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## 4. Bayes' Estimators

### Basic Theory

#### The General Method

Suppose again that we have an observable random variable $$\bs{X}$$ for an experiment, that takes values in a set $$S$$. Suppose also that distribution of $$\bs{X}$$ depends on a parameter $$\theta$$ taking values in a parameter space $$T$$. Of course, our data variable $$\bs{X}$$ is almost always vector-valued, so that typically $$S \subseteq \R^n$$ for some $$n \in \N_+$$. The parameter $$\theta$$ may also be vector-valued, so that typically $$T \subseteq \R^k$$ for some $$k \in \N_+$$.

In Bayesian analysis, named for the famous Thomas Bayes, we model the deterministic, but unknown parameter $$\theta$$ with a random variable $$\Theta$$ that has a specified distribution on the parameter space $$T$$. This distribution is called the prior distribution of $$\Theta$$ and is intended to reflect our knowledge of the parameter $$\theta$$, before we gather data. After observing $$\bs{x} \in S$$, we then use Bayes' theorem, to compute the conditional distribution of $$\Theta$$ given $$\bs X = \bs x$$. This distribution is called the posterior distribution of $$\Theta$$, and is an updated distribution, given the information in the data. Let's go through the derivation of this theorem again. First let $$f(\cdot \mid \theta)$$ denote the probability density function of $$\bs{X}$$ for a given $$\theta \in T$$, so now we can interpret $$f( \cdot \mid \theta)$$ as the conditional probability density function of $$\bs X$$ given $$\Theta = \theta$$. Next let $$h$$ denote the probability density function of the prior distribution of $$\Theta$$. Recall that he joint probability density function of $$(\bs{X}, \Theta)$$ is the mapping on $$S \times T$$ given by $(\bs{x}, \theta) \mapsto h(\theta) f(\bs{x} \mid \theta)$ Next recall that the (marginal) probability density function $$f$$ of $$\bs{X}$$ is given by $f(\bs{x}) = \sum_{\theta \in T} h(\theta) f(\bs{x} | \theta), \quad \bs{x} \in S$ if $$\Theta$$ has a discrete distribution on $$T$$, or $f(\bs{x}) = \int_T h(\theta) f(\bs{x} \mid \theta) \, d\theta, \quad \bs{x} \in S$ if $$\Theta$$ has a continuous distribution on $$T \subseteq \R^k$$. Finally, the conditional probability density function of $$\Theta$$ given $$\bs{X} = \bs{x} \in S$$ is $h(\theta \mid \bs{x}) = \frac{h(\theta) f(\bs{x} \mid \theta)}{f(\bs{x})}, \quad \theta \in T$ Note that $$f(\bs{x})$$ is simply the normalizing constant for the function $$\theta \mapsto h(\theta) f(\bs{x} \mid \theta)$$. It may not be necessary to explicitly compute $$f(\bs{x})$$, if one can recognize the functional form of $$\theta \mapsto h(\theta) f(\bs{x} \mid \theta)$$ as that of a known distribution. This will indeed be the case in several of the examples explored below.

#### Random Samples

Of course, an important and essential special case occurs when $$\bs{X} = (X_1, X_2, \ldots, X_n)$$ is a random sample of size $$n$$ from the distribution of a basic variable $$X$$. Specifically, suppose that $$X$$ takes values in a set $$R$$ and has probability density function $$g(\cdot \mid \theta)$$ for a given $$\theta \in T$$. In this case, $$S = R^n$$ and the probability density function $$f(\cdot \mid \theta)$$ of $$\bs{X}$$ given $$\theta$$ is $f(x_1, x_2, \ldots, x_n \mid \theta) = g(x_1 \mid \theta) g(x_2 \mid \theta) \cdots g(x_n \mid \theta), \quad (x_1, x_2, \ldots, x_n) \in S$

#### Real Parameters

Suppose that $$\theta$$ is a real-valued parameter, so that $$T \subseteq \R$$. Here is our main definition.

The conditional expected value $$\E(\Theta \mid \bs{X})$$ is the Bayesian estimator of $$\theta$$.

1. If $$\Theta$$ has a discrete distribution on $$T$$ then $\E(\Theta \mid \bs X = \bs x) = \sum_{\theta \in T} \theta h(\theta \mid \bs x), \quad \bs x \in S$
2. If $$\Theta$$ has a continuous distribution on $$T$$ then $\E(\Theta \mid \bs X = \bs x) = \int_T \theta h(\theta \mid \bs x) d\theta, \quad \bs x \in S$

Recall that $$\E(\Theta \mid \bs{X})$$ is a function of $$\bs{X}$$ and, among all functions of $$\bs{X}$$, is closest to $$\Theta$$ in the mean square sense. Of course, once we collect the data and observe $$\bs{X} = \bs{x}$$, the Bayesian estimate of $$\theta$$ is $$\E(\Theta \mid \bs{X} = \bs{x})$$. As always, the term estimator refers to a random variable, before the data are collected, and the term estimate refers to an observed value of the random variable after the data are collected. The definitions of bias and mean square error are as before, but now conditioned on $$\Theta = \theta$$ for a given $$\theta \in T$$.

Suppose that $$U$$ is the Bayes estimator of $$\theta$$.

1. The bias of $$U$$ is $$\bias(U \mid \theta) = \E(U - \theta \mid \theta)$$ for $$\theta \in T$$.
2. The mean square error of $$U$$ is $$\mse(U \mid \theta) = \E[(U - \theta)^2 \mid \theta]$$ for $$\theta \in T$$.

