Uniform Prior

Using a uniform prior, the MAP becomes the MLE

Suppose we want to find the chance of head $p$ of a possibly unfair coin. However we have no reason to expect the coin to be unfair as most coin tosses are fair-ish. We take the uniform prior.

$$
p  \sim \text{Uniform}(0,1)
$$

We perform $n$ trials with the coin and got $H=k$ heads. The likelihood is binomial. The posterior distribution is then,

$$
\begin{align*}
	P(p \mid H = k)  &= \frac{1 \cdot P(H=k \mid p)}{P(H=k)}\\
					 &\propto P(H=k \mid p) \\
	P(p \mid H = k)  &\propto p^{k}(1-p)^{n-k}
\end{align*}
$$

Since $p$ is a random variable the posterior distribution is,

$$
(p \mid H=k) \sim \text{Beta}(k+1, n-k+1)
$$

With expectation and MAP,

$$
E(p \mid H=k) = \frac{k+1}{n+2}
$$

$$
\text{mode} (p \mid H=k) = \frac{k}{n}
$$

At large $n$, these two are nearly equal. We have found that the observed propotion of heads is the MAP.

Disproportionate Prior

Suppose you want to determine the chance of an event \(p\). If you believe the event is disproportionate, that is, you believe on average there is \(x\) events out of \(n\).

\[ E(p) = \frac{x}{n} \]

In general the proposed \(E(p)\) does not have to be a rational number.

\[ p \sim \text{Beta}(x, n-x) \]

Using the unfair coin example above, suppose in life, we have experience a lot of cheating so we think that the coin is unfair. Our prior knowledge motivates us to propose the coin lands head at chance $70\%$,

$$
E(p) = \frac{7}{10}
$$

We take the prior distribution to be $p \sim \text{Beta}(7, 3)$.