5 Bivariate distributions

Upon completion of this chapter you should be able to:

• apply the concept of bivariate random variables.
• compute joint probability functions and the distribution function of two random variables.
• describe and apply the joint density–mass function for mixed bivariate random variables.
• find the marginal and conditional probability functions of random variables in both discrete and continuous cases.
• apply the concept of independence of two random variables.
• compute the expectation and variance of linear combinations of random variables.
• interpret and compute the covariance and the coefficient of correlation between two random variables.
• compute the conditional mean and conditional variance of a random variable for some given value of another random variable.

5.1 Introduction

Not all random processes are sufficiently simple to have the outcome denoted by a single variable \(X\). Many situations require observing two (or more) numerical characteristics simultaneously. This chapter discusses the two-dimensional (bivariate) case.

5.2 Bivariate random variables and probability distributions

The joint probability function simultaneously describes how two random variables vary and interact. The set of all possible pairs of outcomes, the joint range (or support) of \((X, Y)\), denoted \(\mathcal{R}_{X,Y}\), is a subset of the Euclidean plane \(\mathbb{R}^2\). Each outcome \((X(s), Y(s))\) (usually written as \((X, Y)\)) is represented as a point \((x, y)\) in this plane, and is called a random vector. As with univariate distributions, discrete and continuous random variables must be distinguished to determine how probabilities are assigned to these points.

Definition 5.1 (Random vector) Let \(X = X(s)\) and \(Y = Y(s)\) be two functions, each assigning a real number to each sample point \(s \in S\). The pair \((X, Y)\) is called a two-dimensional random variable, or a random vector. More formally, write \((X, Y): S \to \mathbb{R}^2\).

Example 5.1 (Bivariate discrete) Consider a random process where:

1. two coins are tossed, and \(X\) is the number of heads that show on the two coins; and
2. one die is rolled repeatedly, and \(Y\) is the number of rolls needed to roll a .

\(X\) is discrete with \(\mathcal{R}_X = \{0, 1, 2\}\). \(Y\) is discrete with a countably infinite range \(\mathcal{R}_Y = \{ 1, 2, 3, \dots\}\).

The range is \(\mathcal{R}_{X\times Y} = \{ (x, y)\mid x = 0, 1, 2; y = 1, 2, 3, \dots\}\).

As with the univariate case, the description and language for the probability function are different, depending on whether the random variables \(X\) and \(Y\) are discrete or continuous (though the ideas remain similar).

When \(X\) and \(Y\) are discrete, a joint probability mass function (joint PMF) is defined.

Definition 5.2 (Discrete joint probability mass function) Let \((X, Y)\) be a \(2\)-dimensional random variable. The joint probability mass function (or joint PMF), denoted \(p_{X, Y}(x, y)\), assigns a probability to each pair of outcomes \((x_i, y_j)\) such that \(p_{X, Y}(x_i, y_j) = \Pr(X = x_i \cap Y = y_j)\). This function must satisfy: \[\begin{align} p_{X, Y}(x, y) &\geq 0 \quad \text{for all } (x, y) \in \mathcal{R}_{X\times Y}\notag \\ \sum_{(x_i, y_j)\in \mathcal{R}_{X\times Y}} p_{X, Y}(x_i, y_j) &= 1. \tag{5.1} \end{align}\] The collection of all triples \(\{(x_i, y_j, p_{X,Y}(x_i, y_j))\}\) for all \(i, j\) constitutes the joint probability distribution of \((X, Y)\).

When both \(X\) and \(Y\) are continuous, a joint probability density function (joint PDF) is defined.

Definition 5.3 (Continuous joint probability density function) The joint probability density function (joint PDF) function of the pair of continuous random variables \(X\) and \(Y\) is a function \(f_{X, Y}(x, y)\) such that:

• \(f_{X, Y}(x, y) \ge 0\) for all \((x, y) \in \mathcal{R}_{X\times Y}\), and
• The total volume under the surface is \(1\): \[ \iint_{\mathcal{R}_{(X, Y)}} f_{X,Y}(x, y) \, dx \, dy = 1. \] The probability that \((X, Y)\) falls within a specific region \(A\) in the plane is the volume above that region: \[ \Pr((X, Y) \in A) = \iint_A f_{X,Y}(x, y) \, dx \, dy. \]

While not common, the case where one variable, say \(X\), is continuous and the other, say \(Y\), is discrete also occurs. This gives a mixed joint distribution. The total probability is still \(1\) by using a hybrid approach: \[ \sum_{x \in \mathcal{R}_X} \int_{-\infty}^{\infty} f_{X,Y}(x, y) \, dy = 1 \] where \(f_{X,Y}(x, y)\) is the joint density-mass function. To fully understand the mixed joint distribution case, conditional distributions must first be studied (Sect. 5.4), so this case is deferred until later (Sect. 5.5).

Example 5.2 (Bivariate discrete) Suppose the joint probability mass function for \((X, Y)\) is defined as \[ p_{X, Y}(x, y) = \frac{6(x + 1)}{47(y + 1)}\quad\text{for $(x, y) \in\mathbb{Z}$, $0 \le |x + y| \le 2$.} \] The joint PMF can also be described using a table (Table 5.1) or a graph (Fig. 5.1). Then, for example, \[ \Pr( \{X = 1, Y = 0, 1\} ) = \frac{12}{47} + \frac{6}{47} = \frac{18}{47}. \]

TABLE 5.1: The joint probability function for Example 5.2.

	\(x = 0\)	\(x = 1\)	\(x = 2\)
\(y = 0\)	\(6/47\)	\(12/47\)	\(18/47\)
\(y = 1\)	\(3/47\)	\(6/47\)	\(0\)
\(y = 2\)	\(2/47\)	\(0\)	\(0\)

FIGURE 5.1: A bivariate discrete probability function.

Example 5.3 (Bivariate uniform distribution) Consider the following continuous bivariate distribution with joint PDF \[ f_{X, Y}(x, y) = 1, \quad \text{for $0 \leq x \leq 1$ and $0 \leq y \leq 1$}. \] This is sometimes called the bivariate continuous uniform distribution (Fig. 5.2). The volume under the surface is one.

The probability \(\Pr(0 \leq x \leq \frac{1}{2}, 0 \leq y \leq \frac{1}{2})\) is the volume above the square with vertices \((0, 0), (0, 1/2), (1/2, 0), (1/2, 1/2)\). Hence the probability is \(1/4\).

