# Chapter 1 Introduction

The central idea we will explore in this book is: given some data, how to use Bayes’ theorem to quantify uncertainty about our belief regarding a scientific question of interest. Before we get into the details of the underlying theory and its application, some familiarity with the following topics needs to be in place: the basic concepts behind probability, the concept of random variables, probability distributions, and the concept of likelihood. We therefore turn to these topics first.

## 1.1 Probability

Informally, we all understand what the term probability means. We routinely talk about things like the probability of it raining today. However, there are two distinct ways to think about probability. One can think of the probability of an event with reference to the frequency with which it might occur in repeated observations. Such a conception of probability is easy to imagine in cases where an event can, at least in principle, occur repeatedly. An example would be obtaining a 6 when tossing a die again and again. However, this frequentist view of probability is difficult to justify when talking about certain one-of-a-kind events, such as earthquakes. In such situations, probability is expressing our uncertainty about the event happening. Moreover, we could even be uncertain about exactly how probable the event in question is; for example, we might say something like “I am 90% certain that the probability of an earthquake happening in the next year is between 10 and 40%”. In this book, we will be particularly interested in quantifying uncertainty in this way: we will always want to know how unsure we are of the estimate we are interested in.

Both the frequency-based and the uncertain-belief perspective have their place in statistical inference, and depending on the situation, we are going to rely on both ways of thinking. Regardless of these differences in perspective, the probability of an event happening is defined to be constrained in the following way. The statements below are not formal statements of the axioms of probability theory; for more details (and more precise formulations), see Ross (2002) or Kolmogorov (1933).

• The probability of an event must lie between 0 and 1, where 0 means that the event is impossible and cannot happen, and 1 means that the event is certain to happen.
• For any two mutually exclusive events, the probability that one or the other occurs is the sum of their individual probabilities.
• Two events are independent if and only if the probability of both events happening is equal to the product of the probabilities of each event happening.
• The probabilities of all possible events in the entire sample space must sum up to 1.

The above definitions are based on the axiomatic definition of probability by Kolmogorov (1933).

In the context of data analysis, we will talk about probability in the following way. Consider some data that we might have collected. This could be discrete 0,1 responses in a question-response accuracy task, or continuous measurements of reading times in milliseconds from an eyetracking study, etc. In any such cases, we will say that the data are being generated from a random variable, which we will designate with a capital letter such as $$Y$$.1

The actually observed data will be distinguished from the random variable that generated it by using lower case $$y$$. We can call $$y$$ an instance of $$Y$$; every new set of data will be slightly different due to random variability.

We can summarize the above informal concepts relating to random variables very compactly if we re-state them in mathematical form. A mathematical statement has the advantage not only of brevity but also of reducing ambiguity.

Somewhat more formally (following Blitzstein and Hwang 2014), we define a random variable $$Y$$ as a function from a sample space of possible outcomes $$S$$ to the real number system:2

$\begin{equation} Y : S \rightarrow \mathbb{R} \end{equation}$

The random variable associates to each outcome $$\omega \in S$$ exactly one number $$Y(\omega) = y$$. Suppose that $$y$$ represents all the possible values that the random variable generates; these values are taken to belong to the support of S: $$y \in S_Y$$.

As a concrete example, consider an experiment where we ask subjects to respond to 10 questions that can either have a correct or incorrect answer. We will say that the number of correct responses from a subject is generated from a random variable $$Y$$. Because only discrete responses are possible (the number of correct responses can be 0, 1, 2, …, 10), this is an example of a discrete random variable.

This random variable will be assumed to have a parameter $$\theta$$ that represents the probability of producing a correct response. In statistics, given some observed data, typically our goal is to obtain an estimate of this parameter’s true (unknown) value.

This discrete random variable $$Y$$ has associated with it a function called a probability mass function or PMF. This function, which is written $$p(y)$$, gives us the probability of obtaining each of these $$11$$ possible outcomes (from 0 correct responses to 10). We are using lower-case $$p(\cdot)$$ here, and this is distinct from $$P(\cdot)$$, which we will use to talk about probabilities.

We will write that this PMF $$p(y)$$ depends on, or is conditional on, a particular fixed but unknown value for $$\theta$$; the PMF will be written $$p(y|\theta)$$.3

In frequentist approaches to data analysis, only the observed data $$y$$ are used to draw inferences about $$\theta$$. A typical question that we ask in the frequentist paradigm is: does $$\theta$$ have a particular value $$\theta_0$$? One can obtain estimates of the unknown value of $$\theta$$ from the observed data $$y$$, and then draw inferences about how different–or more precisely how far away–this estimate is from the hypothesized $$\theta_0$$. This is the essence of null hypothesis significance testing. The conclusions from such a procedure are framed in terms of either rejecting the hypothesis that $$\theta$$ has value $$\theta_0$$, or failing to reject this hypothesis. Here, rejecting the null hypothesis is the primary goal of the statistical hypothesis test.

Bayesian data analysis begins with a different question. What is common to the frequentist paradigm is the assumption that the data are generated from a random variable $$Y$$ and that there is a function $$p(y|\theta)$$ that depends on the parameter $$\theta$$. Where the Bayesian approach diverges from the frequentist one is that an important goal is to express our uncertainty about $$\theta$$. In other words, we treat the parameter $$\theta$$ itself as a random variable, which means that we assign a probability distribution $$p(\theta)$$ to this random variable. This distribution $$p(\theta)$$ is called the prior distribution on $$\theta$$; such a distribution could express our belief about the probability of correct responses, before we observe the data $$y$$.

In later chapters, we will spend some time trying to understand how such a prior distribution can be defined for a range of different research problems.

Given such a prior distribution and some data $$y$$, the end-product of a Bayesian data analysis is what is called the posterior distribution of the parameter (or parameters) given the data: $$p(\theta | y)$$. This posterior distribution is the probability distribution of $$\theta$$ after conditioning on $$y$$, i.e., after the data has been observed and is therefore known. All our statistical inference is based on this posterior distribution of $$\theta$$; we can even carry out hypothesis tests that are analogous (but not identical) to the likelihood ratio based frequentist hypothesis tests.

We already mentioned conditional probability above when discussing the probability of the data given some parameter $$\theta$$, which we wrote as the PMF $$p(y|\theta)$$. Conditional probability is an important concept in Bayesian data analysis, not least because it allows us to derive Bayes’ theorem. Let’s look at the definition of conditional probability next.

## 1.2 Conditional probability

Suppose that $$A$$ stands for some discrete event; an example would be “the streets are wet.” Suppose also that $$B$$ stands for some other discrete event; an example is “it has been raining.” We can talk about the probability of the streets being wet given that it has been raining; or more generally, the probability of $$A$$ given that $$B$$ has happened.

This kind of statement is written as $$Prob(A|B)$$ or more simply $$P(A|B)$$. This is the conditional probability of event $$A$$ given $$B$$. Conditional probability is defined as follows.

$\begin{equation} P(A|B)= \frac{P(A,B)}{P(B)} \hbox{ where } P(B)>0 \end{equation}$

The conditional probability of A given B is thus defined to be the joint probability of A and B, divided by the probability of B. We can rearrange the above equation so that we can talk about the joint probability of both events $$A$$ and $$B$$ happening. This joint probability can be computed by first taking $$P(B)$$, the probability that event $$B$$ (it has been raining) happens, and multipling this by the probability that $$A$$ happens conditional on $$B$$, i.e., the probability that the streets are wet given it has been raining. This multiplication will give us $$P(A,B)$$, the joint probability of $$A$$ and $$B$$, i.e., that it has been raining and that the streets are wet. We will write the above description as: $$P(A,B)=P(A|B)P(B)$$.

