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.

เนื้อหา

1 Method of moments estimators
2 Maximum likelihood estimators
- 2.1 Examples
3 Bayes estimators

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 $Wikimedia Error * { margin: 0; padding: 0; } body { background: #fff; font: 15px/1.6 sans-serif; color: #333; } .content { margin: 7% auto 0; padding: 2em 1em 1em; max-width: 640px; display: flex; flex-direction: row; flex-wrap: wrap; } .footer { clear: both; margin-top: 14%; border-top: 1px solid #e5e5e5; background: #f9f9f9; padding: 2em 0; font-size: 0.8em; text-align: center; } img { margin: 0 2em 2em 0; } a img { border: 0; } h1 { margin-top: 1em; font-size: 1.2em; } .content-text { flex: 1; } p { margin: 0.7em 0 1em 0; } a { color: #0645ad; text-decoration: none; } a:hover { text-decoration: underline; } code { font-family: sans-serif; } summary { font-weight: bold; cursor: pointer; } details[open] { background: #970302; color: #dfdedd; } .text-muted { color: #777; } @media (prefers-color-scheme: dark) { a { color: #9e9eff; } body { background: transparent; color: #ddd; } .footer { border-top: 1px solid #444; background: #060606; } #logo { filter: invert(1) hue-rotate(180deg); } .text-muted { color: #888; } }$

Error

Too many requests (f061ab2)

k=1,\ldots ,r

. Our estimate

{\hat {\theta }}

is thus

$Wikimedia Error * { margin: 0; padding: 0; } body { background: #fff; font: 15px/1.6 sans-serif; color: #333; } .content { margin: 7% auto 0; padding: 2em 1em 1em; max-width: 640px; display: flex; flex-direction: row; flex-wrap: wrap; } .footer { clear: both; margin-top: 14%; border-top: 1px solid #e5e5e5; background: #f9f9f9; padding: 2em 0; font-size: 0.8em; text-align: center; } img { margin: 0 2em 2em 0; } a img { border: 0; } h1 { margin-top: 1em; font-size: 1.2em; } .content-text { flex: 1; } p { margin: 0.7em 0 1em 0; } a { color: #0645ad; text-decoration: none; } a:hover { text-decoration: underline; } code { font-family: sans-serif; } summary { font-weight: bold; cursor: pointer; } details[open] { background: #970302; color: #dfdedd; } .text-muted { color: #777; } @media (prefers-color-scheme: dark) { a { color: #9e9eff; } body { background: transparent; color: #ddd; } .footer { border-top: 1px solid #444; background: #060606; } #logo { filter: invert(1) hue-rotate(180deg); } .text-muted { color: #888; } }$

Error

Too many requests (f061ab2)

{\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 $Wikimedia Error * { margin: 0; padding: 0; } body { background: #fff; font: 15px/1.6 sans-serif; color: #333; } .content { margin: 7% auto 0; padding: 2em 1em 1em; max-width: 640px; display: flex; flex-direction: row; flex-wrap: wrap; } .footer { clear: both; margin-top: 14%; border-top: 1px solid #e5e5e5; background: #f9f9f9; padding: 2em 0; font-size: 0.8em; text-align: center; } img { margin: 0 2em 2em 0; } a img { border: 0; } h1 { margin-top: 1em; font-size: 1.2em; } .content-text { flex: 1; } p { margin: 0.7em 0 1em 0; } a { color: #0645ad; text-decoration: none; } a:hover { text-decoration: underline; } code { font-family: sans-serif; } summary { font-weight: bold; cursor: pointer; } details[open] { background: #970302; color: #dfdedd; } .text-muted { color: #777; } @media (prefers-color-scheme: dark) { a { color: #9e9eff; } body { background: transparent; color: #ddd; } .footer { border-top: 1px solid #444; background: #060606; } #logo { filter: invert(1) hue-rotate(180deg); } .text-muted { color: #888; } }$

Error

Too many requests (f061ab2)

