I'm working on a Bayesian analysis of a coin-toss scenario and have a conceptual question to clarify my understanding.


Given a uniform prior on the probability that a coin lands tails over $[1-b,1-a]$ (and therefore, the probability of landing heads is uniformly distributed over $[a, b]$), with some sample evidence $D$, I aimed to determine the posterior distribution for the probability $\theta$ of landing on heads. I approached this using Bayes' theorem combined with a binomial likelihood. Specifically the posterior can be calculated as: $$ f_H(\theta \mid D) = \frac{\theta^s(1-\theta)^f \cdot \mathbb{1}_{a \leq \theta \leq b}}{\beta_b(s+1, f+1)-\beta_a(s+1, f+1)} $$ Where:

  • $s$ is the number of observed heads.

  • $f$ is the number of observed tails.

  • $\beta_b(s+1, f+1)$ and $\beta_a(s+1, f+1)$ are evaluations of the incomplete beta function at points $b$ and $a$, respectively.

New Scenario:

Now, I'm considering a twist: the coin can also land on its side. The probability of this side landing is fixed at 10%. Given a uniform prior on the probability of tails over $[1-b,1-a]$, the prior on the probability of heads is now over $[a-0.1, b-0.1]$. I aim to find the posterior on the probability of heads given new sample evidence.

Main Question:

Given the fixed 10% probability for the side landing, can I essentially employ a similar approach as my original scenario, only adjusting the boundaries of the prior distributions, incomplete beta function, the indicator function and the ($1-\theta$) to $(0.9-\theta)$? Specifically, can I use: $$ f_H(\theta \mid D) = \frac{\theta^s(0.90-\theta)^f \cdot \mathbb{1}_{a-0.1 \leq \theta \leq b-0.1}}{\beta_{b-0.1}(s+1, f+1)-\beta_{a-0.1}(s+1, f+1)} $$ In effect, I ignore the sample evidence associated with the side-flip is its distribution is known with certiainty. This means i treat the process effectively as a binomial with reduced bounds on the probability. Id greatly appreciate any insights or corrections to my approach.


1 Answer 1


You are correct to point out that the posterior distribution of $\theta$ will not depend on the number of times the coin landed on its side, as the probability of the side landing is known.

The denominator in your final expression is not quite right, because the integral you need to evaluate is $$ \int_{a-0.1}^{b-0.1}\theta^s (0.9-\theta)^f d\theta $$ whereas the difference of incomplete beta functions you wrote down corresponds to the integral with $(1-\theta)^f$ in the integrand instead of $(0.9-\theta)^f$.

I think the easiest way to get the solution is to reparameterise the problem. If $\theta$ is the probability of heads (the quantity of interest), you could define $\theta^*$ to be the probability of heads conditional on the coin not landing on its side, i.e. $\theta^*=\theta/0.9$. You can now discard all the trials where the coin landed on its side and write down the posterior for $\theta^*$.


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