As before, $$\bias(U \mid \theta) = \E(U \mid \theta) - \theta$$ and $$\mse(U \mid \theta) = \var(U \mid \theta) + \bias^2(U \mid \theta)$$. Suppose now that we observe the random variables $$(X_1, X_2, X_3, \ldots)$$ sequentially, and we compute the Bayes estimator $$U_n$$ of $$\theta$$ based on $$(X_1, X_2, \ldots, X_n)$$ for each $$n \in \N_+$$. Again, the most common case is when we are sampling from a distribution, so that the sequence is independent and identically distributed. We have the natural asymptotic properties that we have seen before.

Let $$\bs U = (U_n: n \in \N_+)$$ be the sequence of Bayes estimators of $$\theta$$ as above.

1. $$\bs U$$ is asymptotically unbiased if $$\bias(U_n \mid \theta) \to 0$$ as $$n \to \infty$$ for each $$\theta \in T$$.
2. $$\bs U$$ is mean-square consistent if $$\mse(U_n \mid \theta) \to 0$$ as $$n \to \infty$$ for each $$\theta \in T$$.

Often we cannot construct unbiased Bayesian estimators, but we do hope that our estimators are at least asymptotically unbiased and consistent. It turns out that the sequence of Bayesian estimators $$\bs U$$ is a martingale. The theory of martingales provides some powerful tools for studying these estimators.

#### Conjugate Families

Often, the prior distribution of $$\Theta$$ is itself a member of a parametric family, with the parameters specified to reflect our prior knowledge of $$\theta$$. In many important special cases, the parametric family can be chosen so that the posterior distribution of $$\Theta$$ given $$\bs{X} = \bs{x}$$ belongs to the same family for each $$\bs x \in S$$. In such a case, the family of distributions of $$\Theta$$ is said to be conjugate to the family of distributions of $$\bs{X}$$. Conjugate families are nice from a computational point of view, since we can often compute the posterior distribution through a simple formula involving the parameters of the family, without having to use Bayes' theorem directly. Similarly, in the case that the parameter is real valued, we can often compute the Bayesian estimator through a simple formula involving the parameters of the conjugate family.

### Special Distributions

#### The Bernoulli Distribution

Suppose that $$\bs X = (X_1, X_2, \ldots)$$ is sequence of independent variables, each having the Bernoulli distribution with unknown success parameter $$p \in (0, 1)$$. In short, $$\bs X$$ is a sequence of Bernoulli trials, given $$p$$. In the usual language of reliability, $$X_i = 1$$ means success on trial $$i$$ and $$X_i = 0$$ means failure on trial $$i$$. Recall that given $$p$$, the Bernoulli distribution has probability density function $g(x \mid p) = p^x (1 - p)^{1-x}, \quad x \in \{0, 1\}$ Note that the number of successes in the first $$n$$ trials is $$Y_n = \sum_{i=1}^n X_i$$. Given $$p$$, random variable $$Y_n$$ has the binomial distribution with parameters $$n$$ and $$p$$.

Suppose now that we model $$p$$ with a random variable $$P$$ that has a prior beta distribution with left parameter $$a \in (0, \infty)$$ and right parameter $$b \in (0, \infty)$$, where $$a$$ and $$b$$ are chosen to reflect our initial information about $$p$$. So $$P$$ has probability density function $h(p) = \frac{1}{B(a, b)} p^{a-1} (1 - p)^{b-1}, \quad p \in (0, 1)$ For example, if we know nothing about $$p$$, we might let $$a = b = 1$$, so that the prior distribution is uniform on the parameter space $$(0, 1)$$. On the other hand, if we believe that $$p$$ is about $$\frac{2}{3}$$, we might let $$a = 4$$ and $$b = 2$$, so that the prior distribution is unimodal, with mean $$\frac{2}{3}$$. As a random process, the sequence $$\bs X$$ with $$p$$ randomized by $$P$$, is known as the beta-Bernoulli process, and is very interesting on its own, outside of the context of Bayesian estimation.

For $$n \in \N_+$$, the posterior distribution of $$P$$ given $$\bs{X}_n = (X_1, X_2, \ldots, X_n)$$ is beta with left parameter $$a + Y_n$$ and right parameter $$b + (n - Y_n)$$.

Proof:

Fix $$n \in \N_+$$. Let $$\bs x = (x_1, x_2, \ldots, x_n) \in \{0, 1\}^n$$, and let $$y = \sum_{i=1}^n x_i$$. Then $f(\bs x \mid p) = g(x_1 \mid p) g(x_2 \mid p) \cdots g(x_n \mid p) = p^y (1 - p)^{n-y}$ Hence $h(p) f(\bs x \mid p) = \frac{1}{B(a, b)} p^{a-1} (1 - p)^{b-1} p^y (1 - p)^{n-y} = \frac{1}{B(a, b)}p^{a + y - 1} (1 - p)^{b + n - y - 1}, \quad p \in (0, 1)$ As a function of $$p$$ this expression is proportional to the beta PDF with parameters $$a + y$$, $$b + n - y$$. Note that it's not necessary to compute the normalizing factor $$f(\bs x)$$.

Thus, the beta distribution is conjugate to the Bernoulli distribution. Note also that the posterior distribution depends on the data vector $$\bs{X}_n$$ only through the number of successes $$Y_n$$. This is true because $$Y_n$$ is a sufficient statistic for $$p$$. In particular, note that the left beta parameter is increased by the number of successes $$Y_n$$ and the right beta parameter is increased by the number of failures $$n - Y_n$$.

