# Invert Bayesian Probability: P(S|D,L) =?

First time posting, apologies for any convention missteps. I've been trying to tackle bayesian probability and bayes networks for the past few days, and I'm trying to figure out what appears to be something very simple, but I don't know that I'm doing it right.

• I have a node L with parents S and D.
• P(S) = 0.1
• P(D) = 0.1
• P(L|D,S) = 0.95

(This network is from http://hugin.com/wp-content/uploads/2016/05/Building-a-BN-Tutorial.pdf, CPT given on page 3, visualized on page 4 and following)

What I'm trying to find is an algorithm for is:

P(S|L,D)

I know basic bayes rule means P(S|L) = ( P(L|S) * P(S) ) / P(L), but I'm not sure how to modify that basic algorithm for more than one dependent. In other words, I know how to get P(S|L), but can I extend that to give me P(S|L,D) ?

Two applications of this:

• This would be used for the situation where I observe L as true, so I want to calculate the maximum likely explanation for L. The possible explanations being: SD, S!D, !SD, or !S!D. (Naive question, but could that just be the max value in the CPT table for L=true then...?)
• Additionally, I want to update the probability of S given that I observed L.
• Now, I'm wondering since S is d-separated from D by L, is P(S|L) simply the basic bayes formula?
• If that's the case, it leads to another question, in the basic bayes formula, what would I then use for P(L|S) after I've observed it?
• Since L has two parents, S and D, P(L|S) could be one of 2 possible probabilities from the CPT (P(L|S,D) or P(L|S,!D)) - or, since I observed L, do I just consider P(L|anything) = 1, for the purpose of P(S|L) ?

It is highly recommended to start solving problems like that from drawing a tree.

So you want to find $P(S|L\cap D)$. By definition of conditional probability $$P(S|L\cap D)=\frac{P(S\cap L\cap D)}{P(L\cap D)}$$

The numerator $P(S\cap L\cap D)=P(S)P(D|S)P(L|D\cap S)=0.1\cdot 0.1 \cdot 0.95$

The denominator $P(L\cap D)=P(L\cap D\cap S)+P(L\cap D\cap S^c)=\\ 0.1\cdot 0.1 \cdot 0.95+0.9\cdot 0.1\cdot 0.85$

Which is important for solving the problem.

We have that:

$$\begin{split} P(L,D,S) & = P(L|D,S)P(D)P(S)\\ \Rightarrow P(L) & = \sum_S\sum_D P(L,D,S) = \sum_S\sum_D P(L|D,S)P(D)P(S) \end{split}$$

Once these marginals are estimated, we have that:

$$\begin{split} P(S|D,L) & = \frac{P(L,S)|D }{P(L|D)}\\ & = \frac{P(L|D,S)P(S)}{P(L|D)} \end{split}$$ Calculating:

$$\begin{split} P(L=\textrm{"yes"}|D = \textrm{"dry"}) & = P(L=\textrm{"yes"},S=\textrm{"sick"}| D = \textrm{"dry"})+ P(L=\textrm{"yes"},S=\textrm{"not"}| D = \textrm{"dry"})\\ & = .1*.1*.95 + .9*.1*.85\\ & = .086\\ P(L=\textrm{"yes"}|D = \textrm{"not"}) & = P(L=\textrm{"yes"},S=\textrm{"sick"}| D = \textrm{"not"})+ P(L=\textrm{"yes"},S=\textrm{"not"}| D = \textrm{"not"})\\ & = .9*.1*.9 + .9*.9*.02\\ & = .0972 \end{split}$$

$$\begin{split} P(S|D,L) & = \frac{P(L,S|D) }{P(L|D)}\\ & = \frac{P(L|D,S)P(S)}{P(L|D)}\\ & = \frac{P(L|D,S)P(S)}{P(L|D)} \end{split}$$

Which we have all the required quantities to compute for any configuration of $S$, $D$, and $L$.

• I believe this is wrong because the events $D$ and $L$ are dependent. – kludg Dec 22 '17 at 18:38
• Why does that make it incorrect? I'm not following – Ryan Warnick Dec 22 '17 at 18:58
• I think $P(S|D,L) = \frac{P(L,D,S) }{P(D)P(L)}$ can't be true if $L$ and $D$ are dependent. – kludg Dec 22 '17 at 19:01
• So if I made it $P(S|D,L)= \frac{P(L,S|D)}{P(L|D)} = \frac{P(L|D,S)P(S)}{P(L|D)}$ I think you'd get the same answer and because you're conditioning on $D$ you take into account the dependence. – Ryan Warnick Dec 22 '17 at 19:13
• I think you were right, but the last comment fixes it I believe, and I put that approach in the answer. – Ryan Warnick Dec 22 '17 at 19:28