0
$\begingroup$

Let's say I have an 'unfair' coin, for which I'm interested in estimating the 'heads' likelihood or 'p' value.

Knowing nothing about the coin, the distribution of probable 'p' values is a uniform distribution from 0 to 1.

How can I update this distribution of probable 'p' values after observing X heads out of Y trials?

$\endgroup$
0
$\begingroup$

Seems like Bayesian is the natural approach here. Since prior of p is unif(0,1), posterior is Beta(X+1, Y-X+1) as already answered by jbowman.

So I'll add the details here. Let $x_i$, $i=1,\ldots,Y$ be the outcomes from coin tosses. Given: $\sum_{i=1}^Y x_i = X$, the number of heads. Let $\theta = Pr(X_i = 1)$, where we'll treat 'head' as 1. Therefore, the random variable $X_i \sim Bernoulli(\theta)$, with 'head' as the 'success'.

Given prior for $\theta$: $f(\theta) = 1 I(\theta \in (0,1))$, which is unif(0,1). Since $\sum_{i=1}^Y x_i$ is $Binomial(Y,\theta)$, the likelihood is $Pr(\sum_{i=1}^Y x_i = X) = K\theta^X(1-\theta)^{Y-X}$, where $K$ is $Y$ choose $X$.

Now, posterior of $\theta$: $f(\theta\mid D) = \frac{f(D\mid\theta)f(\theta)}{\int_{0}^{1}f(D\mid\theta)f(\theta)d\theta}$, where $D$ is $\sum_{i=1}^Y x_i$, the 'data'. Therefore, $f(\theta\mid D) = \frac{K\theta^X(1-\theta)^{Y-X}}{K\int_{0}^{1}\theta^{X}(1-\theta)^{Y-X}d\theta} = \frac{\theta^X(1-\theta)^{Y-X}}{Beta(X+1, Y-X+1)}$, which is the density function of the Beta distribution with parameters $X+1$ and $Y-X+1$. $Beta(,)$ is the Beta function.

$\endgroup$
0
$\begingroup$

I have gone through an example of updating the probability of $p$ after each observation, starting with a flat prior - with a graph of the posterior probability of $p$ after each update. Basically the same as the other 2 answers, but graphed it, and done sequentially after each new flip.

https://stats.stackexchange.com/a/371520/223056

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.