The Bayesian estimator of $$p$$ given $$\bs{X}_n$$ is $U_n = \frac{a + Y_n}{a + b + n}$

Proof:

Recall that the mean of the beta distribution is the left parameter divided by the sum of the parameters, so this result follows from the previous result.

In the beta coin experiment, set $$n = 20$$ and $$p = 0.3$$, and set $$a = 4$$ and $$b = 2$$. Run the simulation 100 times and note the estimate of $$p$$ and the shape and location of the posterior probability density function of $$p$$ on each run.

Next let's compute the bias and mean-square error functions.

For $$n \in \N_+$$, $\bias(U_n \mid p) = \frac{a(1 - p) - b p}{a + b + n}, \quad p \in (0, 1)$ The sequence $$\bs U = (U_n: n \in \N_+)$$ is asymptotically unbiased.

Proof:

Given $$p$$, $$Y_n$$ has the binomial distribution with parameters $$n$$ and $$p$$ so $$E(Y_n \mid p) = n p$$. Hence $\bias(U_n \mid p) = \E(U_n \mid p) - p = \frac{a + n p}{a + b + n} - p$ Simplifying gives the formula above. Clearly $$\bias(U_n \mid p) \to 0$$ as $$n \to \infty$$.

Note also that we cannot choose $$a$$ and $$b$$ to make $$U_n$$ unbiased, since such a choice would involve the true value of $$p$$, which we do not know.

In the beta coin experiment, vary the parameters and note the change in the bias. Now set $$n = 20$$ and $$p = 0.8$$, and set $$a = 2$$ and $$b = 6$$. Run the simulation 1000 times. Note the estimate of $$p$$ and the shape and location of the posterior probability density function of $$p$$ on each update. Compare the empirical bias to the true bias.

For $$n \in \N_+$$, $\mse(U_n \mid p) = \frac{p [n - 2 \, a (a + b)] + p^2[(a + b)^2 - n] + a^2}{(a + b + n)^2}, \quad p \in (0, 1)$ The sequence $$(U_n: n \in \N_+)$$ is mean-square consistent.

Proof:

Once again, given $$p$$, $$Y_n$$ has the binomail distribution with parameters $$n$$ and $$p$$ so $\var(U_n \mid p) = \frac{n p (1 - p)}{(a + b + n)^2}$ Hence $\mse(U_n \mid p) = \frac{n p (1 - p)}{(a + b + n)^2} + \left[\frac{a (1 - p) - b p}{a + b + n}\right]^2$ Simplifying gives the result. Clearly $$\mse(U_n \mid p) \to 0$$ as $$n \to \infty$$.

In the beta coin experiment, vary the parameters and note the change in the mean square error. Now set $$n = 10$$ and $$p = 0.7$$, and set $$a = b = 1$$. Run the simulation 1000 times. Note the estimate of $$p$$ and the shape and location of the posterior probability density function of $$p$$ on each update. Compare the empirical mean square error to the true mean square error.

Interestingly, we can choose $$a$$ and $$b$$ so that $$U$$ has mean square error that is independent of the unknown parameter $$p$$:

Let $$n \in \N_+$$ and let $$a = b = \sqrt{n} / 2$$. Then

$\mse(U_n \mid p) = \frac{n}{4 \left(n + \sqrt{n}\right)^2}, \quad p \in (0, 1)$

In the beta coin experiment, set $$n = 36$$ and $$a = b = 3$$. Vary $$p$$ and note that the mean square error does not change. Now set $$p = 0.8$$ and run the simulation 1000 times. Note the estimate of $$p$$ and the shape and location of the posterior probability density function on each update. Compare the empirical bias and mean square error to the true values.

Recall that the method of moments estimator and the maximum likelihood estimator of $$p$$ (on the interval $$(0, 1)$$) is the sample mean (the proportion of successes):

$M_n = \frac{Y}{n} = \frac{1}{n} \sum_{i=1}^n X_i$

This estimator has mean square error $$\mse(M_n \mid p) = \frac{1}{n} p (1 - p)$$. The two estimators are closely related since $U = \frac{a / n + M}{a / n + b / n + 1}$

#### Another Bernoulli Distribution

Bayesian estimation, like other forms of parametric estimation, depends critically on the parameter space. Suppose again that $$(X_1, X_2, \ldots)$$ is a sequence of Bernoulli trials, given the unknown success parameter $$p$$, but suppose now that the parameter space is $$\left\{\frac{1}{2}, 1\right\}$$. This setup corresponds to the tossing of a coin that is either fair or two-headed, but we don't know which. We model $$p$$ with a random variable $$P$$ that has the prior probability density function $$h$$ given by $$h(1) = a$$, $$h\left(\frac{1}{2}\right) = 1 - a$$, where $$a \in (0, 1)$$ is chosen to reflect our prior knowledge of the probability that the coin is two-headed. If we are completely ignorant, we might let $$a = \frac{1}{2}$$. If with think the coin is more likely to be two-headed, we might let $$a = \frac{3}{4}$$. Again let $$Y_n = \sum_{i=1}^n X_i$$ for $$n \in \N_+$$.