Now, since the probability $$A$$ and $$B$$ happening is the same as the probability of $$B$$ and $$A$$ happening, i.e., since $$P(B,A)=P(A,B)$$, we can equate the expansions of these two terms:

$\begin{equation} P(A,B) = P(A|B) P(B) \hbox{ and } P(B,A) = P(B|A)P(A) \end{equation}$

Equating the two expansions, we get:

$\begin{equation} P(A|B) P(B) = P(B|A)P(A) \end{equation}$

Dividing both sides by $$P(B)$$:

$\begin{equation} P(A|B)=\frac{P(B|A)P(A)}{P(B)} \end{equation}$

The above statement is Bayes’ rule, and is the basis for all the statistical inference we will do in this book.

## 1.3 The law of total probability

Related to the above discussion of conditional probability is the law of total probability. Suppose that we have $$A_1,\dots,A_n$$ distinct events that are pairwise disjoint which together make up the entire sample space $$S$$; see Figure 1.1. Then, $$P(B)$$, the probability of an event $$B$$, will be the sum of the probabilities $$P(B\cap A_i)$$, i.e., the sum of the joint probabilities of $$B$$ and each $$A$$ occurring (the symbol $$\cap$$ is the and’’ operator used in set theory.)

Formally:

$\begin{equation} P(B) = \sum_{i=1}^n P(B \cap A_i) \end{equation}$

Because of the conditional probability rule, we can rewrite this as:

$\begin{equation} P(B) = \sum_{i=1}^n P(B | A_i) P(A_i) \end{equation}$

Thus, the probability of $$B$$ is the sum of the conditional probabilities $$P(B|A_i)$$ weighted by the probability $$P(A_i)$$. We will see the law of total probability in action below when we talk about marginal likelihood. FIGURE 1.1: An illustration of the law of total probability.

For now, this is all the probability theory we need to know!

The next sections expand on the idea of a random variable, the probability distributions associated with the random variable, what it means to specify a prior distribution on a parameter, and how the prior and data can be used to derive the posterior distribution of $$\theta$$.

To make the discussion concrete, we will use an example of a discrete random variable, the binomial. After discussing this discrete random variable, we present another example, this time involving a continuous random variable, the normal random variable.

The binomial and normal cases serve as the canonical examples that we will need in the initial stages of this book. We will introduce other random variables as needed: in particular, we will need the Uniform and Beta distributions. In other textbooks, you will encounter distributions like the Poisson, gamma, and the exponential. The most commonly used distributions and their properties are discussed in most textbooks on statistics (see Further Readings at the end of this chapter).

## 1.4 Discrete random variables: An example using the binomial distribution

Consider the following sentence:

“It’s raining, I’m going to take the ….”

Suppose that our research goal is to estimate the probability, call it $$\theta$$, of the word “umbrella” appearing in this sentence, versus any other word. If the sentence is completed with the word “umbrella”, we will refer to it as a success; any other completion will be referred to as a failure. This is an example of a binomial random variable: given $$n$$ trials, there can be only two possible outcomes in each trial, a success or a failure, and there is some true unknown probability $$\theta$$ of success that we want to estimate. When the number of trials $$n=1$$ is one, the random variable is called a Bernoulli distribution. So the Bernoulli distribution is the binomial distribution with $$n=1$$.

One way to empirically estimate this probability of success is to carry out a cloze task. In a cloze task, subjects are asked to complete a fragment of the original sentence, such as “It’s raining, I’m going to take the …”. The predictability or cloze probability of “umbrella” is then calculated as the proportion of times that the target word “umbrella” was produced as an answer by subjects.

Assume for simplicity that $$10$$ subjects are asked to complete the above sentence; each subject does this task only once. This gives us independent responses from $$10$$ trials that are either coded a success (“umbrella” was produced) or as a failure (some other word was produced). We can sum up the number of successes to calculate how many of the 10 trials had “umbrella” as a response. For example, if $$8$$ instances of “umbrella” are produced in $$10$$ trials, we would estimate the cloze probability of producing “umbrella” would be $$8/10$$.

We can repeatedly generate simulated sequences of the number of successes in R (later on we will demonstrate how to generate such random sequences of simulated data). Here is a case where we run the same experiment $$n=10$$ times, and the sample size, is $$\hbox{size}=10$$ each time.

rbinom(n = 20, size = 10, prob = 0.5)
##    7  3  4  7  5  2  5  5  5  4  5  8  4  4  8  4  2  5 10  9

The number of successes in each of the $$20$$ simulated experiments above is being generated by a discrete random variable $$Y$$ with a probability distribution $$p(y|\theta)$$ called the binomial distribution.

For discrete random variables such as the binomial, the probability distribution $$p(y|\theta)$$ is called a probability mass function (PMF). The PMF defines the probability of each possible outcome. In the above example, with $$n=10$$ trials, there are $$11$$ possible outcomes: $$y=0,1,2,...,10$$ successes. Which of these outcomes is most probable depends on the parameter $$\theta$$ in the binomial distribution that represents the probability of success.

The left-hand side plot in Figure 1.2 shows an example of a binomial PMF with $$10$$ trials, with the parameter $$\theta$$ fixed at $$0.5$$. Setting $$\theta$$ to $$0.5$$ leads to a PMF where the most probable outcome is $$5$$ successes out of $$10$$. If we had set $$\theta$$ to, say 0.1, then the most probable outcome would be $$1$$ success out of $$10$$; and if we had set $$\theta$$ to $$0.9$$, then the most probable outcome would be $$9$$ successes out of $$10$$.   FIGURE 1.2: Probability mass functions of a binomial distribution assuming 10 trials, with 50%, 10%, and 90% probability of success.

The probability mass function for the binomial is written as follows.

$\begin{equation} \mathit{Binomial}(k|n,\theta) = \binom{n}{k} \theta^{k} (1-\theta)^{n-k} \end{equation}$

Here, $$n$$ represents the total number of trials, $$k$$ the number of successes (this could range from 0 to 10), and $$\theta$$ the probability of success. The term $$\binom{n}{k}$$, pronounced n-choose-k, represents the number of ways in which one can choose $$k$$ successes out of $$n$$ trials. For example, 1 success out of 10 can occur in 10 possible ways: the very first trial could be a 1, the second trial could be a 1, etc. The term $$\binom{n}{k}$$ expands to $$\frac{n!}{k!(n-k)!}$$. In R, it is computed using the function choose(n,k), with $$n$$ and $$k$$ representing positive integer values.

When we want to express the fact that the data is assumed to be generated from a binomial random variable, we will write $$Y \sim \mathit{Binomial}(n,\theta)$$. If the data is generated from a random variable that has some other probability distribution $$f(\theta)$$, we will write $$Y\sim f(\theta)$$. In this book, we use $$f(\cdot)$$ synonymously with $$p(\cdot)$$ to represent a probability distribution.

### 1.4.1 The mean and variance of the binomial distribution

It is possible to analytically compute the mean (expectation) and variance of the PMF associated with the binomial random variable $$Y$$.

The expectation of a discrete random variable $$Y$$ with probability mass function f(y), is defined as

$\begin{equation} E[Y] = \sum_y y \cdot f(y) \end{equation}$

As a simple example, suppose that we toss a fair coin once. The possible outcomes are Tails (represented as 0) and Heads (represented as 1), each with equal probability, 0.5. The expectation is:

$\begin{equation} E[Y] = \sum_{y=0}^{1} y \cdot f(y) = 0\cdot 0.5 + 1\cdot 0.5 = 0.5 \end{equation}$

The expectation has the interpretation that if we were to do the experiment a large number of times and calculate the sample mean of the observations, in the long run we would approach the value 0.5. Another way to look at the above definition is that the expectation gives us the weighted mean of the possible outcomes, weighted by the respective probabilities of each outcome.