FIGURE 5.2: The bivariate continuous uniform distribution.

Example 5.4 (Bivariate discrete) Consider a random process where two coins are tossed, and one die is rolled simultaneously (Example 5.1). Let \(X\) be the number of heads that show on the two coins, and \(Y\) the number on the die.

Since the toss of the coin and the roll of the die are independent, the probabilities are computed as follows: \[\begin{align*} \Pr(X = 0, Y = 1) &= \Pr(X = 0) \times \Pr(Y = 1) = \frac{1}{4}\times\frac{1}{6} = \frac{1}{24};\\ \Pr(X = 1, Y = 2) &= \Pr(X = 1) \times \Pr(Y = 2) = \frac{1}{2}\times\frac{1}{6} = \frac{1}{12}; \end{align*}\] and so on. The complete joint probability function can be displayed in a graph (often tricky), a function, or a table (Table 5.2). Here, the joint probability function could be given as the function \[ p_{X, Y}(x, y) = \begin{cases} \left(\frac{1}{12}\right) 0.5^{|x - 1|} & \text{for $(x, y)\in S$ defined earlier};\\ 0 & \text{elsewhere.} \end{cases} \]

TABLE 5.2: The joint PMF for Example 5.4.

	\(y = 1\)	\(y = 2\)	\(y = 3\)	\(y = 4\)	\(y = 5\)	\(y = 6\)	Total
\(x = 0\)	\(1/24\)	\(1/24\)	\(1/24\)	\(1/24\)	\(1/24\)	\(1/24\)	\(1/4\)
\(x = 1\)	\(1/12\)	\(1/12\)	\(1/12\)	\(1/12\)	\(1/12\)	\(1/12\)	\(1/2\)
\(x = 2\)	\(1/24\)	\(1/24\)	\(1/24\)	\(1/24\)	\(1/24\)	\(1/24\)	\(1/4\)
Total	\(1/6\)	\(1/6\)	\(1/6\)	\(1/6\)	\(1/6\)	\(1/6\)	\(1\)

Example 5.5 (Two dice) Consider the bivariate discrete distribution which results when two dice are thrown.

Let \(X\) be the number of times a appears, and \(Y\) the number of times a appears. The ranges of \(X\) and \(Y\) are \(\mathcal{R}_X = \{0, 1 ,2 \}\), \(\mathcal{R}_Y = \{0, 1, 2\}\) and the range for the random process is the Cartesian product of \(\mathcal{R}_X\) and \(\mathcal{R}_Y\), understanding that some of the resulting points may have probability zero. The probabilities in Table 5.3 are \(\Pr(X = x, Y = y)\) for the \((x, y)\) pairs in the range.

The probabilities are found by realising we really have two repetitions of a simple random process with three possible outcomes, \(\{5, 6, (\text{$5$ or $6$})^c \}\), with probabilities \(\frac{1}{6}, \frac{1}{6}, \frac{2}{3}\), the same for each repetition. (Recall: \((\text{$5$ or $6$})^c\) means ‘not (\(5\) or \(6\))’; see Def. 2.6.) Of course the event \(X = 2, Y = 1\) cannot occur in two trials, so has probability zero.

TABLE 5.3: Joint probability distribution for Example 5.5.

	\(x = 0\)	\(x = 1\)	\(x = 2\)
\(y = 0\)	\((2/3)^2\)	\(2(1/6)(2/3)\)	\((1/6)^2\)
\(y = 1\)	\(2(1/6)(2/3)\)	\(2(1/6)(1/6)\)	\(0\)
\(y = 2\)	\((1/6)^2\)	\(0\)	\(0\)

Example 5.5 is a special case of the multinomial distribution (a generalisation of the binomial distribution), described later (Sect. 11.9).

Example 5.6 (Banks) A bank operates both an ATM and an in-branch teller. On a randomly selected day, let \(X_1\) be the proportion of time the ATM is in use (at least one customer is being served or waiting to be served), and \(X_2\) is the proportion of time the teller is busy.

The set of possible values for \(X_1\) and \(X_2\) is the rectangle \(R = \{(x_1, x_2)\mid 0 \le x_1 \le 1, 0 \le x_2 \le 1\}\). From experience, the joint PDF of \((X_1, X_2)\) is \[ f_{X_1, X_2}(x_1, x_2) = \begin{cases} c(x_1 + x_2^2) & \text{for $0\le x_1\le 1$; $0\le x_2\le 1$};\\ 0 & \text{elsewhere.} \end{cases} \]

To determine a value for \(c\), first see that if \(f_{X_1, X_2}(x_1, x_2) \ge 0\) for all \(x_1\) and \(x_2\), then \(c > 0\); and \[ \int_{-\infty}^{\infty}\!\int_{-\infty}^{\infty} f_{X_1, X_2}(x_1, x_2)\, dx_1\,dx_2 = 1. \] Hence, \[\begin{align*} \int_{-\infty}^{\infty}\!\int_{-\infty}^{\infty} f_{X_1, X_2}(x_1, x_2)\, dx_1\,dx_2 &= \int_{0}^{1}\!\!\!\int_{0}^{1} f_{X_1, X_2}(x_1, x_2)\, dx_1\,dx_2 \\ &= c \int_{x_2 = 0}^{1}\left\{\int_{x_1=0}^{1} (x_1 + x_2^2)\, dx_1\right\} dx_2\\ &= c (1/2 + 1/3) = 5c/6, \end{align*}\] and so \(c = 6/5\).

Consider the probability neither facility is busy more than half the time. Mathematically, the question is asking to find \(\Pr( 0\le X_1\le 0.5, 0\le X_2\le 0.5)\); call this event \(A\). Then, \[\begin{align*} \Pr(A) &= \int_{0}^{0.5}\,\,\, \int_{0}^{0.5} f_{X_1, X_2}(x_1, x_2)\, dx_1\, dx_2 \\ &= \frac{6}{5} \int_{0}^{0.5}\left\{\int_{0}^{0.5} x_1 + x_2^2\, dx_1\right\} dx_2 \\ &= \frac{6}{5} \int_{0}^{0.5} (1/8 + x_2^2/2) \, dx_2 = 1/10. \end{align*}\]

5.3 Marginal distributions

With each two-dimensional random variable \((X, Y)\), two one-dimensional random variables (\(X\) and \(Y\)) can be described. The probability distributions of each of \(X\) and \(Y\) can be found separately.

