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I know that my prior distribution is Beta(3,3) and that after tossing 12 coins, the number of 'heads' is less than 4 but I don't know the exact number. How do I calculate the posterior density?

What I've tried to do is:

If $X=\#$ of heads in $n=12$ tosses then $X\sim Bin(12,\theta)$

$$\pi(\theta|x)=\frac{p(x|\theta)\pi(\theta)}{f(x)}$$

Where $f(x)$ is the marginal density of the Likelihood.

I tried using 4 cases for $X={0,1,2,3}$

And calculate 4 different values of the posterior density. But is that correct?

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1 Answer 1

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What use is four posterior densities? I would have thought you wanted one. It would be a weighted average of them, but perhaps difficult to find the weights.

If $p$ is the probability of a head then, if my calculations are correct,

  • The prior density is proportional to $p^2(1-p)^2$ from your Beta distribution;

  • The likelihood given $3$ or fewer heads from $12$ attempts is proportional to ${12 \choose 0}(1-p)^{12}+{12 \choose 1}p(1-p)^{11}+{12 \choose 2}p^2(1-p)^{10}+{12 \choose 3}p^3(1-p)^9$; and

  • the posterior density is proportional to the product of these, but needs to integrate to $1$ over $[0,1]$

which suggests to me a posterior density of $$\dfrac{185640}{1271} p^2(1-p)^{11}\left(1+9p+45p^2+165p^3\right)$$

which is in a sense a weighted average of $\operatorname{Beta}(3,15)$, $\operatorname{Beta}(4,14)$, $\operatorname{Beta}(5,13)$ and $\operatorname{Beta}(6,12)$ densities, but I do not see a quicker way which involves calculating the weights required

In the chart below, the information from the observation has shifted the red prior density to the blue posterior density

enter image description here

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