Without getting into the details of how these are derived mathematically (Kerns 2014), we just state here that the mean of $$Y$$ (the expectation $$E[Y]$$) and the variance of $$Y$$ (written $$Var(Y)$$) of a binomial distribution with parameter $$\theta$$ and $$n$$ trials are $$E[Y] = n\theta$$ and $$Var(Y) = n\theta (1-\theta)$$.

In the binomial example above, $$n$$ is a fixed number because we decide on the total number of trials before running the experiment. In the PMF, $$\binom{n}{k} \theta^{k} (1-\theta)^{n-k}$$, $$\theta$$ is also a fixed value; the only variable in a PMF is $$k$$. In real experimental situations, we never know the true value of $$\theta$$. But $$\theta$$ can be estimated from the data. From the observed data, we can compute the estimate of $$\theta$$, $$\hat \theta=k/n$$. The quantity $$\hat \theta$$ is the observed proportion of successes, and is called the maximum likelihood estimate of the true (but unknown) parameter $$\theta$$. Once we have estimated $$\theta$$ in this way, we can also obtain an estimate of the variance by computing $$n\hat\theta (1-\hat\theta)$$. These estimates are then used for statistical inference.

What does the term “maximum likelihood estimate” mean? In order to understand this term, it is necessary to first understand what a likelihood function is. Recall that in the discussion above, the PMF $$p(k|n,\theta)$$ assumes that $$\theta$$ and $$n$$ are fixed, and $$k$$ will vary from 0 to 10 when the experiment is repeated multiple times.

The likelihood function refers to the PMF $$p(k|n,\theta)$$, treated as a function of $$\theta$$. Once we have observed a particular value for $$k$$, this value is now fixed, along with $$n$$. Once $$k$$ and $$n$$ are fixed, the function $$p(k|n,\theta)$$ only depends on $$\theta$$. Thus, the likelihood function is the same function as the PMF, but assumes that the data is fixed and only the parameter $$\theta$$ varies (from 0 to 1). The likelihood function is written $$\mathcal{L}(\theta| k,n)$$, or simply $$\mathcal{L}(\theta)$$.

For example, suppose that we record $$n=10$$ trials, and observe $$k=7$$ successes. The likelihood function is:

$\begin{equation} \mathcal{L}(\theta|k=7,n=10)= \binom{10}{7} \theta^{7} (1-\theta)^{10-7} \end{equation}$

If we now plot the likelihood function for all possible values of $$\theta$$ ranging from $$0$$ to $$1$$, we get the plot shown in Figure 1.3. FIGURE 1.3: The likelihood function for 7 successes out of 10.

What is important about this plot is that it shows that, given the data, the maximum point is at the point $$0.7$$, which corresponds to the estimated mean using the formula shown above: $$k/n = 7/10$$. Thus, the maximum likelihood estimate (MLE) gives us the most likely value that the parameter $$\theta$$ has, given the data. In the binomial, the proportion of successes $$k/n$$ can be shown to be the maximum likelihood estimate of the parameter $$\theta$$ (e.g., see p. 339-340 of Miller and Miller 2004).

A crucial point: the “most likely” value of the parameter is with respect to the data at hand. The goal is to estimate an unknown parameter value from the data. This parameter value is chosen such that the probability (discrete case) or probability density (continuous case) of getting the sample values (i.e., the data) is a maximum. This parameter value is the maximum likelihood estimate (MLE).

This MLE from a particular sample of data need not invariably give us an accurate estimate of $$\theta$$. For example, if we run our experiment for $$10$$ trials and get $$1$$ success out of $$10$$, the MLE is $$0.10$$. We could have just happened to observe only one success out of ten by chance, even if the true $$\theta$$ were $$0.7$$. If we were to repeatedly run the experiment with increasing sample sizes, as the sample size increases, the MLE would converge to the true value of the parameter. Figure 1.4 illustrates this point. The key point here is that with a smaller sample size, the MLE from a particular data-set may or may not point to the true value. FIGURE 1.4: The plot shows the estimate of the mean proportion of successes sampled from a binomial distribution with true probability of success 0.7, with increasing sample sizes. As the sample size increases, the estimate converges to the true value of 0.7.

### 1.4.2 What information does a probability distribution provide?

In Bayesian data analysis, we will constantly be asking the question: what information does a probability distribution give us? In particular, we will treat each parameter $$\theta$$ as a random variable; this will raise questions like: “what is the probability that the parameter $$\theta$$ lies between two values $$a$$ and $$b$$”; and “what is the range over which we can be 95% certain that the true value of the parameter lies”? In order to be able to answer questions like these, we need to know what information we can obtain once we have decided on a probability distribution that is assumed to have generated the data, and how to extract this information using R. We therefore discuss the different kinds of information we can obtain from a probability distribution. For now we focus only on the binomial random variable introduced above.

#### 1.4.2.1 Compute the probability of a particular outcome (discrete case only)

The binomial distribution shown in Figure 1.2 already shows the probability of each possible outcome under a different value for $$\theta$$. In R, there is a built-in function that allows us to calculate the probability of $$k$$ successes out of $$n$$, given a particular value of $$k$$ (this number constitutes our data), the number of trials $$n$$, and given a particular value of $$\theta$$; this is the dbinom function. For example, the probability of 5 successes out of 10 when $$\theta$$ is 0.5 is:

dbinom(5, size = 10, prob = 0.5)
##  0.246

The probabilities of success when $$\theta$$ is 0.1 or 0.9 can be computed by replacing 0.5 above by each of these probabilities. One can just do this by giving dbinom a vector of probabilities:

dbinom(5, size = 10, prob = c(0.1, 0.9))
##  0.00149 0.00149

The probability of a particular outcome like $$k=5$$ successes is only computable in the discrete case. In the continuous case, the probability of obtaining a particular point value will always be zero (we discuss this when we turn to continuous probability distributions below).

#### 1.4.2.2 Compute the cumulative probability of k or less (more) than k successes

Using the dbinom function, we can compute the cumulative probability of obtaining 1 or less, 2 or less successes etc. This is done through a simple summation procedure:

## the cumulative probability of obtaining
## 0, 1, or 2 successes out of 10,
## with theta=0.5:
dbinom(0, size = 10, prob = 0.5) +
dbinom(1, size = 10, prob = 0.5) +
dbinom(2, size = 10, prob = 0.5)
##  0.0547

Mathematically, we could write the above summation as:

$\begin{equation} \sum_{k=0}^2 \binom{n}{k} \theta^{k} (1-\theta)^{n-k} \end{equation}$

An alternative to the cumbersome addition in the R code above is this more compact statement, which is identical to the above mathematical expression:

sum(dbinom(0:2, size = 10, prob = 0.5))
##  0.0547

R has a built-in function called pbinom that does this summation for us. If we want to know the probability of $$2$$ or fewer than 2 successes as in the above example, we can write:

pbinom(2, size = 10, prob = 0.5, lower.tail = TRUE)
##  0.0547

The specification lower.tail = TRUE (the default value) ensures that the summation goes from $$2$$ to numbers smaller than $$2$$ (which lie in the lower tail of the distribution in Figure 1.2). If we wanted to know what the probability is of obtaining $$3$$ or more successes out of $$10$$, we can set lower.tail to FALSE:

pbinom(2, size = 10, prob = 0.5, lower.tail = FALSE)
##  0.945
## equivalently:
## sum(dbinom(3:10,size = 10, prob = 0.5))

The cumulative distribution function or CDF can be plotted by computing the cumulative probabilities for any value $$k$$ or less than $$k$$, where $$k$$ ranges from $$0$$ to $$10$$ in our running example. The CDF is shown in Figure 1.5. FIGURE 1.5: The cumulative distribution function for a binomial distribution assuming 10 trials, with 50% probability of success.