The posterior distribution of $$P$$ given $$\bs{X}_n = (X_1, X_2, \ldots, X_n)$$ is

1. $$h(1 \mid \bs{X}_n) = \frac{2^n a}{2^n a + (1 - a)}$$ if $$Y_n = n$$ and $$h(1 \mid \bs{X}_n) = 0$$ if $$Y_n \lt n$$
2. $$h\left(\frac{1}{2} \mid \bs{X}_n\right) = \frac{1 - a}{2^n a + (1 - a)}$$ if $$Y_n = n$$ and $$h\left(\frac{1}{2} \mid \bs{X}_n\right) = 1$$ if $$Y_n \lt n$$
Proof:

Fix $$n \in \N_+$$. Let $$\bs x = (x_1, x_2, \ldots, x_n) \in \{0, 1\}^n$$, and let $$y = \sum_{i=1}^n x_i$$. As before, $f(\bs x \mid p) = p^y (1 - p)^{n-y}$ We adopt the usual conventions (which gives the correct mathematics) that $$0^k = 0$$ if $$k \in \N_+$$ but $$0^0 = 1$$. So from Bayes' theorem, \begin{align} h(1 \mid \bs x) & = \frac{h(1) f(\bs x \mid 1)}{h(1/2) f(\bs x \mid 1/2) + h(1) f(\bs x \mid 1)} \\ & = \frac{a 1^y 0^{n-y}}{(1 - a)(1/2)^n + a 1^y 0^{n-y}} \end{align} So if $$y \lt n$$ then $$h(1 \mid \bs x) = 0$$ while if $$y = n$$ $h(1 \mid \bs x) = \frac{a}{(1 - a)(1/2)^n + a}$ Of course, $$h\left(\frac{1}{2} \mid \bs x\right) = 1 - h(1 \mid \bs x)$$. The results now follow after a bit of algebra.

Now let $p_n = \frac{2^{n+1} a + (1 - a)}{2^{n+1} a + 2 (1 - a)}$

The Bayes' estimator of $$p$$ given $$\bs{X}_n$$ the statistic $$U_n$$ defined by

1. $$U_n = p_n$$ if $$Y_n = n$$
2. $$U_n = \frac{1}{2}$$ if $$Y_n \lt n$$
Proof:

By definition, the Bayes' estimator is $$U_n = E(P \mid \bs{X}_n)$$. From the previous result, if $$Y_n = n$$ then $U_n = 1 \cdot \frac{2^n a}{2^n a + (1 - a)} + \frac{1}{2} \cdot \frac{1 - a}{2^n a + (1 - a)}$ which simplifies to $$p_n$$. If $$Y_n \lt n$$ then $$U = 1 \cdot 0 + \frac{1}{2} \cdot 1 = \frac{1}{2}$$.

If we observe $$Y_n \lt n$$ then $$U_n$$ gives the correct answer $$\frac{1}{2}$$. This certainly makes sense since we know that we do not have the two-headed coin. On the other hand, if we observe $$Y_n = n$$ then we are not certain which coin we have, and the Bayesian estimate $$p_n$$ is not even in the parameter space! But note that $$p_n \to 1$$ as $$n \to \infty$$ exponentially fast. Next let's compute the bias and mean-square error for a given $$p \in \left\{\frac{1}{2}, 1\right\}$$.

For $$n \in \N_+$$,

1. $$\bias(U_n \mid 1) = p_n - 1$$
2. $$\bias\left(U_n \mid \frac{1}{2}\right) = \left(\frac{1}{2}\right)^n \left(p_n - \frac{1}{2}\right)$$

The sequence of estimators $$(U_n: n \in \N_+)$$ is asymptotically unbiased.

Proof:

By definition, $$\bias(U_n \mid p) = E(U - p \mid p)$$. Hence from the previous result, \begin{align} \bias(U \mid p) & = (p_n - p) \P(Y = n \mid p) + \left(\frac{1}{2} - p\right) \P(Y \lt n \mid p) \\ & = (p_n - p) p^n + \left(\frac{1}{2} - p\right) (1 - p^n) \end{align} Substituting $$p = 1$$ and $$p = \frac{1}{2}$$ gives the results. In both cases, $$\bias(U_n \mid p) \to 0$$ as $$n \to \infty$$ since $$p_n \to 1$$ and $$\left(\frac{1}{2}\right)^n \to 0$$ as $$n \to \infty$$.

If $$p = 1$$, the estimator $$U_n$$ is negatively biased; we noted this earlier. If $$p = \frac{1}{2}$$, then $$U_n$$ is positively biased for sufficiently large $$n$$ (depending on $$a$$).

For $$n \in \N_+$$,

1. $$\mse(U_n \mid 1) = (p_n - 1)^2$$
2. $$\mse\left(U_n \mid \frac{1}{2}\right) = \left(\frac{1}{2}\right)^n \left(p_n - \frac{1}{2}\right)^2$$

The sequence of estimators $$\bs U = (U_n: n \in \N_+)$$ is mean-square consistent.

Proof:

By definition, $$\mse(U_n \mid p) = \E[(U_n - p)^2 \mid p]$$. Hence \begin{align} \mse(U_n \mid p) & = (p_n - p)^2 \P(Y_n = n \mid p) + \left(\frac{1}{2} - p\right)^2 \P(Y_n \lt n \mid p) \\ & = (p_n - p)^2 p^n + \left(\frac{1}{2} - p\right)^2 (1 - p^n) \end{align} Substituting $$p = 1$$ and $$p = \frac{1}{2}$$ gives the results. In both cases, $$\mse(U_n \mid p) \to 0$$ as $$n \to \infty$$ since $$p_n \to 1$$ and $$\left(\frac{1}{2}\right)^n \to 0$$ as $$n \to \infty$$.

