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Given the Naive Bayes graphical representation below, I want to calculate $P(X|Y_1,Y_2)$.

Are the calculations below correct?

enter image description here

The factored joint distribution regarding the system is: $$P(X,Y_1,Y_2)=P(X) \cdot P(Y_1|X) \cdot P(Y_2|X)$$ $$\frac{P(X,Y_1,Y_2)}{P(X)}=P(Y_1|X) \cdot P(Y_2|X)$$ $$P(Y_1,Y_2|X)= P(Y_1|X) \cdot P(Y_2|X)$$ Using Bayes rule to flip the left hand side: $$\frac{P(X|Y_1,Y_2) \cdot P(Y_1,Y_2)}{P(X)}= P(Y_1|X) \cdot P(Y_2|X)$$ $$P(X|Y_1,Y_2) = \frac{P(Y_1|X) \cdot P(Y_2|X) \cdot P(X)}{P(Y_1,Y_2)}$$ $$P(X|Y_1,Y_2) = \frac{P(Y_1|X) \cdot P(Y_2|X) \cdot P(X)}{P(Y_1) \cdot P(Y_2)}$$

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3 Answers 3

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No, the move from your penultimate equation to the final equation has an error.

In general you should not say $P(Y_1,Y_2)= P(Y_1)P(Y_2)$.

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Up to the last line, like Henry already pointed out, is all correct. What you could write is,

$$ P(X|Y_{1},Y_{2}) = \frac{P(Y_{1},Y_{2}|X)P(X)}{\sum_{X}P(Y_{1},Y_{2}|X)P(X)} = \frac{P(Y_{1},|X)P(Y_{2},|X)P(X)}{\sum_{X}P(Y_{1},|X)P(Y_{2},|X)P(X)} $$

$Y_{1}$ and $Y_{2}$ are only independent given $X$.

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According to the Naive Bayesian graph, the covariate $Y_1$ is independent of the covariate $Y_2$.

Therefore $P(Y_1|X) \cdot P(Y_2|X)=P(Y_1,Y_2|X)$ and $P(Y_1,Y_2) = P(Y_1) \cdot P(Y_2) $

Implementing these into your last equation, you can see that:

$$P(X|Y_1,Y_2) = \frac{P(Y_1,Y_2|X) \cdot P(X)}{P(Y_1,Y_2)} $$

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    $\begingroup$ Don't know much about Naive Bayes. However, looking at the graph, Y1 and Y2 are only conditionally independent given X. Think of them as points uniformly distributed on a frisbee, and X as the co-ordinates of where the frisbee lands $\endgroup$ Commented Nov 8, 2013 at 14:57

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