In the case of a discrete random vector \((X, Y)\), the event \(X = x_i\) is the union of the mutually exclusive events \[ \{X = x_i, Y = y_1\}, \{\ X = x_i, Y = y_2\}, \{X = x_i, Y = y_3\}, \dots \] Thus, \[\begin{align*} \Pr(X = x_i) &= \Pr(X = x_i, Y = y_1) + \Pr(X = x_i, Y = y_2) + \dots \\ &= \sum_jp_{X, Y}(x_i, y_j), \end{align*}\] where the notation means to sum over all values given under the summation sign. Hence, the marginal distributions can be defined when \((X, Y)\) is a discrete random vector.

Definition 5.4 (Bivariate discrete marginal distributions) Given \((X, Y)\) with joint discrete probability function \(p_{X, Y}(x, y)\), the marginal probability functions of \(X\) and \(Y\) are, respectively \[\begin{equation} p_X(x) = \sum_{y}p_{X, Y}(x, y) \quad\text{and}\quad p_Y(y) = \sum_{x}p_{X, Y}(x, y). \tag{5.2} \end{equation}\]

An analogous definition exists when the random vector \((X,Y)\) is continuous.

Definition 5.5 (Bivariate continuous marginal distributions) If \((X, Y)\) has joint continuous PDF \(f_{X, Y}(x, y)\), the marginal PDFs of \(X\) and \(Y\), denoted by \(f_X(x)\), \(f_Y(y)\) respectively, are \[ f_X(x) = \int_{-\infty}^{\infty}f_{X, Y}(x,y) \, dy \quad\text{and}\quad f_Y(y) = \int_{-\infty}^{\infty}f_{X, Y}(x,y) \, dx. \]

Example 5.7 (Bivariate continuous marginal distribution) The joint probability density functions of \(X\) and \(Y\) is \[ f_{X, Y}(x, y) = \left\{ \begin{array}{ll} \frac{1}{3} (3x^2 + xy), & 0 \leq x \leq 1, \, 0 \leq y \leq 2;\\ 0 & \text{ elsewhere.} \end{array} \right. \] The marginal probability density function for \(X\) is \[\begin{align*} f_X(x) = \int_0^2\left(3x^2 + \frac{xy}{3}\right) dy &= \left.x^2y + \frac{xy^2}{6}\right|_{y = 0}^2\\ &= 2x^2 + \frac{2x}{3}\quad\text{for $0 \leq x \leq 1$}. \end{align*}\] Also, \[ f_Y(y) = \int_0^1\left(x^2 + \frac{xy}{3}\right)dx = \left.\frac{1}{3}x^3 + \frac{1}{6}x^2y\right|_{x = 0}^1. \] So \(\displaystyle f_Y(y) = \frac{1}{6}(2 + y)\), for \(0 \leq y \leq 2\).

Consider computing \(\Pr(Y < X)\); see Fig. 5.3. Then \[\begin{align*} \Pr(Y < X) &= \int \!\!\int_{\substack{(x, y) \in A\\ y < x}} f(x,y) \, dx \, dy \\ &= \frac{1}{3}\int_0^1 \int_y^1(3x^2 + xy) \, dx \, dy\\ &= \frac{1}{3} \int_0^1\left. x^3 + \frac{1}{2}x^2y\right|_y^1 dy\\ &= \frac{1}{3} \int_0^1(1 + \frac{1}{2}y - \frac{3}{2}y^3) \, dy = \frac{7}{24}. \end{align*}\]

FIGURE 5.3: The region where \(Y < X\).

Example 5.8 (Bivariate discrete marginal distributions) Recall Example 5.5, where two dice are rolled. We can find the marginal distributions of \(X\) and \(Y\) (Table 5.4). The probabilities in the first row (where \(Y = 0\)), for instance, are summed and appear as the first term in the final column; this is the marginal distribution for \(Y = 0\). Similarly for the other rows.

Recalling that \(X\) is the number of times a is rolled when two dice are thrown, the distribution of \(X\) should be \(\text{Bin}(2, 1/6\)); the probabilities given in the last row of the table agree with this. That is, \[ p_X(x) = \binom{2}{x}\left(\frac{1}{6}\right)^x \left(\frac{5}{6}\right)^{2 - x} \] for \(x = 0, 1, 2\). Of course, the distribution of \(Y\) is the same since each die shows a with the same probability \(1/6\) as a .

TABLE 5.4: The marginal distributions.

	\(x = 0\)	\(x = 1\)	\(x = 2\)	\(\Pr(Y = y)\)
\(y = 0\)	\(4/9\)	\(2/9\)	\(1/36\)	\(25/36\)
\(y = 1\)	\(2/9\)	\(1/18\)	\(0\)	\(10/36\)
\(y = 2\)	\(1/36\)	\(0\)	\(0\)	\(1/36\)
\(\Pr(X = x)\)	\(25/36\)	\(10/36\)	\(1/36\)	\(1\)

Example 5.9 (Bivariate discrete marginal distributions) Consider again the random process in Example 5.15. From Table 5.2, the marginal distribution for \(X\) is found simply by summing over the values for \(Y\) in the table. When \(x = 0\), \[ p_{X}(0) = \sum_{y=1}^6 p_{X, Y}(0, y) = 6 \times \frac{1}{24} = \frac{6}{24} = \frac{1}{4}. \] Likewise, \[\begin{align*} p_{X}(1) &= \sum_{y} p_{X, Y}(1, y) = 6/12;\\ p_{X}(2) &= \sum_{y} p_{X, Y}(2, y) = 6/24. \end{align*}\] So the marginal distribution of \(X\) is \[ p_{X}(x) = \begin{cases} 1/4 & \text{if $x = 0$};\\ 1/2 & \text{if $x = 1$};\\ 1/4 & \text{if $x = 2$};\\ 0 & \text{otherwise}.\\ \end{cases} \] In this example, the marginal distribution is easily found from the total column of Table 5.2.

5.4 Conditional distributions

Consider \((X, Y)\) with joint probability function as in Example 5.2, with marginal distributions of \(X\) and \(Y\) as shown in Table 5.5.

