Chapter 8 (Bayesian Statistical Inference): Bayesian Inference and the Posterior Distribution

本文为 $Introduction$ $to$ $Probability$ 的读书笔记

Statistical Inference 统计推断

• Statistical inference is the process of extracting information about an unknown variable or an unknown model from available data. We aim to:
  • $(a)$ Develop an appreciation of the main two approaches (Bayesian (贝叶斯统计推断) and classical (经典统计推断)), their differences, and similarities.
  • $(b)$ Present the main categories of inference problems (parameter estimation (参数估计), hypothesis testing (假设检验), and significance testing (显著性检验)).
  • $(c)$ Discuss the most important methodologies (maximum a posteriori probability rule (MAP 最大后验概率准则), least mean squares estimation (LMS 最小均方估计), maximum likelihood (最大似然估计), regression (回归), likelihood ratio tests (似然比检验), etc).

Bayesian versus Classical Statistics 贝叶斯统计和经典统计

• Their fundamental difference relates to the nature of the unknown models or variables.
  • In the Bayesian view, they are treated as random variables with known prior distributions.
    • In particular, when trying to infer the nature of an unknown model, the Bayesian approach views the model as chosen randomly from a given model class. This is done by introducing a random variable $\Theta$ that characterizes the model, and by postulating a prior probability distribution (先验概率分布) $p_{\Theta }(\theta )$. Given the observed data $x$, one can, in principle, use Bayes’ rule to derive a posterior probability distribution (后验概率分布) $p_{\Theta |X}(\theta |x)$. This captures all the information that $x$ can provide about $\theta$.
  • By contrast, the classical approach views the unknown quantity $\theta$ as a constant that happens to be unknown. It then strives to develop an estimate of $\theta$ that has some performance guarantees. This introduces an important conceptual difference from other methods: we are not dealing with a single probabilistic model, but rather with multiple candidate probabilistic models, one for each possible value of $\theta$.

Model versus Variable Inference 模型推断和变量推断

• Applications of statistical inference tend to be of two different types.
  • In model inference, the object of study is a real phenomenon or process for which we wish to construct or validate a model on the basis of available data (e.g., do planets follow elliptical trajectories?). Such a model can then be used to make predictions about the future, or to infer some hidden underlying causes.
  • In variable inference, we wish to estimate the value of one or more unknown variables by using some related, possibly noisy information (e.g., what is my current position, given a few GPS readings?).
• The distinction between model and variable inference is not sharp; for example, by describing a model in terms of a set of variables, we can cast a model inference problem as a variable inference problem.
  • In any case, we will not emphasize this distinction in the sequel, because the same methodological principles apply to both types of inference.
  • In some applications, both model and variable inference issues may arise. For example, we may collect some initial data, use them to build a model, and then use the model to make inferences about the values of certain variables.

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Example 8.1. A Noisy Channel.

• A transmitter sends a sequence of binary messages s i ∈ { 0 , 1 } s_i \in \{ 0, 1\} si​∈{ 0,1}, and a receiver observes
  $$X_i=a{s}_i+W_i,i=1,...,n$$where the $W_i$ are zero mean normal random variables that model channel noise, and $a$ is a scalar that represents the channel attenuation (信道衰减率).
• In a model inference setting, $a$ is unknown. The transmitter sends a pilot signal consisting of a sequence of messages $s_1,...,s_n$, whose values are known by the receiver. On the basis of the observations $X_1,...,X_n$, the receiver wishes to estimate the value of $a$, that is, build a model of the channel.
• Alternatively, in a variable inference setting, $a$ is assumed to be known (possibly because it has already been inferred using a pilot signal, as above). The receiver observes $X_1,...,X_n$, and wishes to infer the values of $s_1,...,s_n$.

A Rough Classification of Statistical Inference Problems

• In an estimation problem (估计问题), a model is fully specified, except for an unknown, possibly multidimensional, parameter $\theta$, which we wish to estimate. This parameter can be viewed as either a random variable (Bayesian approach) or as an unknown constant (classical approach). The usual objective is to arrive at an estimate of $\theta$ that is close to the true value in some sense. For example:
  • (a) In the noisy transmission problem of Example 8.1, use the knowledge of the pilot sequence and the observations to estimate $a$.
  • (b) On the basis of historical stock market data, estimate the mean and variance of the daily movement in the price of a particular stock.
• In a binary hypothesis testing problem (二重假设检验问题), we start with two hypotheses and use the available data to decide which of the two is true. For example:
  • (a) In the noisy transmission problem of Example 8.1, use the knowledge of $a$ and $X_i$ to decide whether $s_i$ was 0 or 1.
  • (b) Given a noisy picture, decide whether there is a person in the picture or not.
  • $(c)$ Given a set of trials with two alternative medical treatments, decide which treatment is most effective.
• More generally. in an $m$-ary hypothesis testing problem ($m$ 重假设检验问题), there is a finite number $m$ of competing hypotheses. The performance of a particular method is typically judged by the probability that it makes an erroneous decision.

Bayesian Inference and the Posterior Distribution

贝叶斯推断 和 后验分布

• In Bayesian inference, the unknown quantity of interest, which we denote by $\Theta$, is modeled as a random variable or as a finite collection of random variables.