#### The Geometric distribution

Suppose that $$\bs{X} = (X_1, X_2, \ldots)$$ is a sequence of independent random variables, each having the geometric distribution on $$\N_+$$ with unknown success parameter $$p \in (0, 1)$$. Recall that these variables can be interpreted as the number of trials between successive successes in a sequence of Bernoulli trials. Given $$p$$, the geometric distribution has probability density function $g(x \mid p) = p (1 - p)^{x-1}, \quad x \in \N_+$ Once again for $$n \in \N_+$$, let $$Y_n = \sum_{i=1}^n X_i$$. In this setting, $$Y_n$$ is the trial number of the $$n$$th success, and given $$p$$, has the negative binomial distribution with parameters $$n$$ and $$p$$.

Suppose now that we model $$p$$ with a random variable $$P$$ having a prior beta distribution with left parameter $$a \in (0, \infty)$$ and right parameter $$b \in (0, \infty)$$. As usual, $$a$$ and $$b$$ are chosen to reflect our prior knowledge of $$p$$.

The posterior distribution of $$P$$ given $$\bs{X}_n = (X_1, X_2, \ldots, X_n)$$ is beta with left parameter $$a + n$$ and right parameter $$b + (Y_n - n)$$.

Proof:

Fix $$n \in \N_+$$. Let $$\bs x = (x_1, x_2, \ldots, x_n) \in \N_+^n$$ and let $$y = \sum_{i=1}^n x_i$$. Then $f(\bs x \mid p) = g(x_1 \mid p) g(x_2 \mid p) \cdots g(x_n \mid p) = p^n (1 - p)^{y - n}$ Hence $h(p) f( \bs x \mid p) = \frac{1}{B(a, b)} p^{a-1} (1 - p)^{b-1} p^n (1 - p)^{y - n} = \frac{1}{B(a, b)} p^{a + n - 1} (1 - p)^{b + y - n - 1}, \quad p \in (0, 1)$ As a function of $$p \in (0, 1)$$ this expression is proportional to the beta PDF with parameters $$a + n$$ and $$b + y - n$$. Note that it's not necessary to compute the normalizing constant $$f(\bs{x})$$.

Thus, the beta distribution is conjugate to the geometric distribution. Moreover, note that in the posterior beta distribution, the left parameter is increased by the number of successes $$n$$ while the right parameter is increased by the number of failures $$Y - n$$, just as in the Bernoulli model. In particular, the posterior left parameter is deterministic and depends on the data only through the sample size $$n$$.

The Bayesian estimator of $$p$$ based on $$\bs{X}_n$$is $V_n = \frac{a + n}{a + b + Y_n}$

Proof:

By definition, the Bayesian estimator is the mean of the posterior distribution. Recall again that the mean of the beta distribution is the left parameter divided by the sum of the parameters, so the result follows from our previous theorem.

Recall that the method of moments estimator of $$p$$, and the maximum likelihood estimator of $$p$$ on the interval $$(0, 1)$$ are both $$1 / M_n = n / Y_n$$. To compare, note that $V_n = \frac{a/n + 1}{a/n + b/n + M_n}$ Since $$V_n$$ is a nonlinear function of $$Y_n$$, it's not easy to compute the bias or mean-square error functions.

#### The Poisson Distribution

Suppose that $$\bs{X} = (X_1, X_2, \ldots)$$ is a sequence of random variable each having the Poisson distribution with unknown parameter $$\lambda \in (0, \infty)$$. Recall that the Poisson distribution is often used to model the number of random points in a region of time or space and is studied in more detail in the chapter on the Poisson Process. The distribution is named for the inimitable Simeon Poisson and given $$\lambda$$, has probability density function $g(x \mid \lambda) = e^{-\lambda} \frac{\lambda^x}{x!}, \quad x \in \N$ Once again, for $$n \in \N_+$$, let $$Y_n = \sum_{i=1}^n X_i$$. Given $$\lambda$$, random variable $$Y_n$$ also has a Poisson distribution, but with parameter $$n \lambda$$.

Suppose now that we model $$\lambda$$ with a random variable $$\Lambda$$ having a prior gamma distribution with shape parameter $$k \in (0, \infty)$$ and rate parameter $$r \in (0, \infty)$$. As usual $$k$$ and $$r$$ are chosen to reflect our prior knowledge of $$\lambda$$. Thus the prior probability density function of $$\Lambda$$ is $h(\lambda) = \frac{r^k}{\Gamma(k)} \lambda^{k-1} e^{-r \lambda}, \quad \lambda \in (0, \infty)$ The scale parameter of the gamma distribution is $$b = 1/r$$, but the formulas will work out nicer if we use the rate parameter.

The posterior distribution of $$\Lambda$$ given $$\bs{X}_n = (X_1, X_2, \ldots, X_n)$$ is gamma with shape parameter $$k + Y_n$$ and rate parameter $$r + n$$.

Proof:

Fix $$n \in \N_+$$. Let $$\bs x = (x_1, x_2, \ldots, x_n) \in \N^n$$ and $$y = \sum_{i=1}^n x_i$$. Then $f(\bs x \mid \lambda) = g(x_1 \mid \lambda) g(x_2 \mid \lambda) \cdots g(x_n \mid \lambda) = e^{-n \lambda} \frac{\lambda^y}{x_1! x_2! \cdots x_n!}$ Hence \begin{align} h(\lambda) f( \bs x \mid \lambda) & = \frac{r^k}{\Gamma(k)} \lambda^{k-1} e^{-r \lambda} e^{-n \lambda} \frac{\lambda^y}{x_1! x_2! \cdots x_n!} \\ & = \frac{r^k}{\Gamma(k) x_1! x_2! \cdots x_n!} e^{-(r + n) \lambda} \lambda^{k + y - 1}, \quad \lambda \in (0, \infty) \end{align} As a function of $$\lambda \in (0, \infty)$$ the last expression is proportional to the gamma PDF with shape parameter $$k + y$$ and rate parameter $$r + n$$. Note again that it's not necessary to compute the normalizing constant $$f(\bs{x})$$.