TABLE 5.5: A joint distribution with the marginal distributions.

	\(x = 0\)	\(x = 1\)	\(x = 2\)	\(\Pr(Y = y)\)
\(y = 0\)	\(6/47\)	\(12/47\)	\(18/47\)	\(36/47\)
\(y = 1\)	\(3/47\)	\(6/47\)	\(0\)	\(9/47\)
\(y = 2\)	\(2/47\)	\(0\)	\(0\)	\(2/47\)
\(\Pr(X = x)\)	\(11/47\)	\(18/47\)	\(18/47\)	\(1\)

Suppose we want to evaluate the conditional probability \(\Pr(X = 1 \mid Y = 1)\). We use that \(\Pr(A \mid B) = \Pr(A \cap B)/\Pr(B)\). So \[ \Pr(X = 1 \mid Y = 1) = \frac{\Pr(X = 1, Y = 1)}{\Pr(Y = 1)} = \frac{6/47}{9/47} = \frac{2}{3}. \] So, for each \(x\in \mathcal{R}_X\) we could find \(\Pr(X = x, Y = 1)\) and this is then the conditional distribution of \(X\) given that \(Y = 1\).

Definition 5.6 (Bivariate discrete conditional distributions) For a discrete random vector \((X, Y)\) with probability function \(p_{X, Y}(x, y)\) the conditional probability distribution of \(X\) given \(Y = y\) is defined by \[\begin{align*} p_{X \mid Y=y}(x) &= \Pr(X = x \mid Y = y)\\ &= \frac{\Pr(X = x, Y = y)}{\Pr(Y = y)}\\ &= \frac{p_{X, Y}(x, y)}{p_Y(y)} \end{align*}\] for \(x \in \mathcal{R}_X\) and provided \(p_Y(y) > 0\).

The continuous case is analogous.

Definition 5.7 (Bivariate continuous conditional distributions) If \((X, Y)\) is a continuous \(2\)-dimensional random variable with joint PDF \(f_{X, Y}(x, y)\) and respective marginal probability density functions \(f_X(x)\), \(f_Y(y)\), then the conditional probability distribution of \(X\) given \(Y = y\) is defined by \[ f_{X \mid Y=y}(x) = \frac{f_{X, Y}(x, y)}{f_Y(y)} \] for \(x \in \mathcal{R}_X\) and provided \(f_Y(y) > 0\).

The above conditional probability density functions satisfy the requirements for a univariate PDF; that is, \(f_{X \mid Y=y}(x) \ge 0\) for all \(x\) and \(\int_{\mathcal{R}_X} f_{X\mid Y=y}(x)\,dx = 1\).

Example 5.10 (Bivariate continuous conditional distributions) The joint PDF from Example 5.7 is \[ f_{X,Y}(x,y) = \frac{1}{3}(3x^2 + xy) \quad \text{for $0 \leq x \leq 1$ and $0 \leq y \leq 2$}. \] The marginal probability density functions of \(X\) and \(Y\) are \[\begin{align*} f_X(x) &= 2x^2 + \frac{2}{3}x \quad\text{for $0 \leq x \leq 1$}; \\ f_Y(y) &= \frac{1}{6}(2 + y) \quad \text{for $0 \leq y \leq 2$}. \end{align*}\] Hence, the conditional distribution of \(X \mid Y = y\) is \[ f_{X\mid Y=y}(x) = \frac{(3x^2 + xy)/3}{(2 + y)/6} = \frac{2x(3x + y)}{2 + y} \quad\text{for $0 \leq x \leq 1$}, \] and the conditional distribution of \(Y \mid X = x\) is \[ f_{Y \mid X=x}(y) = \frac{3x + y}{2(3x + 1)}\quad\text{for $0 \leq y \leq 2$}. \] Both these conditional density functions are valid density functions (verify!).

The marginal distribution for \(Y\), and two conditional distributions of \(Y\) (given \(X = 0.1\) and \(X = 0.9\)) are shown in Fig. 5.4.

FIGURE 5.4: The marginal distribution of \(Y\) (left panel), and the conditional distribution of \(Y\) for \(X = 0.1\) (centre panel) and \(X = 0.9\) (right panel).

To interpret the conditional distribution, for example \(f_{X \mid Y = y}(x)\), consider slicing through the surface \(f_{X, Y}(x, y)\) with the plane \(y = 0.5\) say (as shown in Fig. 5.5). The intersection of the plane with the surface is proportional to a \(1\)-dimensional PDF. This is \(f_{X, Y}(x, c)\), which will not, in general, be a density function since the area under this curve will be \(f_Y(c)\). Dividing by the constant \(f_Y(c)\) ensures the area under \(\displaystyle\frac{f_{X,Y}(x,c)}{f_Y(c)}\) is one. This is a one-dimensional PDF, of \(X\) given \(Y = c\); that is \(f_{X \mid Y = c}(x)\).

FIGURE 5.5: Joint PDF.

Example 5.11 (Bivariate discrete conditional distributions) Consider again the joint distribution in Example 5.2, with marginal distributions shown in Table 5.5. The conditional distribution of \(X\) given \(Y = 0\) is \[ p_{X\mid Y=0}(x) = \frac{p_{X,Y}(x,0)}{p_Y(0)} = \frac{p_{X,Y}(x,0)}{36/47} \] giving \(p_{X\mid Y=0}(0) = 6/36\), \(p_{X\mid Y=0}(1) = 12/36 = 1/3\), \(p_{X\mid Y=0}(2) = 18/36 = 1/2\).

Example 5.12 (Bivariate discrete conditional distributions) Consider again the random process in Example 5.4. The conditional distribution of \(Y\) given \(X = 0\) is \[\begin{align*} p_{Y\mid X=0}(y) &= \frac{p_{X, Y}(0, y)}{p_{X}(0)} = \frac{p_{X, Y}(0, y)}{1/4} = \frac{1}{6}\quad\text{for $y = 1, 2, 3, 4, 5, 6$}. \end{align*}\] Since \(Y\) is the number on the top face of the die, this result is exactly as expected—knowing the number of heads on the coins tells us nothing about the die.

5.5 Joint density–mass function

When one variable is discrete and the other is continuous, the joint probability function combines a PMF and a PDF into a joint density–mass function. As shown in Sect. 5.2, the total probability is one via a hybrid sum–integral. Using conditional distributions (Sect. 5.4), the joint function can be factored into a marginal probability and a conditional density.