For simplicity, unless the contrary is explicitly stated, we view $\Theta$ as a single random variable.

• We aim to extract information about $\Theta$, based on observing a collection $X=(X_1,...,X_n)$ of related random variables, called observations, measurements, or an observation vector. For this, we assume that we know the joint distribution of $\Theta$ and $X$. Equivalently, we assume that we know:
  • (a) A prior distribution $p_{\Theta }$ or $f_{\Theta }$. depending on whether $\Theta$ is discrete or continuous.
  • (b) A conditional distribution $p_{X|\Theta }$ or $f_{X|\Theta }$, depending on whether $X$ is discrete or continuous.
• Once a particular value $x$ of $X$ has been observed, a complete answer to the Bayesian inference problem is provided by the posterior distribution $p_{\Theta |X}(\theta |x)$ or $f_{\Theta |X}(\theta |x)$ of $\theta$. This distribution is determined by the appropriate form of Bayes rule. It encapsulates everything there is to know about $\Theta$, given the available information, and it is the starting point for further analysis.
  

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Example 8.2.
Romeo and Juliet start dating, but Juliet will be late on any date by a random amount $X$, uniformly distributed over the interval $[0,\theta ]$. The parameter $\theta$ is unknown and is modeled as the value of a random variable $\Theta$, uniformly distributed between zero and one hour. Assuming that Juliet was late by an amount $x$ on their first date, how should Romeo use this information to update the distribution of $\Theta$?

SOLUTION

• Here the prior PDF is
  and the conditional PDF of the observation is
  
• Using Bayes’ rule, and taking into account that $f_{\Theta }(\theta )f_{X|\Theta }(x|\theta )$ is nonzero only if $0\le x\le \theta \le 1$, we find that for any $x\in [0,1]$, the posterior PDF is
  $$f_{\Theta |X}(\theta |x)=\frac{f_{\Theta }(\theta )f_{X|\Theta }(x|\theta )}{{\int }_0^1{f}_{\Theta }({\theta }^{\prime })f_{X|\Theta }(x|{\theta }^{\prime })d{\theta }^{\prime }}=\frac{1/\theta }{{\int }_x^1\frac{1}{{\theta }^{\prime }}d{\theta }^{\prime }}=\frac{1}{\theta \cdot |\ln x|},if0\le x\le \theta \le 1$$and $f_{\Theta |X}(\theta |x)=0$ if $\theta <x$ or $\theta >1$.
• Consider now a variation involving the first $n$ dates. Assume that Juliet is late by random amounts $X_1,...,X_n$ , which given $\Theta =\theta$, are uniformly distributed in the interval $[0,\theta ]$, and conditionally independent. Let $X=(X_1,...,X_n)$ and $x=(x_1,...,x_n)$. Similar to the case where $n=1$, we have
  where
  x ‾ = max ⁡ { x 1 , . . . , x n } \overline x=\max\{x_1,...,x_n\} x=max{ x1​,...,xn​}The posterior PDF is
  where $c(\overline{x})$ is a normalizing constant that depends only on $\overline{x}$：
  $$c(\overline{x})=\frac{1}{{\int }_{\overline{x}}^1\frac{1}{({\theta }^{\prime }{)}^n}d{\theta }^{\prime }}$$

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Example 8.3. Inference of a Common Mean of Normal Random Variables.
We observe a collection $X=(X_1,...,X_n)$ of random variables, with an unknown common mean whose value we wish to infer. We assume that given the value of the common mean, the $X_i$ are normal and independent, with known variances ${\sigma }_1^2,...,{\sigma }_n^2$. In a Bayesian approach to this problem, we model the common mean as a random variable $\theta$, with a given prior. For concreteness, we assume a normal prior, with known mean $x_0$ and known variance ${\sigma }_0^2$.