#### 1.4.2.3 Compute the inverse of the cumulative distribution function (the quantile function)

We can also find out the value of the variable $$k$$ (the quantile) such that the probability of obtaining $$k$$ or less than $$k$$ successes is some specific probability value $$p$$. If we switch the x and y axes of Figure 1.5, we obtain another very useful function, the inverse of the CDF.

The inverse of the CDF (known as the quantile function in R because it returns the quantile, the value $$k$$) is available in R as the function qbinom. The usage is as follows: to find out what the value $$k$$ of the outcome is such that the probability of obtaining $$k$$ or less successes is $$0.37$$, type:

qbinom(0.37, size = 10, prob = 0.5)
##  4

One can visualize the inverse CDF of the binomial as in Figure 1.6. FIGURE 1.6: The inverse CDF for the binomial(size=10,prob=0.5).

#### 1.4.2.4 Generate simulated data from a $$\mathit{Binomial}(n,\theta)$$ distribution

We can generate simulated data from a binomial distribution by specifying the number of trials and the probability of success $$\theta$$. In R, we do this as follows:

rbinom(n = 1, size = 10, prob = 0.5)
##  7

The above code generates the number of successes in an experiment with $$10$$ trials. Repeatedly run the above code; we will get different numbers of successes each time.

As mentioned earlier, if there is only one trial, then instead of the binomial distribution, we have a Bernoulli distribution. For example, if we have 10 observations from a Bernoulli distribution, where the probability of success is 0.5, we can simulate data as follows using the function rbern from the package extraDistr.

rbern(n=10,prob=0.5)
##   0 1 1 0 1 1 0 0 1 1

The above kind of output can also be generated by using the rbinom function: rbinom(n = 10, size = 1, prob = 0.5). When the data are generated using the rbinom function in this way, one can calculate the number of successes by just summing up the vector, or computing its mean and multiplying by the number of trials, here $$10$$:

(y <- rbinom(n = 10, size = 1, prob = 0.5))
##   0 1 1 1 0 1 0 0 0 0
mean(y) * 10
##  4
sum(y)
##  4

## 1.5 Continuous random variables: An example using the normal distribution

We will now revisit the idea of the random variable using a continuous distribution. Imagine that we have a vector of reading time data $$y$$ measured in milliseconds and coming from a normal distribution. The normal distribution is defined in terms of two parameters: the location, its mean value $$\mu$$, which determines its center, and the scale, its standard deviation, $$\sigma$$, which determines how much spread there is around this center point.

The probability density function (PDF) of the normal distribution is defined as follows:

$\begin{equation} \mathit{Normal}(y|\mu,\sigma)=f(y)= \frac{1}{\sqrt{2\pi \sigma^2}} \exp \left(-\frac{(y-\mu)^2}{2\sigma^2} \right) \end{equation}$

Here, $$\mu$$ is some true, unknown mean, and $$\sigma^2$$ is some true, unknown variance of the normal distribution that the reading times have been sampled from. There is a built-in function in R that computes the above function once we specify the mean $$\mu$$ and the standard deviation $$\sigma$$ (in R, this parameter is specified in terms of the standard deviation rather than the variance).

Figure 1.7 visualizes the normal distribution for particular values of $$\mu$$ and $$\sigma$$, as a PDF (using dnorm), a CDF (using pnorm), and the inverse CDF (using qnorm). It should be clear from the figure that these are three different ways of looking at the same information.   FIGURE 1.7: The PDF, CDF, and inverse CDF for the $$\mathit{Normal}(\mu=500,\sigma=100)$$.

As in the discrete example, the PDF, CDF, and inverse of the CDF allow us to ask questions like:

• What is the probability of observing values between $$a$$ and $$b$$ from a normal distribution with mean $$\mu$$ and standard deviation $$\sigma$$? Using the above example, we can ask what the probability of observing values between 200 and 700 ms:
pnorm(700, mean = 500, sd = 100) - pnorm(200, mean = 500, sd = 100)
##  0.976

The probability of any point value in a PDF is always $$0$$. This is because the probability in a continuous probability distribution is the area under the curve, and the area at any point on the x-axis is always $$0$$. The implication here is that it is only meaningful to ask about probabilities between two different point values; e.g., the probability that $$Y$$ lies between $$a$$ and $$b$$, or $$P(a<Y<b)$$. Notice that $$P(a<Y<b)$$ is the same statement as $$P(a\leq Y\leq b)$$.

• What is the quantile $$q$$ such that the probability is $$p$$ of observing that value $$q$$ or a value more extreme than $$q$$? For example, we can work out the quantile $$q$$ such that the probability of observing $$q$$ or some value less than it is 0.975, in the normal(500,100) distribution. Formally, we would write this as $$P(Y<a)$$.
qnorm(0.975, mean = 500, sd = 100)
##  696

The above output says that the quantile value $$q$$ such that $$Prob(X<q)=0.975$$ is $$q=696$$.

• Generate simulated data. Given a vector of $$n$$ independent and identically distributed data $$y$$, i.e., given that each data point is being generated independently from $$Y \sim \mathit{Normal}(\mu,\sigma)$$ for some values of the parameters, the sample mean and standard deviation4 are:

$\begin{equation} \bar{y} = \frac{\sum_{i=1}^n y_i}{n} \end{equation}$

$\begin{equation} sd(y) = \sqrt{\frac{\sum_{i=1}^n (y_i- \bar{y})^2}{n}} \end{equation}$

For example, we can generate $$10$$ data points using the rnorm function, and then use the simulated data to compute the mean and standard deviation:

y <- rnorm(10, mean = 500, sd = 100)
mean(y)
##  532
sd(y)
##  80.6

Again, the sample mean and sample standard deviation computed from a particular (simulated or real) data set need not necessarily be close to the true values of the respective parameters. Especially when sample size is small, one can end up with mis-estimates of the mean and standard deviation.

Incidentally, simulated data can be used to generate all kinds of statistics. For example, we can compute the lower and upper quantiles such that 95% of the simulated data are contained within these quantiles:

quantile(y, probs = c(0.025, 0.975))
##  2.5% 97.5%
##   427   673

Later on, we will be using samples to produce summary statistics like the ones shown above.

### 1.5.1 An important distinction: probability vs. density in a continuous random variable

In continuous distributions like the normal discussed above, it is important to understand that the probability density function or PDF, $$p(y| \mu, \sigma)$$ defines a mapping from the $$y$$ values (the possible values that the data can have) to a quantity called the density of each possible value. We can see this function in action when we use dnorm to compute, say, the density value corresponding to $$y=1$$ and $$y=2$$ in the standard normal distribution.

## density:
dnorm(1, mean = 0, sd = 1)
##  0.242
dnorm(2, mean = 0, sd = 1)
##  0.054

If the density at a particular point value like 1 is high compared to some other value – 2 in the above example – then this point value 1 has a higher likelihood than 2 in the standard normal distribution.

The quantity computed for the values 1 and 2 are not the probability of observing 1 or 2 in this distribution. As mentioned earlier, probability in a continuous distribution is defined as the area under the curve, and this area will always be zero at any point value (because there are infinitely many different possible values). If we want to know the probability of obtaining values between an upper and lower bound $$b$$ and $$a$$, i.e., $$P(a<Y<b)$$ where these are two distinct values, we must use the cumulative distribution function or CDF: in R, for the normal distribution, this is the pnorm function. For example, the probability of observing a value between +2 and -2 in a normal distribution with mean 0 and standard deviation 1 is:

pnorm(2, mean = 0, sd = 1) - pnorm(-2, mean = 0, sd = 1)
##  0.954

The situation is different in discrete random variables. These have a probability mass function (PMF) associated with them—an example is the binomial distribution that we saw earlier. There, the PMF maps the possible $$y$$ values to the probabilities of those values occurring. That is why, in the binomial distribution, the probability of observing exactly 2 successes when sampling from a $$\mathit{Binomial}(n=10,\theta=0.5)$$ can be computed using either dbinom or pbinom:

dbinom(2, size = 10, prob = 0.5)
##  0.0439
pbinom(2, size = 10, prob = 0.5) - pbinom(1, size = 10, prob = 0.5)
##  0.0439

In the second line of code above, we are computing the cumulative probability of observing two or less successes, minus the probability of observing one or less successes. This gives us the probability of observing exactly two successes. The dbinom gives us this same information.