It follows that the gamma distribution is conjugate to the Poisson distribution. Note that the posterior rate parameter is deterministic and depends on the data only through the sample size $$n$$.

The Bayesian estimator of $$\lambda$$ based on $$\bs{X}_n = (X_1, X_2, \ldots, X_n)$$ is $V_n = \frac{k + Y_n}{r + n}$

Proof:

By definition, the Bayes estimator is the mean of the posterior distribution. Recall that mean of the gamma distribution is the shape parameter divided by the rate parameter.

Since $$V_n$$ is a linear function of $$Y_n$$, and we know the distribution of $$Y_n$$ given $$\lambda \in (0, \infty)$$, we can compute the bias and mean-square error functions.

For $$n \in \N_+$$, $\bias(V_n \mid \lambda) = \frac{k - r \lambda}{r + n}, \quad \lambda \in (0, \infty)$ The sequence of estimators $$\bs V = (V_n: n \in \N_+)$$ is asymptotically unbiased.

Proof:

The computation is simple, since the distribution of $$Y_n$$ given $$\lambda$$ is Poisson with parameter $$n \lambda$$. $\bias(V_n \mid \lambda) = \E(V_n \mid \lambda) - \lambda = \frac{k + n \lambda}{r + n} - \lambda = \frac{k - r \lambda}{r + n}$ Clearly $$\bias(V_n \mid \lambda) \to 0$$ as $$n \to \infty$$.

Note that, as before, we cannot choose $$k$$ and $$r$$ to make $$V_n$$ unbiased, without knowledge of $$\lambda$$.

For $$n \in \N_+$$, $\mse(V_n \mid \lambda) = \frac{n \lambda + (k - r \lambda)^2}{(r + n)^2}, \quad \lambda \in (0, \infty)$ The sequence of estimators $$\bs V = (V_n: n \in \N_+)$$ is mean-square consistent.

Proof:

Again, the computation is easy since the distribution of $$Y_n$$ given $$\lambda$$ is Poisson with parameter $$n \lambda$$. $\mse(V \mid \lambda) = \var(V_n \mid \lambda) + \bias^2(V_n \mid \lambda) = \frac{n \lambda}{(r + n)^2} + \left(\frac{k - r \lambda}{r + n}\right)^2$ Clearly $$\mse(V_n \mid \lambda) \to 0$$ as $$n \to \infty$$.

Recall that the method of moments estimator of $$\lambda$$ and the maximum likelihood estimator of $$\lambda$$ on the interval $$(0, \infty)$$ are both $$M_n = Y_n / n$$. This estimator is unbiased and has mean square error $$\lambda / n$$. To compare, note that $V_n = \frac{k/n + M_n}{r/n + 1}$

#### The Normal Distribution

Suppose that $$\bs X = (X_1, X_2, \ldots)$$ is a sequence of independent random variables, each having the normal distribution with unknown mean $$\mu \in \R$$ but known variance $$\sigma^2 \in (0, \infty)$$. Of course, the normal distribution plays an especially important role in statistics, in part because of the central limit theorem. The normal distribution is widely used to model physical quantities subject to numerous small, random errors. In many statistical applications, the variance of the normal distribution is more stable than the mean, so the assumption that the variance is known is not entirely artificial. Recall that the normal probability density function (given $$\mu$$) is $g(x \mid \mu) = \frac{1}{\sqrt{2 \, \pi} \sigma} \exp\left[-\frac{1}{2}\left(\frac{x - \mu}{\sigma}\right)^2 \right], \quad x \in \R$ Again, for $$n \in \N_+$$ let $$Y_n = \sum_{i=1}^n X_i$$. Recall that $$Y_n$$ also has a normal distribution (given $$\mu$$) but with mean $$n \mu$$ and variance $$n \sigma^2$$.

Suppose now that $$\mu$$ is modeled by a random variable $$\Psi$$ that has a prior normal distribution with mean $$a \in \R$$ and variance $$b^2 \in (0, \infty)$$. As usual, $$a$$ and $$b$$ are chosen to reflect our prior knowledge of $$\mu$$. An interesting special case is when we take $$b = \sigma$$, so the variance of the prior distribution of $$\Psi$$ is the same as the variance of the underlying sampling distribution.

For $$n \in \N_+$$, the posterior distribution of $$\Psi$$ given $$\bs{X}_n = (X_1, X_2, \ldots, X_n)$$ is normal with mean and variance given by \begin{align} \E(\Psi \mid \bs{X}_n) & = \frac{Y_n b^2 + a \sigma^2}{n b^2 + \sigma^2}\\ \var(\Psi \mid \bs{X}_n) & = \frac{\sigma^2 b^2}{n b^2 + \sigma^2} \end{align}

Proof:

