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$P$ and $S$ are the common cause of $c$. If $P(C=true| P,S )$ is given as the table below, and $P(S=true) =0.3$, $P(P=true) =0.9$ how can I eliminate $S$ and calculate $P(C=true | P=true )$ and $P(C= false| P=true)$ ? Many thanks for your help.

$P(C=true | P,S)$:

+p  +s +c  0.05
+p  -s +c  0.02
-p  +s +c  0.03
-p  -s +c  0.001
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1 Answer 1

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You need to sum out a variable using the law of total probability. So $$ P(c=true | p=true) = P(c=true | p=true, s=true) * P(p=true, s=true) + P(c=true | p=true, s=false) * P(p=true, s=false) $$ This can intuitively be understood as summing probabilities across mutually exclusive spaces. This Stanford pdf explains where the equation comes from well.

Because p and s are independent

$$ P(p=true, s=true) = P(p=true) * P(s=true) $$ $$ P(p=true, s=false) = P(p=true) * P(s=false) $$ So do the substitution in the above equation and you're good to go.

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  • $\begingroup$ I know I can sum out P(C, P,S) , can I sum out P(C | P, S) as well? How to calculate P(C=false | P=true) then? I am confused as if I use P(C,P,S) = P(P,S)*P(C|P,S) , and sum out P(C, P,S), the results are different from suming up P(C, P,S) directly. Which one is correct? Thanks! $\endgroup$
    – user4046073
    Commented May 26, 2019 at 3:58
  • $\begingroup$ Sorry, read your question too quickly and assumed your table was the joint distribution. Have updated the answer $\endgroup$
    – Brent
    Commented May 26, 2019 at 4:29
  • $\begingroup$ Thanks. could you explain a bit more why the first equation ? $\endgroup$
    – user4046073
    Commented May 26, 2019 at 8:31
  • $\begingroup$ should the last bit be P(s= false, p= true)? $\endgroup$
    – user4046073
    Commented May 26, 2019 at 8:32
  • $\begingroup$ I mean should it be +𝑃(𝑐=𝑡𝑟𝑢𝑒|𝑝=𝑡𝑟𝑢𝑒,𝑠=𝑓𝑎𝑙𝑠𝑒)∗𝑃(𝑠=𝑓𝑎𝑙𝑠𝑒,𝑝=true) ? $\endgroup$
    – user4046073
    Commented May 26, 2019 at 10:51

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