### 1.5.2 Truncating a normal distribution

In the above discussion, the support for the normal distribution ranges from minus infinity to plus infinity. One can define PDFs with a more limited support; an example would be a normal distribution whose PDF $$f(x)$$ is such that the lower bound is truncated at $$0$$ to allow only positive values. In such a case, the area under the range minus infinity to zero ($$\int_{-\infty}^0 f(x) \, dx$$) will be $$0$$ because the range lies outside the support of the truncated normal distribution. Also, if one truncates a probability density function like the standard normal ($$\mathit{Normal}(0,1)$$) at $$0$$, in order to make the area between zero and plus infinity sum up to $$1$$, we would have to multiply the truncated distribution f(x) by some factor $$k$$ such that the following integral sums to $$1$$:

$\begin{equation} k \int_{0}^{\infty} f(x)\, dx = 1 \tag{1.1} \end{equation}$

Clearly, this factor is $$k = \frac{1}{\int_{0}^{\infty} f(x)\, dx}$$. For the standard normal, this integral is easy to compute using R; we just calculate the complement of the cumulative distribution (CCDF):

pnorm(0, mean = 0, sd = 1, lower.tail = FALSE)
##  0.5
## alternatively:
1 - pnorm(0, mean = 0, sd = 1, lower.tail = TRUE)
##  0.5

Also, if we had truncated the distribution at 0 to the right instead of the left (allowing only negative values), we would have to find the factor $$k$$ in the same way as above, except that we would have to find $$k$$ such that:

$\begin{equation} k \int_{-\infty}^{0} f(x)\, dx = 1 \end{equation}$

For the standard normal case, in R, this factor would require us to use the CDF:

pnorm(0, mean = 0, sd = 1, lower.tail = TRUE)
##  0.5

Later in this book, we will be using such truncated distributions when doing Bayesian modeling, and when we use them, we will want to multiply the truncated distribution by the factor $$k$$ to ensure that it is still a proper PDF that sums to $$1$$. Truncated normal distributions will be discussed in more detail in Box 4.1.

## 1.6 Bivariate and multivariate distributions

So far, we have only discussed univariate distributions; these are distributions that involve only one variable. For example, when we talk about data generated from a Binomial distribution, or from a Normal distribution, these are univariate distributions.

It is also possible to specify distributions with two or more dimensions. Some examples will make it clear what a bivariate (or more generally, multivariate) distribution is.

### 1.6.1 Example 1: Discrete bivariate distributions

Starting with the discrete case, consider the discrete bivariate distribution shown below. These are data from an experiment where, inter alia, in each trial a Likert acceptability rating and a question-response accuracy were recorded (the data are from a study by Laurinavichyute (2020), used with permission here). Load the data by loading the R package bcogsci.

data("df_discreteagrmt")

Figure 1.8 shows the joint probability mass function of two random variables X and Y. The random variable X consists of 7 possible values (this is the 1-7 Likert response scale), and the random variable Y is question-response accuracy, with 0 representing an incorrect response, and 1 representing a correct response. One can also display Figure 1.8 as a table; see Table 1.1. FIGURE 1.8: Example of a discrete bivariate distribution. In these data, in every trial, two pieces of information were collected: Likert responses and yes-no question responses. The random variable X represents Likert scale responses on a scale of 1-7. and the random variable Y represents 0, 1 (incorrect, correct) responses to comprehension questions.

TABLE 1.1: The joint PMF for two random variables X and Y.
$$x=1$$ $$x=2$$ $$x=3$$ $$x=4$$ $$x=5$$ $$x=6$$ $$x=7$$
$$y=0$$ $$0.018$$ $$0.023$$ $$0.04$$ $$0.043$$ $$0.063$$ $$0.049$$ $$0.055$$
$$y=1$$ $$0.031$$ $$0.053$$ $$0.086$$ $$0.096$$ $$0.147$$ $$0.153$$ $$0.142$$

For each possible pair of values of X and Y, we have a joint probability $$p_{X,Y}(x,y)$$. Given such a bivariate distribution, there are two useful quantities we can compute: the marginal distributions ($$p_{X}$$ and $$p_Y$$), and the conditional distributions ($$p_{X|Y}$$ and $$p_{Y|X}$$). Table 1.1 shows the joint probability mass function $$p_{X,Y}(x,y)$$.

#### 1.6.1.1 Marginal distributions

The marginal distribution $$p_Y$$ is defined as follows. $$S_{X}$$ is the support of X, i.e., all the possible values of X.

$\begin{equation} p_{Y}(y)=\sum_{x\in S_{X}}p_{X,Y}(x,y).\label{eq-marginal-pmf} \end{equation}$

Similarly, the marginal distribution $$p_X$$ is defined as:

$\begin{equation} p_{X}(x)=\sum_{y\in S_{Y}}p_{X,Y}(x,y).\label{eq-marginal-pmf2} \end{equation}$

$$p_Y$$ is computed, by summing up the rows; and $$p_X$$ by summing up the columns. We can see why this is called the marginal distribution; the result appears in the margins of the table. In the code below, the object probs contains bivariate PMF shown in Table 1.1.

# P(Y)
(PY <- rowSums(probs))
##   y=0   y=1
## 0.291 0.709
sum(PY) ## sums to 1
##  1
# P(X)
(PX <- colSums(probs))
##    x=1    x=2    x=3    x=4    x=5    x=6    x=7
## 0.0491 0.0766 0.1257 0.1394 0.2102 0.2020 0.1969
sum(PX) ## sums to 1
##  1

The marginal probabilities sum to 1, as they should. Table 1.2 shows the marginal probabilities.

TABLE 1.2: The joint PMF for two random variables X and Y, along with the marginal distributions of X and Y.
$$x=1$$ $$x=2$$ $$x=3$$ $$x=4$$ $$x=5$$ $$x=6$$ $$x=7$$ $$P(Y)$$
$$y=0$$ $$0.018$$ $$0.023$$ $$0.04$$ $$0.043$$ $$0.063$$ $$0.049$$ $$0.055$$ $$0.291$$
$$y=1$$ $$0.031$$ $$0.053$$ $$0.086$$ $$0.096$$ $$0.147$$ $$0.153$$ $$0.142$$ $$0.709$$
$$P(X)$$ $$0.049$$ $$0.077$$ $$0.126$$ $$0.139$$ $$0.21$$ $$0.202$$ $$0.197$$

To compute the marginal distribution of X, one is summing over all the Ys; and to compute the marginal distribution of Y, one sums over all the X’s. We say that we are marginalizing out the random variable that we are summing over. One can also visualize the two marginal distributions using barplots (Figure 1.9). FIGURE 1.9: The marginal distributions of the random variables X and Y, presented as barplots.