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

$Wikimedia Error * { margin: 0; padding: 0; } body { background: #fff; font: 15px/1.6 sans-serif; color: #333; } .content { margin: 7% auto 0; padding: 2em 1em 1em; max-width: 640px; display: flex; flex-direction: row; flex-wrap: wrap; } .footer { clear: both; margin-top: 14%; border-top: 1px solid #e5e5e5; background: #f9f9f9; padding: 2em 0; font-size: 0.8em; text-align: center; } img { margin: 0 2em 2em 0; } a img { border: 0; } h1 { margin-top: 1em; font-size: 1.2em; } .content-text { flex: 1; } p { margin: 0.7em 0 1em 0; } a { color: #0645ad; text-decoration: none; } a:hover { text-decoration: underline; } code { font-family: sans-serif; } summary { font-weight: bold; cursor: pointer; } details[open] { background: #970302; color: #dfdedd; } .text-muted { color: #777; } @media (prefers-color-scheme: dark) { a { color: #9e9eff; } body { background: transparent; color: #ddd; } .footer { border-top: 1px solid #444; background: #060606; } #logo { filter: invert(1) hue-rotate(180deg); } .text-muted { color: #888; } }$

Error

Too many requests (f061ab2)

{\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.,

$Wikimedia Error * { margin: 0; padding: 0; } body { background: #fff; font: 15px/1.6 sans-serif; color: #333; } .content { margin: 7% auto 0; padding: 2em 1em 1em; max-width: 640px; display: flex; flex-direction: row; flex-wrap: wrap; } .footer { clear: both; margin-top: 14%; border-top: 1px solid #e5e5e5; background: #f9f9f9; padding: 2em 0; font-size: 0.8em; text-align: center; } img { margin: 0 2em 2em 0; } a img { border: 0; } h1 { margin-top: 1em; font-size: 1.2em; } .content-text { flex: 1; } p { margin: 0.7em 0 1em 0; } a { color: #0645ad; text-decoration: none; } a:hover { text-decoration: underline; } code { font-family: sans-serif; } summary { font-weight: bold; cursor: pointer; } details[open] { background: #970302; color: #dfdedd; } .text-muted { color: #777; } @media (prefers-color-scheme: dark) { a { color: #9e9eff; } body { background: transparent; color: #ddd; } .footer { border-top: 1px solid #444; background: #060606; } #logo { filter: invert(1) hue-rotate(180deg); } .text-muted { color: #888; } }$

Error

Too many requests (f061ab2)

{\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

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Error

Too many requests (f061ab2)

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 $Wikimedia Error * { margin: 0; padding: 0; } body { background: #fff; font: 15px/1.6 sans-serif; color: #333; } .content { margin: 7% auto 0; padding: 2em 1em 1em; max-width: 640px; display: flex; flex-direction: row; flex-wrap: wrap; } .footer { clear: both; margin-top: 14%; border-top: 1px solid #e5e5e5; background: #f9f9f9; padding: 2em 0; font-size: 0.8em; text-align: center; } img { margin: 0 2em 2em 0; } a img { border: 0; } h1 { margin-top: 1em; font-size: 1.2em; } .content-text { flex: 1; } p { margin: 0.7em 0 1em 0; } a { color: #0645ad; text-decoration: none; } a:hover { text-decoration: underline; } code { font-family: sans-serif; } summary { font-weight: bold; cursor: pointer; } details[open] { background: #970302; color: #dfdedd; } .text-muted { color: #777; } @media (prefers-color-scheme: dark) { a { color: #9e9eff; } body { background: transparent; color: #ddd; } .footer { border-top: 1px solid #444; background: #060606; } #logo { filter: invert(1) hue-rotate(180deg); } .text-muted { color: #888; } }$

Error

Too many requests (f061ab2)

