0
$\begingroup$

Assume I have the following Junction Tree between random variables $X_1,\dots,X_7$ that exactly describes the sets variables with non-zero Mutual Information (Alternatively it's the last tree in a Truncated Vine Sequence).

enter image description here

Where the nodes with circles are the clusters of the Junction Tree, and the nodes with squares are the separators of the Junction Tree.

According to this paper and formula ($4.1$), I can then calculate their Joint Distribution. However, I'm having difficulties understanding the formula. Would it be correct to say that for the tree above, we can write up this formula:

$$\begin{matrix} & & \mathbb{P}(X_1\le x_1,X_3\le x_3,X_6\le x_6)\mathbb{P}(X_1\le x_1,X_4\le x_4,X_6\le x_6) \\ & & \mathbb{P}(X_1\le x_1,X_4\le x_4,X_5\le x_5)\mathbb{P}(X_1\le x_1,X_2\le x_2,X_4\le x_4) \\ & & \mathbb{P}(X_1\le x_1,X_2\le x_2,X_7\le x_7)\\ \mathbb{P}(X_1\le x_1,\dots,X_7 \le x_7)& = & \frac{\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}{} \\ & & \mathbb{P}(X_1\le x_1,X_6\le x_6)\mathbb{P}(X_1\le x_1,X_4\le x_4)^2\mathbb{P}(X_1\le x_1,X_2\le x_2) \end{matrix}$$

(Everything is multiplied together in the nominator, I just didn't have enough space; and the $\mathbb{P}(X_1\le x_1,X_4\le x_4)$ term in the denominator is squared because it appears twice.)

So my question is: Is the above formula correct? Or is it perhaps only applicable for joint density/weight functions?

$\endgroup$

1 Answer 1

0
$\begingroup$

I have since found the answer for my question:

No, the formula isn't correct, you can generate counterexamples where it doesn't hold. However it is true if we replace every $\le$ sign with an $=$ sign.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .