## Introduction

Today, I will explain easy things in a complex way. I will give a simple example of maximum likelihood estimation of the probabilities of a biased coin toss.

## Bernoulli trial

A Bernoulli trial is a random experiment with two outcomes. The standard example is the flip of a probably biased coin. Let us first denote the outcomes with $$0$$, and $$1$$ instead of head and tail, since that sounds a lot more professional. The coin will land on one side, say $$1$$ with probability of $$p$$. It will land on the other side $$0$$ with a probability of $$1-p$$, since the probabilities need to sum to one. In mathematical formulation we get the probability $$P(X=1)=p$$ and $$P(X=0)=1-p$$. Now we can define the probability mass function $$f(x;p)$$.

$$$f(x;p)=\begin{cases} p, & x=1\\ 1-p, & x=0\\ \end{cases}$$$

In case you only know continuous probability distributions, the probability mass function is similar to the probability density function, but in the discrete case.

## Maximum Likelihood Estimation

Let us now assume, that we have flipped the coin a few times and got the results $$x_1,...,x_n$$ which are either $$0$$ or $$1$$. In stochastic parlance these are called observations. The question is what the probability $$p$$ is. Intuitively one could assume that it is the number of ones we got divided by the total number of coin throws. For example if we threw the coin a hundred times and 30 times we got a one, then we would maybe guess $$\hat{p}=30/100$$. Actually this is the interpretation from a frequentist perspective, but let us not digress into that territory. I will prove in the following that the intuition in this case is correct, by proving that the guess $$\hat{p}=\sum x_i/n$$ is the “most likely” value for the real $$p$$.

First, let us reformulate our probability mass function as $$f(x;p)=p^x(1-p)^{1-x}$$, which makes it easier to calculate with it. If we have the (independent) observations $$x_1,...,x_n$$, then their joint probability mass function is

$$$f(x_1,...,x_n;p)=\prod\limits_i f(x_i;p)=p^{\sum x_i} (1-p)^{n-\sum x_i}.$$$

If we interpret this as a function not in the observations with the fixed parameter $$p$$, but instead as a function in the model parameter with fixed observations, we get what is called the likelihood function $$\mathcal{L}(p;x_1,...,x_n)=f(x_1,...,x_n;p)$$.

We now want to find the $$p$$ with the highest likelihood given the observation $$x_1,...,x_n$$, that is, we want to maximize our likelihood function $$\mathcal{L}(\cdot;x_1,...,x_n)$$. In this case it is easier to maximize the log of the likelihood $$\log\mathcal{L}(p)=\sum x_i\log p + (n-\sum x_i)\log(1-p)$$ , which yields the same result, since the log is monotonously increasing.

As you may remember, maximizing a function means setting its first derivative to zero.

\begin{align*} \frac{\partial\log\mathcal{L}(p)}{\partial p} = \frac{\sum x_i}{p} - \frac{n-\sum x_i}{1-p} = \frac{(1-p)\sum x_i -pn +p \sum x_i}{p(1-p)} = \frac{\sum x_i -pn}{p(1-p)} \stackrel{!}{=}0 \end{align*}

Multiplying by $$p(1-p)$$ yields that the maximum is reached at $$p=\frac{\sum x_i}{n}$$. This is the most likely value for $$p$$ given our observation which confirms our intuition.

20 December 2017

#### Category

Recreational Math