calculate probability using variable elimination

Question

Consider the following Conditional probability for the Bayesian Network:

By using variable elimination, how to calculate the following probability?

I am summing all the terms related to $E$, then will do for $L$, and then $T$, but I am not sure what values to put in and how to proceed further?

Answer 1

$\def\-{\lnot}$You can make things somewhat easier for yourself by cancelling common factors $\def\P{\operatorname{\sf P}}\def\|{\,\vert\,}\P(a),\P(\-s)$, and reordering the series.

The factorisation is:$$ \small\P(a,s,t,l,b,e,d,x)=\P(a)\P(s)\P(t\|a)\P(l\|s)\P(b\|s)\P(e\|t,l)\P(d\|b,e)\P(x\|e)$$

So $$\begin{align} \P(b\|a,\-s,\-x,d)&=\tfrac{\P(a)\P(\-s)\quad\P(b\|\-s)\quad\sum\limits_{\pm t}\P(\pm t\|a)\sum\limits_{\pm\ell}\P(\pm\ell\|\-s)\sum\limits_{\pm e}\P(\pm e\|\pm t,\pm\ell)\P(d\|b,\pm e)\P(x\|\pm e)}{\P(a)\P(\-s)\sum\limits_{\pm b} \P(\pm b\|\-s)\sum\limits_{\pm t}\P(\pm t\|a)\sum\limits_{\pm\ell}\P(\pm\ell\|\-s)\sum\limits_{\pm e}\P(\pm e\|\pm t,\pm\ell)\P(d\|\pm b,\pm e)\P(x\|\pm e)}\\[1ex]&=\tfrac{\quad\P(b\|\-s)\sum\limits_{\pm e}\P(x\|\pm e)\P(d\|b,\pm e)\sum\limits_{\pm t}\P(\pm t\|a)\sum\limits_{\pm\ell}\P(\pm\ell\|\-s)\P(\pm e\|\pm t,\pm\ell)}{\sum\limits_{\pm b}\P(\pm b\|\-s)\sum\limits_{\pm e}\P(x\|\pm e)\P(d\|\pm b,\pm e)\sum\limits_{\pm t}\P(\pm t\|a)\sum\limits_{\pm\ell}\P(\pm\ell\|\-s)\P(\pm e\|\pm t,\pm\ell)}\\[1ex]&=\tfrac{\quad\P(b\|\-s)\sum\limits_{\pm e}\P(x\|\pm e)\P(d\|b,\pm e)\P(\pm e\|a,\-s)}{\sum\limits_{\pm b}\P(\pm b\|\-s)\sum\limits_{\pm e}\P(x\|\pm e)\P(d\|\pm b,\pm e)\P(\pm e\|a,\-s)} \end{align}$$

Where $$\small\begin{align}\P(e\|a,\-s) &= \sum\limits_{\pm t}\P(\pm t\|a)\sum\limits_{\pm\ell}\P(\pm\ell\|\-s)\P(e\|\pm t,\pm\ell)\\[1ex]&={{\P(t\|a)\left(\P(\ell\|\-s)\P(e\|t,\ell)+\P(\-\ell\|\-s)\P(e\|t,\-\ell)\right)}\\+{\P(\-t\|a)\left(\P(\ell\|\-s)\P(e\|\-t,\ell)+\P(\-\ell\|\-s)\P(e\|\-t,\-\ell)\right)}}\\[1ex]&={{\P(t\|a)\left(\P(\ell\|\-s)+\P(\-\ell\|\-s)\right)}+{\P(\-t\|a)\P(\ell\|\-s)}}\tag 1\\[1ex]&=\P(t\|a)+\P(\-t\|a)\P(\ell\|\-s)\tag 2\\[1ex]&=0.05+(1-0.05)(0.01)\tag 3\\[1ex]&=0.0595\end{align}$$

Since

1. $\P(e\mid\-t,\-\ell)=0$ while the other three such terms equal $1$.
2. $\P(\ell\mid\-s)+\P(\-\ell\|\-s)=1$ due to the law of complements
3. By reading the two values from the table, and the third is a complement.

Now you may find $\P(\-e\|a,\-s)$ by the same procedure.

$\qquad$[Note: $\P(\-e\mid\-t,\-\ell)=1$ while the other three such terms equal $0$ ]

... or of course, $\P(\-e\| a,\-s)=1-0.0595=0.9405$

And so on...