#### 1.6.1.2 Conditional distributions

For computing conditional distributions, recall that conditional probability (see section 1.2) is defined as:

$\begin{equation} p_{X\mid Y}(x\mid y) = \frac{p_{X,Y}(x,y)}{p_Y(y)} \end{equation}$

and

$\begin{equation} p_{Y\mid X}(y\mid x) = \frac{p_{X,Y}(x,y)}{p_X(x)} \end{equation}$

The conditional distribution of a random variable $$X$$ given that $$Y=y$$, where $$y$$ is some specific (fixed) value, is:

$\begin{equation} p_{X\mid Y} (x\mid y) = \frac{p_{X,Y}(x,y)}{p_Y(y)} \quad \hbox{provided } p_Y(y)=P(Y=y)>0 \end{equation}$

As an example, let’s consider how $$p_{X\mid Y}$$ would be computed. The possible values of $$y$$ are $$0,1$$, and so we have to find the conditional distribution (defined above) for each of these values. I.e., we have to find $$p_{X\mid Y}(x\mid y=0)$$, and $$p_{X\mid Y}(x\mid y=1)$$.

Let’s do the calculation for $$p_{X\mid Y}(x\mid y=0)$$.

$\begin{equation} \begin{split} p_{X\mid Y} (1\mid 0) =& \frac{p_{X,Y}(1,0)}{p_Y(0)}\\ =& \frac{0.018}{0.291}\\ =& 0.062 \end{split} \end{equation}$

This conditional probability value will occupy the cell X=1, Y=0 in Table 1.3 summarizing the conditional probability distribution $$p_{X|Y}$$. In this way, one can fill in the entire table, which will then represent the conditional distributions $$p_{X|Y=0}$$ and $$p_{X|Y=1}$$. The reader may want to take a few minutes to complete Table 1.3.

TABLE 1.3: A table for listing conditional distributions of X given Y.
$$x=1$$ $$x=2$$ $$x=3$$ $$x=4$$ $$x=5$$ $$x=6$$ $$x=7$$
$$p_{X|Y(x|y=0)}$$ $$0.062$$
$$p_{X|Y(x|y=1)}$$

Similarly, one can construct a table that shows $$p_{Y|X}$$.

#### 1.6.1.3 Covariance and correlation

Here, we briefly define the covariance and correlation of two discrete random variables. For detailed examples and discussion, see the references at the end of the chapter. Informally, if there is a high probability that large values of a random variable $$X$$ are associate with large values of another random variable $$Y$$, we will say that the covariance between the two random variable $$X$$ and $$Y$$, written $$Cov(X,Y)$$, is positive.

The covariance of two (discrete) random variables $$X$$ and $$Y$$ is defined as follows. $$E[\cdot]$$ refers to the expectation of a random variable.

$\begin{equation} Cov(X,Y) = E[(X - E[X])(Y-E[Y])] \end{equation}$

It is possible to show that this is equivalent to:

$\begin{equation} Cov(X,Y) = E[XY] - E[X]E[Y] \end{equation}$

The expectation E[XY] is defined to be:

$\begin{equation} E[XY]=\sum_x \sum_y xy f_{X,Y}(x,y) \end{equation}$

If the standard deviations of the two random variables is $$\sigma_X$$ and $$\sigma_Y$$, the correlation between the two random variables, $$\rho_{XY}$$, is defined as:

$\begin{equation} \rho_{XY} = \frac{Cov(X,Y)}{\sigma_X\sigma_Y} \end{equation}$

### 1.6.2 Example 2: Continuous bivariate distributions

Consider now the continuous bivariate case; this time, we will use simulated data. Consider two normal random variables $$X$$ and $$Y$$, each of which coming from, for example, a $$\mathit{Normal}(0,1)$$ distribution, with some correlation $$\rho_{X,Y}$$ between the two random variables.

A bivariate distribution for two random variables $$X$$ and $$Y$$, each of which comes from a normal distribution, is expressed in terms of the means and standard deviations of each of the two distributions, and the correlation $$\rho_{XY}$$ between them. The standard deviations and correlation are expressed in a special form of a $$2\times 2$$ matrix called a variance-covariance matrix $$\Sigma$$. If $$\rho_{XY}$$ is the correlation between the two random variables, and $$\sigma _{X}$$ and $$\sigma _{Y}$$ the respective standard deviations, the variance-covariance matrix is written as:

$\begin{equation}\label{eq:covmatfoundations} \Sigma = \begin{pmatrix} \sigma _{X}^2 & \rho_{XY}\sigma _{X}\sigma _{Y}\\ \rho_{XY}\sigma _{X}\sigma _{Y} & \sigma _{Y}^2\\ \end{pmatrix} \end{equation}$

The off-diagonals of this matrix contain the covariance between $$X$$ and $$Y$$.

The joint distribution of $$X$$ and $$Y$$ is defined as follows:

$\begin{equation}\label{eq:jointpriordistfoundations} \begin{pmatrix} X \\ Y \\ \end{pmatrix} \sim \mathcal{N}_2 \left( \begin{pmatrix} 0 \\ 0 \\ \end{pmatrix}, \Sigma \right) \end{equation}$

The joint PDF is written with reference to the two variables $$f_{X,Y}(x,y)$$. It has the property that the area under the curve sums to 1—this sum-to-1 property is the same idea that we encountered in the univariate cases (the normal and binomial distributions), except that we are talking about a bivariate distribution here.

Formally, we would write the area under the curve as a double integral: we are summing up the volume under the curve for both $$X$$ and $$Y$$ (hence the two integrals).

$\begin{equation} \iint_{S_{X,Y}} f_{X,Y}(x,y)\, dx dy = 1 \end{equation}$

Here, the terms $$dx$$ and $$dy$$ express the fact that we are summing the area under the curve along the $$X$$ axis and the $$Y$$ axis.

The joint CDF would be written as follows. The equation below gives us the probability of observing a value like $$(u,v)$$ or some value smaller than that (i.e., some $$(u',v')$$, such that $$u'<u$$ and $$v'<v$$).

$\begin{equation} \begin{split} F_{X,Y}(u,v) =& P(X<u,Y<v)\\ =& \int_{-\infty}^u \int_{-\infty}^v f_{X,Y}(x,y)\, dy dx \hbox{ for } (x,y)\in \mathbb{R}^2\\ \end{split} \end{equation}$

Just as in the discrete case, the marginal distributions can be derived by marginalizing out the other random variable:

$\begin{equation} f_X(x) = \int_{S_Y} f_{X,Y}(x,y)\, dy, \quad f_Y(y) = \int_{S_X} f_{X,Y}(x,y)\, dx \end{equation}$

Here, $$S_X$$ and $$S_Y$$ are the respective supports.

Here, the integral sign $$\int$$ is the continuous equivalent of the summation sign $$\sum$$ in the discrete case. Luckily, we will never have to compute such integrals ourselves; but it is important to appreciate how a marginal distribution arises from a bivariate distribution—by integrating out or marginalizing out the other random variable.

A visualization will help. The figures below show a bivariate distribution with zero correlation (Figure 1.10), a negative (Figure 1.11) and a positive correlation (Figure 1.12). FIGURE 1.10: A bivariate normal distribution with zero correlation. Shown are four plots: the top-right plot shows the three-dimensional bivariate density, the top-left plot the contour plot of the distribution (seen from above). The lower plots show the cumulative distribution function from two views, as a three-dimensional plot and as a contour plot. FIGURE 1.11: A bivariate normal distribution with a negative correlation of -0.6. Shown are four plots: the top-right plot shows the three-dimensional bivariate density, the top-left plot the contour plot of the distribution (seen from above). The lower plots show the cumulative distribution function from two views, as a three-dimensional plot and as a contour plot. FIGURE 1.12: A bivariate normal distribution with a positive correlation of 0.6. Shown are four plots: the top-right plot shows the three-dimensional bivariate density, the top-left plot the contour plot of the distribution (seen from above). The lower plots show the cumulative distribution function from two views, as a three-dimensional plot and as a contour plot.

In this book, we will make use of such multivariate distributions a lot, and it will soon become important to know how to generate simulated bivariate or multivariate data that is correlated. So let’s look at that next.

