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The question is about the following slides: enter image description here

I have confusion regarding the second and the third statements. Regarding the second statement ("If I not observed nor is anything else"). Here, we get the probability of I given S, then we can get the probability of G given I and finally we can get the probability of D given G. Why is the flow not allowed then. What's the flaw with this argument? Regarding the third one (I not observed, G is observed): I don't even know how to comprehend this. Can someone explain?

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    $\begingroup$ $S$ tells you something about $I$, but $I$ and $D$ are independent so knowing $I$ does not tell you anything about $D$ unless you also know $G$. The flaw in your argument is that you assume you can get $G$ given $I$. You cannot, because $G$ is also influenced by $D$. The third one is where you observe $G$, so now you can use that additional knowledge to infer $D$ from $S$. $\endgroup$ – Moss Murderer Jan 18 '18 at 21:49
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The best strategy is to first write the factorized form of the graphical model: $$ P(D, G, L, I, S) = P(D)P(I)P(G|D,I)P(L|G)P(S|I) $$

Now, you wonder why it is not allowed to move all the way from S to D. Using the Bayes theorem: $$ P(D|S) = \frac{\sum_{G,I,L} P(D,G,L,I,S)}{P(S)} = \\ \frac{P(D)}{P(S)}\sum_{G,I} P(I)P(G|D,I)P(S|I) = \\ \frac{P(D)}{P(S)}\sum_{I} P(I)P(S|I) = \\ \frac{P(D)}{P(S)}P(S) = P(D) $$ that is, they are independent. If G is known, then it all changes, since then you cannot factor it out anymore (you would not have the sum over G).

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