• Let us note, for future reference, that our model is equivalent to one of the form
  $$X_i=\Theta +W_i,i=1,...,n$$where the random variables $\Theta$, $W_1,...,W_n$ are independent and normal, with known means and variances. In particular, for any value $\theta$,
  $$\begin{gathered} E[W_i]=E[W_i|\Theta =\theta ]=0 \\ var(W_i)=var(X_i|\Theta =\theta )={\sigma }_i^2 \end{gathered}$$A model of this type is common in many engineering applications involving several independent measurements of an unknown quantity.
• According to our assumptions, we have
  $$f_{\Theta }(\theta )=c_1\cdot \exp \{-\frac{(\theta -x_0{)}^2}{2{\sigma }_0^2}\}$$and
  $$f_{X|\Theta }(x|\theta )=c_2\cdot \exp \{-\frac{(x_1-\theta {)}^2}{2{\sigma }_1^2}\}...\exp \{-\frac{(x_n-\theta {)}^2}{2{\sigma }_n^2}\}$$where $c_1$ and $c_2$ are normalizing constants that do not depend on $\theta$. We invoke Bayes’ rule,
  $$f_{\Theta |X}(\theta |x)=\frac{f_{\Theta }(\theta )f_{X|\Theta }(x|\theta )}{\int f_{\Theta }({\theta }^{\prime })f_{X|\Theta }(x|{\theta }^{\prime })d{\theta }^{\prime }}$$and note that the numerator term, $f_{\Theta }(\theta )f_{X|\Theta }(x|\theta )$, is of the form
  $$c_1{c}_2\cdot \exp \{-\sum _{i=0}^n\frac{(x_i-\theta {)}^2}{2{\sigma }_i^2}\}$$After some algebra, which involves completing the square inside the exponent (对指数的肩膀上的求和部分进行配平方), we find that the numerator is of the form
  $$d\cdot \exp \{-\frac{(\theta -m{)}^2}{2v}\}$$where
  $$m=\frac{{\sum }_{i=0}^n{x}_i/{\sigma }_i^2}{{\sum }_{i=0}^{n}1/{\sigma }_i^2},v=\frac{1}{{\sum }_{i=0}^{n}1/{\sigma }_i^2}$$and $d$ is a constant, which depends on the $x_i$ but does not depend on $\theta$. Since the denominator term in Bayes’ rule does not depend on $\theta$ either, we conclude that the posterior PDF has the form
  $$f_{\Theta |X}(\theta |x)=a\cdot \exp \{-\frac{(\theta -m{)}^2}{2v}\}$$for some normalizing constant $a$, which depends on the $x_i$, but not on $\theta$. We recognize this as a normal PDF, and we conclude that the posterior PDF is normal with mean $m$ and variance $v$.
• As a special case suppose that ${\sigma }_0^2,{\sigma }_1^2,...,{\sigma }_n^2$ have a common value ${\sigma }^2$. Then, the posterior PDF of $\Theta$ is normal with mean and variance
  $$m=\frac{x_0+...+x_n}{n+1},v=\frac{{\sigma }^2}{n+1}$$respectively. In this case, the prior mean $x_0$ acts just as another observation, and contributes equally to determine the posterior mean $m$ of $\theta$. Notice also that the standard deviation of the posterior PDF of $\theta$ tends to zero, at the rough rate of $1/\sqrt{n}$，as the number of observations increases.
• If the variances ${\sigma }_i^2$ are different, the formula for the posterior mean $m$ is still a weighted average of the $x_i$, but with a larger weight on the observations with smaller variance.
• The preceding example has the remarkable property that the posterior distribution of $\Theta$ is in the same family as the prior distribution, namely, the family of normal distributions. This is appealing for two reasons:
  • (a) The posterior can be characterized in terms of only two numbers, the mean and the variance.
  • (b) The form of the solution opens up the possibility of efficient recursive inference. Suppose that after $X_1,...,X_n$ are observed, an additional observation $X_{n+1}$ is obtained. Instead of solving the inference problem from scratch, we can view $f_{\Theta |X_1,...,X_n}$ as our prior, and use the new observation to obtain the new posterior $f_{\Theta |X_1,...,X_n,X_{n+1}}$. It is then plausible (and can be formally established) that the new posterior of $\Theta$ will have mean
    $$\frac{(m/v)+(x_{n+1}/{\sigma }_{n+1}^2)}{(1/v)+(1/{\sigma }_{n+1}^2)}$$and variance
    $$\frac{1}{(1/v)+(1/{\sigma }_{n+1}^2)}$$where $m$ and $v$ are the mean and variance of the old posterior $f_{\Theta |X_1,...,X_n}$.

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Example 8.4. Beta Priors on the Bias of a Coin. (贝塔先验)
We wish to estimate the probability of heads, denoted by $\theta$, of a biased coin. We model $\theta$ as the value of a random variable $\Theta$ with known prior PDF $f_{\theta }$. We consider $n$ independent tosses and let $X$ be the number of heads observed.

• From Bayes’ rule, the posterior PDF of $\theta$ has the form, for $\theta \in [0,1]$,
  $$f_{\Theta |X}(\theta |k)=c{f}_{\Theta }(\theta )p_{X|\Theta }(k|\theta )=d{f}_{\Theta }(\theta ){\theta }^k(1-\theta {)}^{n-k}$$where $c$ is a normalizing constant (independent of $\theta$), and $d=c\left(\begin{array}{c}n\\ k\end{array}\right)$.
• Suppose now that the prior is a beta density with integer parameters $\alpha >0$ and $\beta >0$, of the form
  where $B(\alpha ,\beta )$ is a normalizing constant, known as the Beta function, given by
  $$B(\alpha ,\beta )={\int }_0^1{\theta }^{\alpha -1}(1-\theta {)}^{\beta -1}d\theta =\frac{(\alpha -1)!(\beta -1)!}{(\alpha +\beta -1)!}$$the last equality can be obtained from integration by parts, or through a probabilistic argument (see Problem 30). Then, the posterior PDF of $\theta$ is of the form
  $$f_{\Theta |X}(\theta |k)=\frac{d}{B(\alpha ,\beta )}{\theta }^{k+\alpha -1}(1-\theta {)}^{n-k+\beta -1},0\le \theta \le 1$$and hence is a beta density with parameters
  $${\alpha }^{\prime }=k+\alpha ,{\beta }^{\prime }=n-k+\beta$$