### 1.6.3 Generate simulated bivariate (multivariate) data

Suppose we want to generate $$100$$ pairs of correlated data, with correlation $$\rho=0.6$$. The two random variables have mean $$0$$, and standard deviations $$5$$ and $$10$$ respectively.

Here is how we would generate such data. First, define a variance-covariance matrix; then, use the multivariate analog of the rnorm() function, mvrnorm(), from the MASS package to generate $$100$$ data points.

## define a variance-covariance matrix:
Sigma <- matrix(c(5^2, 5 * 10 * .6, 5 * 10 * .6, 10^2),
byrow = FALSE, ncol = 2
)
## generate data:
u <- mvrnorm(
n = 100,
mu = c(0, 0),
Sigma = Sigma
)
head(u, n = 3)
##       [,1]  [,2]
## [1,] -5.32 -9.66
## [2,]  1.73  5.14
## [3,]  2.07  3.16

Figure 1.13 confirms that the simulated data are positively correlated. FIGURE 1.13: The relationship between two positively correlated random variables, generated by simulating data using the R function mvrnorm from the MASS library.

One final useful fact about the variance-covariance matrix—one that we will need later—is that it can be decomposed into the component standard deviations and an underlying correlation matrix. For example, consider the matrix above:

Sigma
##      [,1] [,2]
## [1,]   25   30
## [2,]   30  100

One can decompose the matrix as follows. The matrix can be seen as the product of a diagonal matrix of the standard deviations and the correlation matrix:

$\begin{equation} \begin{pmatrix} 5 & 0\\ 0 & 10\\ \end{pmatrix} \begin{pmatrix} 1.0 & 0.6\\ 0.6 & 1.0\\ \end{pmatrix} \begin{pmatrix} 5 & 0\\ 0 & 10\\ \end{pmatrix} \end{equation}$

One can reassemble the variance-covariance matrix by pre-multiplying and post-multiplying the correlation matrix with the diagonal matrix containing the standard deviations:5

$\begin{equation} \begin{pmatrix} 5 & 0\\ 0 & 10\\ \end{pmatrix} \begin{pmatrix} 1.0 & 0.6\\ 0.6 & 1.0\\ \end{pmatrix} \begin{pmatrix} 5 & 0\\ 0 & 10\\ \end{pmatrix} = \begin{pmatrix} 25 & 30\\ 30 & 100\\ \end{pmatrix} \end{equation}$

Using R (the symbol %*% is the matrix multiplication operator in R):

## sds:
(sds <- c(5, 10))
##   5 10
## diagonal matrix:
(sd_diag <- diag(sds))
##      [,1] [,2]
## [1,]    5    0
## [2,]    0   10
## correlation matrix:
(corrmatrix <- matrix(c(1, 0.6, 0.6, 1), ncol = 2))
##      [,1] [,2]
## [1,]  1.0  0.6
## [2,]  0.6  1.0
sd_diag %*% corrmatrix %*% sd_diag
##      [,1] [,2]
## [1,]   25   30
## [2,]   30  100

## 1.7 An important concept: The marginal likelihood (integrating out a parameter)

Here, we introduce a concept that will turn up many times in this book. The concept we unpack here is called “integrating out a parameter”. We will need this when we encounter Bayes’ rule in the next chapter, and when we use Bayes factors later in the book (chapter 15).

Integrating out a parameter refers to the following situation. The example used here discusses the binomial distribution, but the approach is generally applicable for any distribution.

Suppose we have a binomial random variable $$Y$$ with PMF $$p(Y)$$. Suppose also that this PMF is defined in terms of parameter $$\theta$$ that can have only three possible values, $$0.1, 0.5, 0.9$$, each with equal probability. In other words, the probability that $$\theta$$ is $$0.1, 0.5,$$ or $$0.9$$ is 1/3 each.

We stick with our earlier example of $$n=10$$ trials and $$k=7$$ successes. The likelihood function then is

$\begin{equation} p(k=7,n=10|\theta) = \binom{10}{7} \theta^7 (1-\theta)^{3} \end{equation}$

There is a related concept of marginal likelihood, which we can write here as $$p(k=7,n=10)$$. Marginal likelihood is the likelihood computed by “marginalizing” out the parameter $$\theta$$: for each possible value that the parameter $$\theta$$ can have, we compute the likelihood at that value and multiply that likelihood with the probability/density of that $$\theta$$ value occurring. Then we sum up each of the products computed in this way. Mathematically, this means that we carry out the following operation.

In our example, there are three possible values of $$\theta$$, call them $$\theta_1=0.1$$, $$\theta_2=0.5$$, and $$\theta_3=0.9$$. Each has probability $$1/3$$; so $$p(\theta_1)=p(\theta_2)=p(\theta_3)=1/3$$. Given this information, we can compute the marginal likelihood as follows:

$\begin{equation} \begin{split} p(k=7,n=10) =& \binom{10}{7} \theta_1^7 (1-\theta_1)^{3} \times p(\theta_1) \\ +& \binom{10}{7} \theta_2^7 (1-\theta_2)^{3}\times p(\theta_2) \\ +& \binom{10}{7} \theta_3^7 (1-\theta_3)^{3}\times p(\theta_3) \end{split} \end{equation}$

Writing the $$\theta$$ values and their probabilities, we get:

$\begin{equation} \begin{split} p(k=7,n=10) =& \binom{10}{7} 0.1^7 (1-0.1)^{3} \times \frac{1}{3} \\ +& \binom{10}{7} 0.5^7 (1-0.5)^{3}\times \frac{1}{3} \\ +& \binom{10}{7} 0.9^7 (1-0.9)^{3}\times \frac{1}{3} \end{split} \end{equation}$

Taking the common factors ($$\frac{1}{3}$$ and $$\binom{10}{7}$$) out:

$\begin{equation} \begin{split} p(k=7,n=10) =& \frac{1}{3} \binom{10}{7}[ 0.1^7 (1-0.1)^{3} \\ +& 0.5^8 (1-0.5)^{3} \\ +& 0.9^8 (1-0.9)^{3}] \\ =& 0.058 \end{split} \end{equation}$

Thus, a marginal likelihood is a kind of weighted sum of the likelihood, weighted by the possible values of the parameter.6

The above example was contrived, because we stated that the parameter $$\theta$$ has only three possible values. In reality, because the parameter $$\theta$$ can have all possible values between 0 and 1, the summation has to be done over a continuous space $$[0,1]$$. The way this summation is expressed in mathematics is through the integral symbol:

$\begin{equation} p(k=7,n=10) = \int_0^1 \binom{10}{7} \theta^7 (1-\theta)^{3}\, d\theta \end{equation}$

This statement is computing something similar to what we computed above with the three discrete parameter values, except that the summation is being done over a continuous space ranging from 0 to 1. We say that the parameter $$\theta$$ has been integrated out, or marginalized. Integrating out a parameter will be a very common operation in this book, but we will never have to do the calculation ourselves. For the above case, we can compute the integral in R:

BinLik <- function(theta) {
choose(10, 7) * theta^7 * (1 - theta)^3
}
integrate(BinLik, lower = 0, upper = 1)\$value
##  0.0909

The value that is output by the integrate function above is the marginal likelihood.

This completes our discussion of random variables and probability distributions. We now summarize what we have learned so far.

## 1.8 Summary of useful R functions relating to distributions

Table 1.4 summarizes the different functions relating to PMFs and PDFs, using the binomial and normal as examples.

TABLE 1.4: Important R functions relating to random variables.
Discrete Continuous
Example: $$\mathit{Binomial}(y|n,\theta)$$ $$\mathit{Norma}l(y|\mu,\sigma)$$
Likelihood function dbinom dnorm
Prob Y=y dbinom always 0
Prob $$Y\geq y, Y\leq y, y_1<Y<y_2$$ pbinom pnorm
Inverse CDF qbinom qnorm
Generate simulated data rbinom rnorm

Later on, we will use other distributions, such as the Uniform, Beta, etc., and each of these has their own set of d-p-q-r functions in R. One can look up these different distributions in, for example, Blitzstein and Hwang (2014).