Example 5.13 (Insurance claims) Suppose an insurance company records two variables for each policy-holder.

• \(X \in \{0, 1\}\) indicates whether a claim is made during the year: \(X = 1\) means a claim is made (with probability \(p = 0.30\)), and \(X = 0\) means no claim is made (with probability \(1 - p = 0.70\)).
• \(Y > 0\) is the age of the policy-holder in years, which is recorded for all policy-holders.

Since \(X\) is discrete and \(Y\) is continuous, the joint probability function cannot be written as a single PMF or PDF. Instead, the joint behaviour can be described separately for each value of \(X\). For each value \(x \in \{0, 1\}\), the joint density–mass function is the product of the probability of \(X = x\) and the conditional density of \(Y\) given \(X = x\). Suppose the conditional density of age given no claim (\(X = 0\)) is (see Fig. 5.6) \[ f_{Y\mid X = 0}(y) = \frac{12(y - 20)^2(80 - y)}{60^4}\quad\text{for $20 < y < 80$}. \] and given a claim (\(X = 1\)) is \[ f_{Y\mid X = 1}(y) = \frac{12(y - 20)(80 - y)^2}{60^4}\quad\text{for $20 < y < 80$} \] where both are valid probability density functions on \(y > 0\). The total probability is then \[\begin{align*} &\sum_{x = 0}^{1} \Pr(X = x) \int_0^\infty f_{Y\mid X=x}(y)\,dy\\ = &\Pr(X = 0) \int_0^\infty f_{Y\mid X=0}(y)\,dy + \Pr(X = 1) \int_0^\infty f_{Y\mid X=1}(y)\,dy\\ = &0.70 \times 1 + 0.30 \times 1 = 1 \end{align*}\] as required.

Older policy-holders tend to drive more carefully, so the conditional distribution of age differs between claimants and non-claimants: \(f_{Y\mid X=1}(y)\) is shifted toward younger ages relative to \(f_{Y\mid X=0}(y)\). This dependence between \(X\) and \(Y\) is the key feature of the joint distribution. Note that \(Y\) is observed for all policy-holders regardless of whether a claim is made; this is what makes \((X, Y)\) a genuine bivariate random variable, rather than simply a mixed univariate distribution.

FIGURE 5.6: The two conditional distributions that, when combined, constitute the joint density–mass function.

Definition 5.8 (Mixed bivariate probability function) Let \((X, Y)\) be a random vector where \(X\) is continuous with range \(S_X \subseteq \mathbb{R}\), and \(Y\) is discrete with range \(S_Y = \{y_1, y_2, \dots\}\). A function \(f_{X,Y}(x, y_j)\) defined on \(\mathcal{R}_{X,Y} = S_X \times S_Y\) is a joint density–mass function of \((X, Y)\) if:

1. \(f_{X,Y}(x, y_j) \geq 0\) for all \((x, y_j) \in \mathcal{R}_{X,Y}\), and
2. \(\displaystyle\sum_{j} \int_{S_X} f_{X,Y}(x, y_j)\, dx = 1\).

In practice \(f_{X,Y}\) is constructed via the multiplication rule: \[ f_{X,Y}(x, y_j) = f_{X \mid Y=y_j}(x) \cdot p_Y(y_j) \] which automatically satisfies both conditions above.

The probability of any event \(A \subseteq S_X\) and \(y_j \in S_Y\) is \[ \Pr(X \in A, Y = y_j) = \int_A f_{X,Y}(x, y_j)\, dx. \]

Example 5.14 (Mixed random variable) Suppose a random vector \((X, Y)\) is defined so that \(Y \in \{1, 2\}\) with \[ p_Y(1) = 0.4, \quad p_Y(2) = 0.6. \] Conditional on \(Y = 1\), the probability density function of \(X\) is \[ f_{X \mid Y = 1}(x) = \begin{cases} 1 & 0 \le x \le 1, \\ 0 & \text{otherwise,} \end{cases} \] and conditional on \(Y = 2\), the probability density function of \(X\) is \[ f_{X \mid Y = 2}(x) = \begin{cases} 1/2 & 0 \le x \le 2, \\ 0 & \text{otherwise.} \end{cases} \] Then the joint density–mass function is \[ f_{X, Y}(x, y) = f_{X \mid Y=y}(x) \cdot p_Y(y) \] or, more explicitly: \[ f_{X,Y}(x, 1) = \begin{cases} 0.4 & 0 \le x \le 1 \\ 0 & \text{otherwise} \end{cases} \qquad f_{X,Y}(x, 2) = \begin{cases} 0.3 & 0 \le x \le 2 \\ 0 & \text{otherwise} \end{cases} \] Notice that \[ \sum_{y \in \{1,2\}} \int_{-\infty}^{\infty} f_{X,Y}(x,y)\, dx = \int_0^1 0.4 \, dx + \int_0^2 0.3 \, dx = 0.4 + 0.6 = 1 \] as required.

Then, for instance, we can find \(\Pr(X \le 0.5, Y = 2)\): \[ \Pr(X \le 0.5, Y = 2) = \int_0^{0.5} f_{X, Y}(x, 2)\, dx = \int_0^{0.5} 0.3\, dx = 0.15. \]

The marginal distribution of \(X\) is \[ f_X(x) = \sum_j f_{X, Y} (x, y) = \begin{cases} 0.7 & \text{for $0 \le x \le 1$} \\ 0.3 & \text{for $1 < x \le 2$.} \end{cases} \]

5.6 Joint distribution function

The (cumulative) distribution function represents a sum of probabilities, or a volume under a surface, is denoted by \(F_{X, Y}(x, y)\), and defined as follows.

Definition 5.9 (Bivariate distribution function) The bivariate distribution function is \[\begin{align} F_{X, Y}(x, y) &= \sum_{x_i \le x} \sum_{y_j\le y} p_{X, Y}(x_i, y_j) & \text{for $(X,Y)$ discrete;} \tag{5.3}\\ F_{X, Y}(x, y) &= \iint_{(u, v)\in \mathcal{R}_{\{X\times Y}: u\le x, v\le y\}} f_{X, Y}(u,v) \, du \, dv & \text{for $(X,Y)$ continuous.} \tag{5.4} \end{align}\]

Example 5.15 (Bivariate discrete) Consider the random process in Example 5.4, where two coins are tossed, and one die is rolled (simultaneously). The probability function is given in Table 5.2.

