enter image description hereI am confused in obtaining a term in Bayesian Formula. I have attached my partial solution in the image. How can I calculate the term P(X1/X3)??. Image attached

  • $\begingroup$ I think it should be $0.5$. $\endgroup$ – Ryan Aug 7 '17 at 6:11
  • $\begingroup$ Could you please explain it. Thank you @Ryan $\endgroup$ – Abdul Karim Khan Aug 7 '17 at 6:13
  • $\begingroup$ I have already attached answer for you, please check :-) $\endgroup$ – Ryan Aug 7 '17 at 6:18

Hint: The most important part is to solve $p(A|x_1)$ and $p(\neg A|x_1)$.

$$p(A|x_1) = \frac{p(x_1|A)p(A)}{p(x_1)} = \frac{p(x_1|A)p(A)}{p(x_1|A)p(A) + p(x_1|\neg A)p(\neg A)} = 0.25$$

Then, $p(\neg A|x_1) = 0.75$.

(PS: Based on the Chapman-Kolmogorov equation, the result of $p(x_3|x_1)$ is obvious from $p(A|x_1)$ and $p(\neg A|x_1)$. Here, we still explain it in details.)

Note that $p(x_3|x_1) = p(x_3, A|x_1) + p(x_3, \neg A|x_1)$ (by the law of total probability). We calculate the following two terms

$$p(x_3,A|x_1) = p(x_3|A)p(A|x_1) = 0.05,$$ $$p(x_3,\neg A|x_1) = p(x_3|A)p(\neg A|x_1) = 0.45,$$

where $p(x_3|A,x_1) = p(x_3|A)$ and $p(x_3|\neg A,x_1) = p(x_3|\neg A)$.

Now, $p(x_3|x_1) = p(x_3, A|x_1) + p(x_3, \neg A|x_1) = 0.05 + 0.45 = 0.5$.

  • $\begingroup$ Would you please explain why p(x3|x1)=p(x3,A|x1)+p(x3,¬A|x1)? @Ryan $\endgroup$ – Abdul Karim Khan Aug 7 '17 at 6:27
  • $\begingroup$ Sorry, It would not be that complex. Just the law of total probability en.wikipedia.org/wiki/Law_of_total_probability, which is the Chapman-Kolmogorov equation based on. $\endgroup$ – Ryan Aug 7 '17 at 6:35
  • $\begingroup$ Yeah, Thanks. I solved it using law of total probability.@Ryan $\endgroup$ – Abdul Karim Khan Aug 7 '17 at 6:37
  • $\begingroup$ @AbdulKarimKhan Cheers! $\endgroup$ – Ryan Aug 7 '17 at 6:38

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