Fix $$n \in \N_+$$. Suppose $$\bs x = (x_1, x_2, \ldots, x_n) \in \R$$ and let $$y = \sum_{i=1}^n x_i$$ and $$w^2 = \sum_{i=1}^n x_i^2$$. Then \begin{align} f(\bs x \mid \mu) & = g(x_1 \mid \mu) g(x_2 \mid \mu) \cdots g(x_n \mid \mu) = \frac{1}{(2 \pi)^{n/2} \sigma^n} \exp\left[-\frac{1}{2} \sum_{i=1}^n \left(\frac{x_i - \mu}{\sigma}\right)^2\right] \\ & = \frac{1}{(2 \pi)^{n/2} \sigma^n} \exp\left[-\frac{1}{2 \sigma^2}(w^2 - 2 \mu y + n \mu^2)\right] \end{align} On the other hand, of course $h(\mu) = \frac{1}{\sqrt{2 \pi} b} \exp\left[-\frac{1}{2}\left(\frac{\mu - a}{b}\right)^2\right] = \frac{1}{\sqrt{2 \pi}b} \exp\left[-\frac{1}{2 b^2}(\mu^2 - 2 a \mu + a^2)\right]$ Therefore, $h(\mu) f(\bs x \mid \mu) = C \exp\left\{-\frac{1}{2}\left[\left(\frac{1}{b^2} + \frac{n}{\sigma^2}\right) \mu^2 - 2 \left(\frac{a}{b^2} + \frac{y}{\sigma^2}\right) \mu\right]\right\}$ where $$C$$ depends on $$n$$, $$\sigma$$, $$a$$, $$b$$, $$\bs x$$, but importantly not on $$\mu$$. So we don't really care what $$C$$ is. Completing the square in $$\mu$$ in the expression above gives $h(\mu) f(\bs x \mid \mu) = K \exp\left[-\frac{1}{2}\left(\frac{1}{b^2} + \frac{n}{\sigma^2}\right) \left(\mu - \frac{a / b^2 + y / \sigma^2}{1 / b^2 + n / \sigma^2}\right)^2\right]$ where $$K$$ is yet another factor that depends on lots of stuff, but not $$\mu$$. As a function of $$\mu$$, this expression is proportional to the normal distribution with mean and variance, respectively, given by \begin{align} &\frac{a / b^2 + y / \sigma^2}{1 / b^2 + n / \sigma^2} = \frac{y b^2 + a \sigma^2}{n b^2 + \sigma^2} \\ & \frac{1}{1 / b^2 + n / \sigma^2} = \frac{\sigma^2 b^2}{\sigma^2 + n b^2} \end{align} Once again, it was not necessary to compute the normalizing constant $$f(\bs{x})$$, which would have been yet another factor that we do not care about.

Therefore, the normal distribution is conjugate to the normal distribution with unknown mean and known variance. Note that the posterior variance is deterministic, and depends on the data only through the sample size $$n$$. In the special case that $$b = \sigma$$, the posterior distribution of $$\Psi$$ given $$\bs{X}_n$$ is normal with mean $$(Y_n + a) / (n + 1)$$ and variance $$\sigma^2 / (n + 1)$$. Of course, it follows immediately that the Bayesian estimator of $$\mu$$ is $U_n = \frac{Y_n b^2 + a \sigma^2}{n b^2 + \sigma^2}$ which again reduces to $$U_n = (Y_n + a) / (n + 1)$$ in the special case that $$b = \sigma$$.

For $$n \in \N_+$$, $\bias(U_n \mid \mu) = \frac{\sigma^2 (a - \mu)}{\sigma^2 + n \, b^2}, \quad \mu \in \R$ The sequence of estimators $$\bs U = (U_n: n \in \N_+)$$ is asymptotically unbiased.

Proof:

Recall that $$Y_n$$ has mean $$n \mu$$ given $$\mu$$. Hence $\bias(U_n \mid \mu) = \E(U_n \mid \mu) - \mu = \frac{n b^2 \mu + a \sigma^2}{n b^2 + \sigma^2} - \mu = \frac{(a - \mu) \sigma^2}{n b^2 + \sigma^2}$ Clearly $$\bias(U_n \mid \mu) \to 0$$ as $$n \to \infty$$ for every $$\mu \in \R$$.

When $$b = \sigma$$, $$\bias(U_n \mid \mu) = (a - \mu) / (n + 1)$$.

For $$n \in \N_+$$, $\mse(U_n \mid \mu) = \frac{n \sigma^2 b^4 + \sigma^4 (a - \mu)^2}{(\sigma^2 + n \, b^2)^2}, \quad \mu \in \R$ The sequence of estimators $$\bs U = (U_n: n \in \N_+)$$ is mean-square consistent.

Proof:

Recall that $$Y_n$$ as variance $$n \sigma^2$$. Hence $\mse(U_n \mid \mu) = \var(U_n \mid \mu) + \bias^2(U_n \mid \mu) = \left(\frac{b^2}{n b^2 + \sigma^2}\right)^2 n \sigma^2 + \left(\frac{(a - \mu) \sigma^2}{n b^2 + \sigma^2}\right)^2$ Clearly $$\mse(U_n \mid \mu) \to 0$$ as $$n \to \infty$$ for every $$\mu \in \R$$.

