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My apology for stating my question with an image. The following image shows that the posterior distribution is as follows with the given information:

enter image description here

I wonder how the posterior has been calculated, analytically.

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    $\begingroup$ You should explain why you cannot compute $p(\theta|x)$ from $p(\theta)$ and $p(x|\theta)$. $\endgroup$
    – Xi'an
    Commented May 13, 2022 at 6:57
  • $\begingroup$ I am new to Bayesain Statistics and trying to understand how the above question is derived from the given information. $\endgroup$ Commented May 13, 2022 at 7:07
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    $\begingroup$ You should at least know how to move from the pair prior x likelihood to the posterior $\endgroup$
    – Xi'an
    Commented May 13, 2022 at 7:15
  • $\begingroup$ I gave it a go and solved the question. When I uploaded the question, I didn't have a clue how to calculate it but now I can. Thank you for your comment. $\endgroup$ Commented May 13, 2022 at 7:17

1 Answer 1

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For $ x=1 $

$ \dfrac{p(\theta)p(x=1 | \theta)}{p(\theta_1)p(x=1 | \theta) + p(\theta_2)p(x=1 | \theta_2)} = \dfrac{p(0)p(x=1 | 0)}{p(0)p(x=1|0)+p(1)p(x=1|1)} = \dfrac{0.1665}{0.417} = 0.399280576 \approx 0.4$

So, when $ x=0 $, the probability of $ x=0 $ is $ 0.6 $

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    $\begingroup$ The use of slightly incorrect decimal approximations, written out in excessive precision, makes this calculation somewhat confusing: the answer $3/5$ is exact, not approximate. $\endgroup$
    – whuber
    Commented May 13, 2022 at 14:01

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