections between the mutually independent Nodes.
Assuming that I have no prior information about $ x_{3} $, I want to calculate the following marginal probability (because I want to infer the position of Node 3 given the data I have from the other Nodes and measurements):
\begin{equation}
\begin{array}{l}
P(x_{3}|everything~else) = P(x_{3} \mid x_{1}, x_{2}, ... , x_{N}, y_{1,2}, y_{1,3}, y_{2,1}, y_{2,3}, ... , y_{2, N}, y_{N, 3})
\end{array}
\end{equation}
In hopefully simpler words, I know the following: $x_{1}$, $x_{2}$, $x_{N}$, $y_{1,3}$, $y_{2,3}$, $y_{N, 3}$ and the functions $P(y_{1,3} \mid x_{1}, x_{3}) \cdot P(y_{2,3} \mid x_{2}, x_{3}) \cdot P(y_{N, 3} \mid x_{N}, x_{3})$, and I need to prove that these are enough to calculate the conditional marginal above.
I have a couple of questions about it.
Firstly, is it correct (and why?) that:
\begin{equation}
\begin{array}{l}
P(x_{3}|everything~else) \propto P(x_{3} \mid x_{1}, x_{2}, x_{N}, y_{1,3}, y_{2,3}, y_{N, 3})
\end{array}
\end{equation}
Also, is it correct (and why?) that:
\begin{equation}
\begin{array}{l}
P(x_{3}|everything~else) \propto P(y_{1,3} \mid x_{1}, x_{3}) \cdot P(y_{2,3} \mid x_{2}, x_{3}) \cdot P(y_{N, 3} \mid x_{N}, x_{3})
\end{array}
\end{equation}
Lastly, is it correct (and why?) that:
\begin{equation}
\begin{array}{l}
P(x_{3}|everything~else) \propto P(x_{3} \mid x_{1}, y_{1,3}) \cdot P(x_{3} \mid x_{2}, y_{2,3}) \cdot P(x_{3}, \mid x_{N}, y_{N, 3})
\end{array}
\end{equation}
Thank you.
 A: Say you have $n$ nodes $x_i$ and measurements $y_{j,k}$ that depend on the values of the nodes $x_j$ and $x_k$. Let's for simplicity assume there's exactly $\frac{1}{2} n(n-1)$ measurements one for each combination of $j,k$ where $j$ and $k$ differ.
So we have
$$x_1,x_2,\dots x_{n-1}, x_{n}$$
And
$$\begin{array}{}
y_{1,2}\\
y_{1,3}&y_{2,3}\\
y_{1,4}&y_{2,4}&y_{3,4}\\
\vdots&\vdots&\vdots&\ddots\\
\vdots&\vdots&\vdots&&\ddots\\
y_{1,n-1}&y_{2,n-1}&y_{3,n-1}&\dots&\dots&y_{n-2,n-1}\\
y_{1,n}&y_{2,n}&y_{3,n}&\dots&\dots&y_{n-2,n}&y_{n-1,n}
\end{array}$$
Say we know the dependencies of the $Y$ on pairs of $X$ in terms of probability mass functions $$P(Y_{j,k} = y_{j,k} | X_j = x_j, X_k = x_k)$$
The probability of an $x_i$ (let's for simplicity use $i=1$ instead of $i=3$ in the OP's question) can be expressed with Bayes' rule if we also know an unconditional probability $P(X_1 = x_1)$ (or if we allow some estimate for this) then
$$P(X_1 = x_1 | \text{something else}) \propto P(\text{something else} | X_1 = x_1)P(X_1 = x_1)$$
Anything that can play the role of something else are the values $y_{1,i}$, with $i$ from $2$ to $n$, which depend on $X_1$. And we can make use of the values $x_i$, with $i$ from $2$ to $n$, on which the values $y_{1,i}$ codepend.
Then we can write
$$P(X_1 = x_1 | \text{everything else}) \propto P(\text{everything else}| X_1 = x_1)P(X_1 = x_1)$$
The $P(\text{everything else}| X_1 = x_1)$ can be written as a product
$$P(\text{everything else}| X_1 = x_1) \propto \prod_{j=2}^n P(y_{1,j}| X_1 = x_1, X_j = x_{j})$$
This product is based on the only information that you need for the Bayes's rule. The probability of $\text{everything else}$ will not be exactly equal to it (but differs with a proportional constant, the same for all values of $x_1$). It might be better to write
$$P(y_{1,2}, y_{1,3} \dots y_{1,n-1}, y_{1,n}| X_1 = x_1, X_2 = x_2 ,\dots, X_n = x_n) = \prod_{j=2}^n P(y_{1,j}| X_1 = x_1, X_j = x_{j})$$
The probability of $\text{everything else}$ will be smaller than this but it only ends up as a constant of proportionality in the expression of Bayes's rule.
So we end up with
$$P(X_1 = x_1 | \text{everything else}) \propto P(X_1 = x_1)  \prod_{j=2}^n P(y_{1,j}| X_1 = x_1, X_j = x_{j})$$
And to get the constant of proportionality
$$P(X_1 = x_1 | \text{everything else}) = \frac{P(X_1 = x_1)  \prod_{j=2}^n P(y_{1,j}| X_1 = x_1, X_j = x_{j})}{\sum_{\text{all possible values $x$}} P(X_1 = x)  \prod_{j=2}^n P(y_{1,j}| X_1 = x, X_j = x_{j})}$$

Relating to your formulas
I hope you can follow it and relate it to your own formula. What seems to be lacking in your own formula is

*

*An incorrect application of Bayes's theorem. What if $P(X_3 = 0) = 1$? Then also $P(X_3 = 0|\text{everything else}) = 1$, but your formulas don't lead to that because you are not incorporating $P(X_3 = x_3)$.


*Also some of the formulas are not clear. You are only computing with values for indices that are $1,2,3$ and $N$, and you have no values in-between $3$ and $N$. It is not just three terms that you should use, but all the possible terms that incorporate $x_3$ should be used.