When $$b = \sigma$$, $$\mse(U \mid \mu) = [n \sigma^2 + (a - \mu)^2] / (n + 1)^2$$. Recall that the method of moments estimator of $$\mu$$ and the maximum likelihood estimator of $$\mu$$ on $$\R$$ are both $$M_n = Y_n / n$$. This estimator is unbiased and has mean square error $$\var(M) = \sigma^2 / n$$. To compare the estimators, note that $U_n = \frac{M_n b^2 + a \sigma^2 / n}{b^2 + \sigma^2 / n}$

#### The Beta Distribution

Suppose that $$\bs{X} = (X_1, X_2, \ldots)$$ is a sequence of independent random variables each having the beta distribution with unknown left shape parameter $$a \in (0, \infty)$$ and right shape parameter $$b = 1$$. The beta distribution is widely used to model random proportions and probabilities and other variables that take values in bounded intervals (scaled to take values in $$(0, 1)$$). Recall that the probability density function (given $$a$$) is $g(x \mid a) = a \, x^{a-1}, \quad x \in (0, 1)$ Suppose now that $$a$$ is modeled by a random variable $$A$$ that has a prior gamma distribution with shape parameter $$k \in (0, \infty)$$ and rate parameter $$r \in (0, \infty)$$. As usual, $$k$$ and $$r$$ are chosen to reflect our prior knowledge of $$a$$. Thus the prior probabiltiy density function of $$a$$ is $h(a) = \frac{r^k}{\Gamma(k)} a^{k-1} e^{-r a}, \quad a \in (0, \infty)$

The posterior distribution of $$A$$ given $$\bs{X}_n = (X_1, X_2, \ldots, X_n)$$ is gamma, with shape parameter $$k + n$$ and rate parameter $$r - \ln(X_1 X_2 \cdots X_n)$$.

Proof:

Fix $$n \in \N_+$$. Let $$\bs x = (x_1, x_2, \ldots, x_n) \in (0, 1)^n$$ and let $$z = x_1 x_2 \cdots x_n$$ Then $f(\bs x \mid a) = g(x_1 \mid a) g(x_2 \mid a) \cdots g(x_n \mid a) = a^n z^{a - 1} = \frac{a^n}{z} e^{a \ln z}$ Hence $h(a) f(\bs x \mid a) = \frac{r^k}{z \Gamma(k)} a^{n + k - 1} e^{-a (r - \ln z)}, \quad a \in (0, \infty)$ As a function of $$a \in (0, \infty)$$ this expression is proportional to the gamma PDF with shape parameter $$n + k$$ and scale parameter $$r - \ln z$$. Once again, it's not necessary to compute the normalizing constant $$f(\bs x)$$.

Thus, the gamma distribution is conjugate to the beta distribution with unknown left parameter and right parameter 1. Note that the posterior shape parameter is deterministic and depends on the data only through the sample size $$n$$.

The Bayesian estimator of $$a$$ based on $$\bs{X}_n$$ is $U_n = \frac{k + n}{r - \ln(X_1 X_2 \cdots X_n)}$

Proof:

The mean of the gamma distribution is the shape parameter divided by the rate parameter, so this follows from the previous theorem.

Given the complicated structure, the bias and mean square error of $$U_n$$ given $$a \in (0, \infty)$$ would be difficult to compute explicitly. Recall that the maximum likelihood estimator of $$a$$ is $$-n / \ln(X_1 \, X_2 \cdots X_n)$$.

#### The Pareto Distribution

Suppose that $$\bs{X} = (X_1, X_2, \ldots)$$ is a sequence of independent random variables each having the Pareto distribution with unknown shape parameter $$a \in (0, \infty)$$ and scale parameter $$b = 1$$. The Pareto distribution is used to model certain financial variables and other variables with heavy-tailed distributions, and is named for Vilfredo Pareto. Recall that the probability density function (given $$a$$) is $g(x \mid a) = \frac{a}{x^{a+1}}, \quad x \in [1, \infty)$ Suppose now that $$a$$ is modeled by a random variable $$A$$ that has a prior gamma distribution with shape parameter $$k \in (0, \infty)$$ and rate parameter $$r \in (0, \infty)$$. As usual, $$k$$ and $$r$$ are chosen to reflect our prior knowledge of $$a$$. Thus the prior probabiltiy density function of $$A$$ is $h(a) = \frac{r^k}{\Gamma(k)} a^{k-1} e^{-r a}, \quad a \in (0, \infty)$

For $$n \in \N_+$$, the posterior distribution of $$A$$ given $$\bs{X}_n = (X_1, X_2, \dots, X_n)$$ is gamma, with shape parameter $$k + n$$ and rate parameter $$r + \ln(X_1 X_2 \cdots X_n)$$.

Proof:

Fix $$n \in \N_+$$. Let $$\bs x = (x_1, x_2, \ldots, x_n) \in [1, \infty)^n$$ and let $$z = x_1 x_2 \cdots x_n$$ Then $f(\bs x \mid a) = g(x_1 \mid a) g(x_2 \mid a) \cdots g(x_n \mid a) = \frac{a^n}{z^{a + 1}} = \frac{a^n}{z} e^{- a \ln z}$ Hence $h(a) f(\bs x \mid a) = \frac{r^k}{z \Gamma(k)} a^{n + k - 1} e^{-a (r + \ln z)}, \quad a \in (0, \infty)$ As a function of $$a \in (0, \infty)$$ this expression is proportional to the gamma PDF with shape parameter $$n + k$$ and scale parameter $$r + \ln z$$. Once again, it's not necessary to compute the normalizing constant $$f(\bs x)$$.

Thus, the gamma distribution is conjugate to Pareto distribution with unknown shape parameter. Note that the posterior shape parameter is deterministic and depends on the data only through the sample size $$n$$.

The Bayesian estimator of $$a$$ based on $$\bs{X}_n$$ is $U_n = \frac{k + n}{r + \ln(X_1 X_2 \cdots X_n)}$

Proof:

Once again, the mean of the gamma distribution is the shape parameter divided by the rate parameter, so this follows from the previous theorem.

Given the complicated structure, the bias and mean square error of $$U$$ given $$a \in (0, \infty)$$ would be difficult to compute explicitly. Recall that the maximum likelihood estimator of $$a$$ is $$n / \ln(X_1 \, X_2 \cdots X_n)$$.