## 1.9 Summary

This chapter briefly reviewed some very basic concepts in probability theory, univariate discrete and continuous random variables, and bivariate distributions. An important set of functions we encountered are the d-p-q-r family of functions for different distributions; these are very useful for understanding the properties of commonly used distributions, visualizing distributions, and for simulating data. Distributions will play a central role in this book; for example, knowing how to visualize distributions will be important for deciding on prior distributions for parameters. Other important ideas we learned about were marginal and conditional probability, marginal likelihood, and how to define multivariate distributions; these concepts will play an important role in Bayesian statistics.

A quick review of the mathematical foundations needed for statistics is available in the short book by Fox (2009), as well as Gill (2006). Morin (2016) and Blitzstein and Hwang (2014) are accessible introductions to probability theory. Ross (2002) is a more advanced treatment which discusses random variable theory and illustrates applications of probability theory. A good formal introduction to mathematical statistics (covering classical frequentist theory) is Miller and Miller (2004). The freely available book by Kerns (2014) introduces frequentist and Bayesian statistics from the ground up in a very comprehensive and systematic manner; the source code for the book is available from https://github.com/gjkerns/IPSUR. The open-access book, Probability and Statistics: a simulation-based introduction, by Bob Carpenter is also worth studying: https://github.com/bob-carpenter/prob-stats. A thorough introduction to the matrix algebra needed for statistics, with examples using R, is provided in Fieller (2016). Commonly used probability distributions are presented in detail in Miller and Miller (2004), Blitzstein and Hwang (2014), and Ross (2002).

## 1.11 Exercises

Exercise 1.1 Practice using the pnorm function - Part 1

Given a normal distribution with mean 500 and standard deviation 100, use the pnorm function to calculate the probability of obtaining values between 200 and 800 from this distribution.

Exercise 1.2 Practice using the pnorm function - Part 2

Calculate the following probabilities. Given a normal distribution with mean 800 and standard deviation 150, what is the probability of obtaining:

• a score of 700 or less
• a score of 900 or more
• a score of 800 or more

Exercise 1.3 Practice using the pnorm function - Part 3

Given a normal distribution with mean 600 and standard deviation 200, what is the probability of obtaining:

• a score of 550 or less.
• a score between 300 and 800.
• a score of 900 or more.

Exercise 1.4 Practice using the qnorm function - Part 1

Consider a normal distribution with mean 1 and standard deviation 1. Compute the lower and upper boundaries such that:

• the area (the probability) to the left of the lower boundary is 0.10.
• the area (the probability) to the left of the upper boundary is 0.90.

Exercise 1.5 Practice using the qnorm function - Part 2

Given a normal distribution with mean 650 and standard deviation 125. There exist two quantiles, the lower quantile q1 and the upper quantile q2, that are equidistant from the mean 650, such that the area under the curve of the normal between q1 and q2 is 80%. Find q1 and q2.

Exercise 1.6 Practice getting summaries from samples - Part 1

Given data that is generated as follows:

data_gen1 <- rnorm(1000, 300, 200)

Calculate the mean, variance, and the lower quantile q1 and the upper quantile q2, that are equidistant and such that the range of probability between them is 80%.

Exercise 1.7 Practice getting summaries from samples - Part 2.

This time we generate the data with a truncated normal distribution from the package extraDistr. The details of this distribution will be discussed later in section 4.1 and in the Box 4.1, but for now we can treat it as an unknown generative process:

data_gen1 <- rtnorm(1000, 300, 200, a = 0)

Using the sample data, calculate the mean, variance, and the lower quantile q1 and the upper quantile q2, such that the probability of observing values between these two quantiles is 80%.

Exercise 1.8 Practice with a variance-covariance matrix for a bivariate distribution.

Suppose that you have a bivariate distribution where one of the two random variables comes from a normal distribution with mean $$\mu_X=600$$ and standard deviation $$\sigma_X=100$$, and the other from a normal distribution with mean $$\mu_Y=400$$ and standard deviation $$\sigma_Y=50$$. The correlation $$\rho_{XY}$$ between the two random variables is $$0.4$$. Write down the variance-covariance matrix of this bivariate distribution as a matrix (with numerical values, not mathematical symbols), and then use it to generate $$100$$ pairs of simulated data points. Plot the simulated data such that the relationship between the random variables $$X$$ and $$Y$$ is clear. Generate two sets of new data ($$100$$ pairs of data points each) with correlation $$-0.4$$ and $$0$$, and plot these alongside the plot for the data with correlation $$0.4$$.

### References

Blitzstein, Joseph K, and Jessica Hwang. 2014. Introduction to Probability. Chapman; Hall/CRC.

Fieller, Nick. 2016. Basics of Matrix Algebra for Statistics with R. Boca Raton, FL: CRC Press.

Fox, John. 2009. A Mathematical Primer for Social Statistics. Vol. 159. Sage.

Gill, Jeff. 2006. Essential Mathematics for Political and Social Research. Cambridge University Press Cambridge.

Kerns, G.J. 2014. Introduction to Probability and Statistics Using R. Second Edition.

Kolmogorov, Andreĭ Nikolaevich. 1933. Foundations of the Theory of Probability: Second English Edition. Courier Dover Publications.

Laurinavichyute, Anna. 2020. “Similarity-Based Interference and Faulty Encoding Accounts of Sentence Processing.” Dissertation, University of Potsdam.

Miller, I., and M. Miller. 2004. John E. Freund’s Mathematical Statistics with Applications. Upper Saddle River, NJ: Prentice Hall.

Morin, David J. 2016. Probability: For the Enthusiastic Beginner. Createspace Independent Publishing Platform.

Ross, Sheldon. 2002. A First Course in Probability. Pearson Education.

Steyer, Rolf, and Werner Nagel. 2017. Probability and Conditional Expectation: Fundamentals for the Empirical Sciences. Vol. 5. John Wiley & Sons.

1. Here, we use $$Y$$, but we could have used any letter, such as $$X, Z,...$$. Later on, in some situations we will use Greek letters like $$\theta, \mu, \sigma$$ to represent a random variable.

2. The actual formal definition of random variable is more complex, and it is based on measure theory. A more rigorous definition can be found in, for example, Steyer and Nagel (2017).

3. A notational aside: In frequentist treatments, the PMF would be written $$p(y;\theta)$$, i.e., with a semi-colon rather than the conditional distribution marked by the vertical bar. The semi-colon is intended to indicate that in the frequentist paradigm, the parameters are fixed point values; by contrast, in the Bayesian paradigm, parameters are random variables. This has the consequence that for Bayesian, the distribution of $$y$$, $$p(y)$$ is really a conditional distribution, conditional on a random variable, here $$\theta$$. For the frequentist, $$p(y)$$ requires some point value for $$\theta$$, but it cannot be a conditional distribution because $$\theta$$ is not a random variable. We define conditional distributions later in this section.

4. R will compute the standard deviation by dividing by $$n-1$$, not $$n$$; this is because dividing by $$n$$ gives a biased estimate (chapter 10 of Miller and Miller 2004). This is not an important detail for our purposes, and in any case for large $$n$$ it doesn’t really matter whether one divides by $$n$$ or $$n-1$$.

5. There is a built-in convenience function, sdcor2cov in the SIN package that does this calculation, taking the vector of standard deviations (not the diagonal matrix) and the correlation matrix to yield the variance-covariance matrix: sdcor2cov(stddev = sds, corr = corrmatrix).

6. Where does the above formula come from? It falls out from the law of total probability discussed above!