The complete joint distribution function is given in Table 5.6, and complicated even for this simple case. As an example, the joint df at \((1, 2)\) would be computed as follows: \[\begin{align*} F_{X, Y}(1, 2) &= \displaystyle \sum_{x\le1} \, \sum_{y\le 2} p_{X, Y}(x, y)\\ &= p_{X, Y}(0, 1) + p_{X, Y}(0, 2) + p_{X, Y}(1, 1) + p_{X, Y}(1, 2) \\ &= 1/24 + 1/24 + 1/12 + 1/12 = 6/24. \end{align*}\]

TABLE 5.6: The bivariate distribution function for \(X\) and \(Y\).

	\(y \lt 1\)	\(y \le 1\)	\(y \le 2\)	\(y \le 3\)	\(y \le 4\)	\(y \le 5\)	\(y \le 6\)
\(x\lt 0\)	0	0	0	0	0	0	0
\(x\le 0\)	0	\(1/24\)	\(2/24\)	\(3/24\)	\(4/24\)	\(5/24\)	\(6/24\)
\(x\le 1\)	0	\(3/24\)	\(6/24\)	\(9/24\)	\(12/24\)	\(15/24\)	\(18/24\)
\(x\le 2\)	0	\(4/24\)	\(8/24\)	\(12/24\)	\(16/24\)	\(20/24\)	\(24/24\)

Example 5.16 (Bivariate continuous) From Example 5.6, \[\begin{align*} F_{X, Y}(x, y) &= \frac{6}{5} \int_0^{x} \int_0^{y} (t_1 + t_2^2)\, dt_2\, dt_1 \\ &= \frac{6}{5} \int_0^{x} (t_1 t_2 + t_2^3/3)\Big|_{t_2 = 0}^{t_2 = y} \, dt_1 \\ &= \frac{6}{5} \int_0^{x} (t_1 y + y^3/3)\, dt_1 \\ &= \frac{6}{5} \left( \frac{x^2 y}{2} + \frac{x y^3}{3}\right) \end{align*}\] for \(0 < x < 1\) and \(0 < y < 1\). So \[ F_{X, Y}(x, y) = \begin{cases} 0 & \text{if $x < 0$ or $y < 0$};\\ \frac{6}{5} \left( x^2 y/2 + x y^3/3\right) & \text{if $0 \le x \le 1$ and $0 \le y \le 1$};\\ 1 & \text{if $x > 1$ and $y > 1$}. \end{cases} \]

5.7 Independent random variables

Recall that events \(A\) and \(B\) are independent if, and only if, \[ \Pr(A \cap B) = \Pr(A)\Pr(B). \] An analogous definition applies for random variables.

Definition 5.10 (Independent random variables) The random variables \(X\) and \(Y\) with joint distribution function \(F_{X, Y}\) and marginal distribution functions \(F_X\) and \(F_Y\) are independent if, and only if, \[\begin{equation} F_{X, Y}(x, y) = F_X(x) \times F_Y(y) \end{equation}\] for all \(x\) and \(y\).

If \(X\) and \(Y\) are not independent they are dependent, or not independent.

The following theorem is often used to establish independence or dependence of random variables. The proof is omitted.

Theorem 5.1 The discrete random variables \(X\) and \(Y\) with joint probability function \(p_{X, Y}(x, y)\) and marginal distributions \(p_X(x)\) and \(p_Y(y)\) are independent if, and only if, \[\begin{equation} p_{X, Y}(x, y) = p_X(x) \times p_Y(y) \text{ for every }(x, y) \in \mathcal{R}_{X \times Y}. \tag{5.5} \end{equation}\] The continuous random variables \((X, Y)\) with joint PDF \(f_{X, Y}\) and marginal PDFs \(f_X\) and \(f_Y\) are independent if, and only if, \[\begin{equation} f_{X, Y}(x, y) = f_X(x)\times f_Y(y) \end{equation}\] for all \((x, y)\in \mathcal{R}_{X\times Y}\).

To show independence for continuous random variables (and analogously for discrete random variables) we must show \(f_{X, Y}(x, y) = f_X(x)\times f_Y(y)\) for all pairs \((x, y)\). If \(f_{X, Y}(x, y)\neq f_X(x)\times f_Y(y)\), even for any one particular pair of \((x, y)\), then \(X\) and \(Y\) are dependent.

Example 5.17 (Bivariate discrete: Independence) The random variables \(X\) and \(Y\) have the joint probability distribution shown in Table 5.7. Summing across rows, the marginal probability function of \(Y\) is: \[ p_Y(y) = \begin{cases} 1/6 & \text{for $y = 1$};\\ 1/3 & \text{for $y = 2$};\\ 1/2 & \text{for $y = 3$}. \end{cases} \] To determine if \(X\) and \(Y\) are independent, the marginal probability function of \(X\) is also needed: \[ p_X(x) = \begin{cases} 1/5 & \text{for $x = 1$};\\ 1/5 & \text{for $x = 2$};\\ 2/5 & \text{for $x = 3$};\\ 1/5 & \text{for $x = 4$}. \end{cases} \] Clearly, Eq. (5.5) is satisfied for all pairs \((x, y)\), so \(X\) and \(Y\) are independent.

TABLE 5.7: A joint probability function.

	\(x = 1\)	\(x = 2\)	\(x = 3\)	\(x = 4\)
\(y = 1\)	\(1/30\)	\(1/30\)	\(2/30\)	\(1/30\)
\(y = 2\)	\(2/30\)	\(2/30\)	\(4/30\)	\(2/30\)
\(y = 3\)	\(3/30\)	\(3/30\)	\(6/30\)	\(3/30\)

Example 5.18 (Bivariate continuous: independence) Consider the random variables \(X\) and \(Y\) with joint PDF \[ f_{X, Y}(x, y) = \begin{cases} 4xy & \text{for $0 < x < 1$ and $0 < y < 1 $}\\ 0 & \text{elsewhere}.\\ \end{cases} \] To show that \(X\) and \(Y\) are independent, the marginal distributions of \(X\) and \(Y\) are needed. Now \[ f_X(x) = \int_0^1 4xy \, dy = 2x\quad\text{for $0 < x < 1$}. \] Similarly \(f_Y(y) = 2y\) for \(0 < y < 1\). Thus we have \(f_X(x) \cdot f_Y(y) = f(x,y)\), so \(X\) and \(Y\) are independent.

