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I'm trying to calculate the posterior probability of a coin toss resulting in "heads". We assume the uniform distribution of the prior $p(\theta)=1,\theta\in[0,1]$. Now suppose we toss a coin the first time and see "heads". The posterior distribution after the first toss is:

$p(\theta|heads)=\frac{p(\theta) p(heads|\theta)}{\int_0^1 p(\theta)p(heads|\theta) \, d\theta }=\frac{\theta }{\frac{1}{2}}=2 \theta$

Looks ok.

Now I update my prior to $p(\theta)=2\theta$, make the second toss and see "tails". To calculate the posterior for the second experiment I do:

$p (\theta |tails)=\frac{p(\theta) p (tails|\theta ))}{\int_0^1 p(\theta) p (tails|\theta )) \, d\theta }=-\frac{2 (1-2 \theta ) \theta }{\frac{1}{3}}=-6 (1-2 \theta ) \theta$

Which is incorrect, because $-6 (1-2 \theta ) \theta$ has negative values if $0<\theta <\frac{1}{2}$.

What is wrong with my calculations?

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    $\begingroup$ I'm not sure where that negative sign came from. If the second flip is a tails it should contribute $1-\theta$ from the likelihood and hence the numerator should be $(1-\theta) \times 2\theta$. $\endgroup$ Commented Mar 25, 2023 at 14:52
  • $\begingroup$ @Demetri Pananos. That's it! Thank you. I confused the probability of seeing tails with the probability density of $\theta$ $\endgroup$
    – Max
    Commented Mar 25, 2023 at 15:55

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