2
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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 $X$.

$\endgroup$
0
$\begingroup$

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

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

$\endgroup$
-1
$\begingroup$

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:

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

$\endgroup$
  • 3
    $\begingroup$ 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 $\endgroup$ – conjectures Nov 8 '13 at 14:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.