Example 5.19 (Bivariate discrete: independence) Consider again the random process in Example 5.4. The marginal distribution of \(X\) was found in Example 5.9. The marginal distribution of \(Y\) is \[ p_Y(y) = \sum_{x = 0}^{2} p_{X,Y}(x,y) = \frac{1}{24} + \frac{1}{12} + \frac{1}{24} = \frac{1}{6} \quad \text{for } y = 1, \ldots, 6 \] To determine if \(X\) and \(Y\) are independent, each \(x\) and \(y\) pair must be considered. As an example, we see \[\begin{align*} p_{X}(0) \times p_{Y}(1) = 1/4 \times 1/6 = 1/24 &= p_{X, Y}(0, 1);\\ p_{X}(0) \times p_{Y}(2) = 1/4 \times 1/6 = 1/24 &= p_{X, Y}(0, 2);\\ p_{X}(1) \times p_{Y}(1) = 1/2 \times 1/6 = 1/12 &= p_{X, Y}(1, 1);\\ p_{X}(2) \times p_{Y}(1) = 1/4 \times 1/6 = 1/24 &= p_{X, Y}(2, 1). \end{align*}\] This is true for all pairs, and so \(X\) and \(Y\) are independent random variables. Independence is, however, obvious from the description of the random process (Example 5.1), and is easily seen from Table 5.2.

Example 5.20 (Bivariate continuous: independence) Consider the continuous random variables \(X\) and \(Y\) with joint PDF \[ f_{X, Y}(x, y) = \begin{cases} \frac{2}{7}(x + 2y) & \text{for $0 < x < 1$ and $1 < y < 2$};\\ 0 & \text{elsewhere.} \end{cases} \] The marginal distribution of \(X\) is \[\begin{align*} f_{X}(x) = \int_1^2 \frac{2}{7}(x + 2y)\,dy\\ = \frac{2}{7}(x + 3) \end{align*}\] for \(0 < x < 1\) (and zero elsewhere). Likewise, the marginal distribution of \(Y\) is \[ f_{Y}(y) = \frac{2}{7}(x^2/2 + 2 x y)\Big|_{x = 0}^1 = \frac{1}{7}(1 + 4y) \] for \(1 < y < 2\) (and zero elsewhere). (Both the marginal distributions must be valid density functions; verify!) Since \[ f_{X}(x) \times f_{Y}(y) = \frac{2}{49}(x + 3)(1 + 4y) \ne f_{X, Y}(x, y), \] the random variables \(X\) and \(Y\) are not independent.

The conditional distribution of \(X\) given \(Y = y\) is \[\begin{align*} f_{X \mid Y = y}(x) &= \frac{ f_{X, Y}(x, y)}{ f_{Y}(y)} \\ &= \frac{ (2/7) (x + 2y)}{ (1/7)(1 + 4y)}\\ &= \frac{ 2 (x + 2y)}{ 1 + 4y} \end{align*}\] for \(0 < x < 1\) and any given value of \(1 < y < 2\). (Again, this conditional density must be a valid probability density function.) So, for example, \[ f_{X \mid Y = 1.5}(x) = \frac{ 2 (x + 2\times 1.5)}{ 1 + (4\times 1.5)} = \frac{2}{7}(x + 3) \] for \(0 < x < 1\) and is zero elsewhere. And, \[ f_{X\mid Y = 1}(x) = \frac{ 2 (x + 2\times 1)}{ 1 + (4\times 1)} = \frac{2}{5}(x + 2) \] for \(0 < x < 1\) and is zero elsewhere. The distribution of \(X\) depends on the given value of \(Y\), so \(X\) and \(Y\) are not independent.

Example 5.21 (Bivariate continuous: independence) Consider the two continuous random variables \(X\) and \(Y\) with joint probability function \[ f_{X, Y}(x, y)= \begin{cases} 2(x + y) & \text{for $0 < x < y < 1$};\\ 0 & \text{elsewhere}. \end{cases} \] A diagram of the region \(R\) over which \(X\) and \(Y\) are defined is shown in Fig. 5.7.

When \(X = 0.2\), the range of \(Y\) is \((0.2, 1)\); when \(X = 0.5\), the range of \(Y\) is \((0.5, 1)\). The range of \(Y\) depends on the value of \(X\), which means the support \(\{(x, y) \mid 0 < x < y < 1\}\) is not a rectangle. Thus, \(X\) and \(Y\) cannot be independent.

FIGURE 5.7: The region \(R\) over which \(f_{X, Y}(x, y)\) is defined.

5.8 Statistical computing with bivariate distributions

5.8.1 Discrete bivariate distributions

Consider the bivariate discrete distribution with the joint probability function shown in Table 5.8. The joint PMF can be defined in R as a matrix:

TABLE 5.8: The bivariate probability function for \(X\) and \(Y\).

	\(y = 1\)	\(y = 2\)
\(x = 1\)	0.1	0.2
\(x = 2\)	0.3	0.1
\(x = 3\)	0.2	0.1

# Joint PMF as a matrix (rows = x, cols = y)
x_vals <- c(1, 2, 3)       # The values for the variable X 
y_vals <- c(10, 20)        # The values for the variable Y
P <- matrix(c(0.1, 0.2,    # The matrix contains the probabilities
              0.3, 0.1,
              0.2, 0.1),
            nrow = 3,      # The matrix have 3 rows
            byrow = TRUE,  # Fill the matrix with the data by rows first
            dimnames = list(x = x_vals,    # The names for the rows...
                            y = y_vals))   #... and the columns
P
#>    y
#> x    10  20
#>   1 0.1 0.2
#>   2 0.3 0.1
#>   3 0.2 0.1
sum(P)                     # Verify: the sum of probabilities is 1
#> [1] 1

The marginal distributions can be found by summing over the rows or columns of the matrix as appropriate:

rowSums(P)      # Marginal distribution of X: sum the rows
#>   1   2   3 
#> 0.3 0.4 0.3
colSums(P)      # Marginal distribution of Y: sum the columns
#>  10  20 
#> 0.6 0.4

The conditional distributions can also be found; for example, to find \(\Pr(X = x \mid Y = 10)\):

P[, "10"] / colSums(P)["10"]   # "10" is the name of the column for "y = 10"
#>         1         2         3 
#> 0.1666667 0.5000000 0.3333333

To check if \(X\) and \(Y\) are independent, multiply the marginal distributions using outer():

outer(rowSums(P),            # Element-by-element multiplication
      colSums(P))
#>     10   20
#> 1 0.18 0.12
#> 2 0.24 0.16
#> 3 0.18 0.12
P                            # Compare to the joint PMF
#>    y
#> x    10  20
#>   1 0.1 0.2
#>   2 0.3 0.1
#>   3 0.2 0.1

The probabilities in the joint PMF are not the same: \(X\) and \(Y\) are not independent.

