Probstat/notes/parameter estimation

This is part of probstat.

Previously we tried to estimate the population means and variances using the sample means and variances. In this section, we shall see the justification why what we did makes sense.

There are many ways to estimate parameters.

Method of moments estimators

See also wikipedia article.

This is probably the simplest estimators. However, they are often biased (as we shall show in the example).

Definition: For a random variable $X$, $E[X^{k}]$ is called the k-th moment of $X$. Note that the first moment is the mean $E[X]$. The variance of a random variable depends on the first and the second moments.

If we want to estimate a parameter $\theta$, using the method of moments, we start by writing the parameter as a function of the moments, i.e.,

$\theta =g(E[X],E[X^{2}],E[X^{3}],\ldots ,E[X^{r}])$

We then estimate the sample moments

$M_{k}={\frac {\sum _{i=1}^{n}X^{k}}{n}},$

for $k=1,\ldots ,r$. Our estimate ${\hat {\theta }}$ is thus

${\hat {\theta }}=g(M_{1},M_{2},\ldots ,M_{r})$

EX1: We show how to estimate the variance with the method of moments. Recall that the variance

$Var(X)=g(E[X],E[X^{2}])=E[X^{2}]-E[X]^{2}.$

We first estimate the first moment $M_{1}={\frac {\sum _{i=1}^{n}X}{n}}$ and the second moment $M_{2}={\frac {\sum _{i=1}^{n}X^{2}}{n}}$. The estimator is

${\hat {\theta }}=M_{2}-M_{1}^{2}=\left({\frac {\sum _{i=1}^{n}X^{2}}{n}}\right)-\left({\frac {\sum _{i=1}^{n}X}{n}}\right)^{2}={\frac {\sum _{i=1}^{n}(X_{i}-{\bar {X}})^{2}}{n}}.$

Note that the estimate ${\hat {\theta }}$ is biased, because $E[{\hat {\theta }}]={\frac {n-1}{n}}Var(X)$.

As this example shows, other estimation techniques are usually preferred over the method of moments.

Maximum likelihood estimators

Suppose that we want to estimate parameter $\theta$ based on observations (or sample) $X_{1},X_{2},\ldots ,X_{n}$. If we can find a joint-distribution function

$f(x_{1},x_{2},x_{3},\ldots ,x_{n}|\theta )$,

that gives the probability that we observe $(x_{1},x_{2},\ldots ,x_{n})$ given a particular value $\theta$. The maximum likelihood estimator is ${\hat {\theta }}$ such that $f(x_{1},x_{2},\ldots ,x_{n}|{\hat {\theta }})$ is maximum, i.e.,

${\hat {\theta }}=\arg \max _{\theta }f(x_{1},x_{2},x_{3},\ldots ,x_{n}|\theta )$.

Examples

EX1: A box contains 5 balls; they are blue balls and white balls. We randomly choose a ball from the box 3 times with replacement and get 2 blue balls and 1 white balls. Estimate the number of blue balls in the box using the maximum likelihood method.

Solution: Let $\theta$ denote the number of blue balls. The possible values are 0,1,2,3,4, and 5. For each value of $\theta$ the probability that we observe the outcome of getting 2 blue balls and 1 while balls is

${3 \choose 2}(\theta /5)^{2}(1-\theta /5)$.

We put the probabilities in the following table.

$\theta$	0	1	2	3	4	5
Probability	0	0.096	0.288	0.432	0.384	0

Since the probability is maximum at $\theta =3$, the maximum likelihood estimator for the number of blue balls is 3.

EX2: Maximum likelihood estimator for a Bernoulli parameter

A coin has probability $p$ of turning up heads. We perform $n$ independent trials and obtain random variables $X_{1},X_{2},\ldots ,X_{n}$ such that $X_{i}=1$ if in the i-th trial, we get a Head, and $X_{i}=0$ otherwise. What is the maximum likelihood estimator for $p$?

Solution: Note that $P\{X_{i}=1\}=p=1-P\{X_{i}=0\}$. We rewrite the probability that the random variable $X_{i}$ takes value $x_{i}$ for $x_{i}=0,1$ as

$P\{X_{i}=x_{i}\}=p^{x_{i}}(1-p)^{1-x_{i}}$.

(Remarks: The random variable is $X_{i}$. We denote the actual value of the variable with $x_{i}$ (i.e., with a small x).)

Therefore, the probability of obtaining the data $x_{1},x_{2},\ldots ,x_{n}$ given parameter $p$ is

$f(x_{1},x_{2},\ldots ,x_{n}|p)=p^{x_{1}}(1-p)^{1-x_{1}}\cdot p^{x_{2}}(1-p)^{1-x_{2}}\cdots p^{x_{n}}(1-p)^{1-x_{n}}=p^{\left(\sum _{i}x_{i}\right)}(1-p)^{\left(\sum _{i}(1-x_{i})\right)}.$

Since $f(x_{1},x_{2},\ldots ,x_{n}|p)$ maximizes also when $\ln f(x_{1},x_{2},\ldots ,x_{n}|p)$ is maximum, we can take a logarithm of the above term and get

$\ln f(x_{1},x_{2},\ldots ,x_{n}|p)=\left(\sum _{i}x_{i}\right)\ln p+\left(\sum _{i}(1-x_{i})\ln(1-p)\right)=\left(\sum _{i}x_{i}\right)\ln p+\left(n-\sum _{i}x_{i}\right)\ln(1-p).$

To find its maximum, we differentiate the term and get

${\frac {d}{dp}}\ln f(x_{1},x_{2},\ldots ,x_{n}|p)={\frac {\sum _{i}x_{i}}{p}}-{\frac {n-\sum _{i}x_{i}}{1-p}}.$

Setting ${\frac {d}{d{\hat {p}}}}\ln f(x_{1},x_{2},\ldots ,x_{n}|{\hat {p}})=0$, we solve the equation and get that

${\frac {\sum _{i}x_{i}}{\hat {p}}}={\frac {n-\sum _{i}x_{i}}{1-{\hat {p}}}},$

implying that

${\hat {p}}={\frac {\sum _{i}x_{i}}{n}},$

which is the maximum likelihood estimator for $p$.

EX3: maximum likelihood estimator for the Poisson mean

To be added...

Bayes estimators

Let the observed data be $x_{1},x_{2},\ldots ,x_{n}$. We fist compute the conditional probability

$f(p|x_{1},x_{2},\ldots ,x_{n})={\frac {f(x_{1},x_{2},\ldots ,x_{n},p)}{f(x_{1},x_{2},\ldots ,x_{n})}}={\frac {f(x_{1},x_{2},\ldots ,x_{n}|p)f(p)}{f(x_{1},x_{2},\ldots ,x_{n})}}.$

With that distribution function, the Bayes estimator is

$E[p|X_{1}=x_{1},X_{2}=x_{2},\ldots ,X_{n}=x_{n}].$