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 $Wikimedia Error * { margin: 0; padding: 0; } body { background: #fff; font: 15px/1.6 sans-serif; color: #333; } .content { margin: 7% auto 0; padding: 2em 1em 1em; max-width: 640px; display: flex; flex-direction: row; flex-wrap: wrap; } .footer { clear: both; margin-top: 14%; border-top: 1px solid #e5e5e5; background: #f9f9f9; padding: 2em 0; font-size: 0.8em; text-align: center; } img { margin: 0 2em 2em 0; } a img { border: 0; } h1 { margin-top: 1em; font-size: 1.2em; } .content-text { flex: 1; } p { margin: 0.7em 0 1em 0; } a { color: #0645ad; text-decoration: none; } a:hover { text-decoration: underline; } code { font-family: sans-serif; } summary { font-weight: bold; cursor: pointer; } details[open] { background: #970302; color: #dfdedd; } .text-muted { color: #777; } @media (prefers-color-scheme: dark) { a { color: #9e9eff; } body { background: transparent; color: #ddd; } .footer { border-top: 1px solid #444; background: #060606; } #logo { filter: invert(1) hue-rotate(180deg); } .text-muted { color: #888; } }$

Error

Too many requests (f061ab2)

f(x_{1},x_{2},\ldots ,x_{n}|p)

maximizes also when

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is maximum, we can take a logarithm of the above term and get

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Error

Too many requests (f061ab2)

\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

$Wikimedia Error * { margin: 0; padding: 0; } body { background: #fff; font: 15px/1.6 sans-serif; color: #333; } .content { margin: 7% auto 0; padding: 2em 1em 1em; max-width: 640px; display: flex; flex-direction: row; flex-wrap: wrap; } .footer { clear: both; margin-top: 14%; border-top: 1px solid #e5e5e5; background: #f9f9f9; padding: 2em 0; font-size: 0.8em; text-align: center; } img { margin: 0 2em 2em 0; } a img { border: 0; } h1 { margin-top: 1em; font-size: 1.2em; } .content-text { flex: 1; } p { margin: 0.7em 0 1em 0; } a { color: #0645ad; text-decoration: none; } a:hover { text-decoration: underline; } code { font-family: sans-serif; } summary { font-weight: bold; cursor: pointer; } details[open] { background: #970302; color: #dfdedd; } .text-muted { color: #777; } @media (prefers-color-scheme: dark) { a { color: #9e9eff; } body { background: transparent; color: #ddd; } .footer { border-top: 1px solid #444; background: #060606; } #logo { filter: invert(1) hue-rotate(180deg); } .text-muted { color: #888; } }$

Error

Too many requests (f061ab2)

{\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 $Wikimedia Error * { margin: 0; padding: 0; } body { background: #fff; font: 15px/1.6 sans-serif; color: #333; } .content { margin: 7% auto 0; padding: 2em 1em 1em; max-width: 640px; display: flex; flex-direction: row; flex-wrap: wrap; } .footer { clear: both; margin-top: 14%; border-top: 1px solid #e5e5e5; background: #f9f9f9; padding: 2em 0; font-size: 0.8em; text-align: center; } img { margin: 0 2em 2em 0; } a img { border: 0; } h1 { margin-top: 1em; font-size: 1.2em; } .content-text { flex: 1; } p { margin: 0.7em 0 1em 0; } a { color: #0645ad; text-decoration: none; } a:hover { text-decoration: underline; } code { font-family: sans-serif; } summary { font-weight: bold; cursor: pointer; } details[open] { background: #970302; color: #dfdedd; } .text-muted { color: #777; } @media (prefers-color-scheme: dark) { a { color: #9e9eff; } body { background: transparent; color: #ddd; } .footer { border-top: 1px solid #444; background: #060606; } #logo { filter: invert(1) hue-rotate(180deg); } .text-muted { color: #888; } }$

Error

Too many requests (f061ab2)

{\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

$Wikimedia Error * { margin: 0; padding: 0; } body { background: #fff; font: 15px/1.6 sans-serif; color: #333; } .content { margin: 7% auto 0; padding: 2em 1em 1em; max-width: 640px; display: flex; flex-direction: row; flex-wrap: wrap; } .footer { clear: both; margin-top: 14%; border-top: 1px solid #e5e5e5; background: #f9f9f9; padding: 2em 0; font-size: 0.8em; text-align: center; } img { margin: 0 2em 2em 0; } a img { border: 0; } h1 { margin-top: 1em; font-size: 1.2em; } .content-text { flex: 1; } p { margin: 0.7em 0 1em 0; } a { color: #0645ad; text-decoration: none; } a:hover { text-decoration: underline; } code { font-family: sans-serif; } summary { font-weight: bold; cursor: pointer; } details[open] { background: #970302; color: #dfdedd; } .text-muted { color: #777; } @media (prefers-color-scheme: dark) { a { color: #9e9eff; } body { background: transparent; color: #ddd; } .footer { border-top: 1px solid #444; background: #060606; } #logo { filter: invert(1) hue-rotate(180deg); } .text-muted { color: #888; } }$

Error

Too many requests (f061ab2)

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

which is the maximum likelihood estimator for .

EX3: maximum likelihood estimator for the Poisson mean

To be added...

Bayes estimators

Let the observed data be $Wikimedia Error * { margin: 0; padding: 0; } body { background: #fff; font: 15px/1.6 sans-serif; color: #333; } .content { margin: 7% auto 0; padding: 2em 1em 1em; max-width: 640px; display: flex; flex-direction: row; flex-wrap: wrap; } .footer { clear: both; margin-top: 14%; border-top: 1px solid #e5e5e5; background: #f9f9f9; padding: 2em 0; font-size: 0.8em; text-align: center; } img { margin: 0 2em 2em 0; } a img { border: 0; } h1 { margin-top: 1em; font-size: 1.2em; } .content-text { flex: 1; } p { margin: 0.7em 0 1em 0; } a { color: #0645ad; text-decoration: none; } a:hover { text-decoration: underline; } code { font-family: sans-serif; } summary { font-weight: bold; cursor: pointer; } details[open] { background: #970302; color: #dfdedd; } .text-muted { color: #777; } @media (prefers-color-scheme: dark) { a { color: #9e9eff; } body { background: transparent; color: #ddd; } .footer { border-top: 1px solid #444; background: #060606; } #logo { filter: invert(1) hue-rotate(180deg); } .text-muted { color: #888; } }$

Error

Too many requests (f061ab2)

x_{1},x_{2},\ldots ,x_{n}

. We fist compute the conditional probability

$Wikimedia Error * { margin: 0; padding: 0; } body { background: #fff; font: 15px/1.6 sans-serif; color: #333; } .content { margin: 7% auto 0; padding: 2em 1em 1em; max-width: 640px; display: flex; flex-direction: row; flex-wrap: wrap; } .footer { clear: both; margin-top: 14%; border-top: 1px solid #e5e5e5; background: #f9f9f9; padding: 2em 0; font-size: 0.8em; text-align: center; } img { margin: 0 2em 2em 0; } a img { border: 0; } h1 { margin-top: 1em; font-size: 1.2em; } .content-text { flex: 1; } p { margin: 0.7em 0 1em 0; } a { color: #0645ad; text-decoration: none; } a:hover { text-decoration: underline; } code { font-family: sans-serif; } summary { font-weight: bold; cursor: pointer; } details[open] { background: #970302; color: #dfdedd; } .text-muted { color: #777; } @media (prefers-color-scheme: dark) { a { color: #9e9eff; } body { background: transparent; color: #ddd; } .footer { border-top: 1px solid #444; background: #060606; } #logo { filter: invert(1) hue-rotate(180deg); } .text-muted { color: #888; } }$

Error

Too many requests (f061ab2)

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

Probstat/notes/parameter estimation

เนื้อหา

Method of moments estimators

Error

Error

Error

Error

Maximum likelihood estimators

Error

Examples

Error

Error

Error

Error

Error

Error

Error

Bayes estimators

Error

Error

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