5.8.2 Continuous bivariate distributions

Consider the joint PDF in Example 5.21. The joint PDF can be written as an R function:

# Joint PDF: f(x,y) = 2(x + y) for 0 < x < y < 1
fXY <- function(x, y) {
  ifelse( (x > 0) & (x < y) & (y < 1),  # This defines the support, R
          2 * (x + y),    # If the values of x and y are within the defined region
          0)              # If the values are outside the defined region
}

In R, the function integrate() can be used to evaluate one-dimensional definite integrals. Using integrate() twice, we can verify that the joint PDF is valid: we use integrate() once in the \(x\)-direction and once in the \(y\)-direction:

# Verify integrates to 1 (integrate: y inner integral, x outer integral)
integrate(function(x) {           # Integrate over x
  sapply(x,                       # To each of these values...
         function(xi) {           # ... apply this function
              integrate(function(y) fXY(xi, y), 
                 lower = xi,      # The lower y-integration limit
                 upper = 1)$value # The lower y-integration limit
           })
  }, lower = 0,                   # The lower x-integration limit 
     upper = 1)$value             # The upper x-integration limit
#> [1] 1

(sapply(X, FUN) is an R function that applies the function FUN to the elements of X.) Marginal distributions can be found using integrate() also:

# Marginal of X at a specific value, say x = 0.3
integrate(function(y) {fXY(0.3, y)},   # The function fXY, with x set to 0.3 
          lower = 0.3, 
          upper = 1)$value
#> [1] 1.33

# Create a function: the marginal distn of X
fX <- Vectorize(function(x) {
  integrate(function(y) fXY(x, y), 
            lower = x, 
            upper = 1)$value
})

(Vectorize() allows the newly-created function fX() to accept vector inputs and return vector outputs.) Probabilities can be found similarly; for example, to find \(\Pr(X < 0.5, Y < 0.8)\):

integrate(function(x) {
  sapply(x, function(xi) {
    integrate(function(y) fXY(xi, y),
              lower = xi, 
              upper = 0.8)$value
  })
}, lower = 0, 
   upper = 0.5)$value
#> [1] 0.395

5.9 Exercises

Selected answers appear in Sect. F.4.

Exercise 5.1 The discrete random variables \(X\) and \(Y\) have the joint probability function shown in Table 5.9.

1. Determine \(\Pr(X = 1, Y = 2)\)
2. Determine \(\Pr(X + Y \le 1)\).
3. Determine \(\Pr(X > Y)\).
4. Find the marginal probability function of \(X\).
5. Find the probability function of \(Y \mid X = 1\).

TABLE 5.9: The joint probability mass function of \(X\) and \(Y\).

	\(y = 0\)	\(y = 1\)	\(y = 2\)
\(x = 0\)	\(1/12\)	\(1/6\)	\(1/24\)
\(x = 1\)	\(1/4\)	\(1/4\)	\(5/24\)

Exercise 5.2 The discrete random variables \(X\) and \(Y\) have the joint probability function shown in Table 5.10.

1. Determine \(\Pr(X < 3, Y = 0)\)
2. Determine \(\Pr(X + Y > 3)\).
3. Determine \(\Pr(X > (Y/2) )\).
4. Find the marginal probability function of \(Y\).
5. Find the marginal probability function of \(X\).
6. Find the probability function of \(Y \mid X = 1\).

TABLE 5.10: The joint probability mass function of \(X\) and \(Y\).

	\(y = 0\)	\(y = 1\)	\(y = 2\)	\(y = 3\)	\(y = 4\)
\(x = 1\)	\(0\)	\(2/15\)	\(1/15\)	\(1/15\)	\(0\)
\(x = 2\)	\(0\)	\(2/15\)	\(1/15\)	\(1/15\)	\(1/15\)
\(x = 3\)	\(1/15\)	\(2/15\)	\(2/15\)	\(1/15\)	\(0\)

Exercise 5.3 The continuous random variables \(X\) and \(Y\) have the joint probability function \[ f_{X, Y}(x, y) = k\,(x + y^2) \] for \(0 < x < 1\) and \(0 < y < 2\).

1. Determine the value of \(k\).
2. Compute \(\Pr(X > 1/2, Y > 1)\).
3. Compute \(\Pr(X + Y > 1)\).
4. Find the marginal probability function of \(X\).
5. Find the marginal probability function of \(Y\).
6. Find the probability function of \(Y \mid X\).
7. Find the probability function of \(Y \mid X = 1\).
8. Find the probability function of \(X \mid Y\).
9. Find the probability function of \(X \mid Y = 1\).
10. Are \(X\) and \(Y\) independent random variables? Explain.

Exercise 5.4 The continuous random variables \(X\) and \(Y\) have the joint probability function \[ f_{X, Y}(x, y) = k\,(2 + x - y) \] for \(1 < x < 2\) and \(-1 < y < 1\).

1. Determine the value of \(k\).
2. Compute \(\Pr(X > 1, Y > 0)\).
3. Compute \(\Pr(X + Y \ge 1)\).
4. Find the marginal probability function of \(X\).
5. Find the marginal probability function of \(Y\).
6. Find the probability function of \(Y \mid X\).
7. Find the probability function of \(Y \mid X = 1\).
8. Find the probability function of \(X \mid Y\).
9. Find the probability function of \(X \mid Y = 1\).
10. Are \(X\) and \(Y\) independent random variables? Explain.

Exercise 5.5 A random vector \((X, Y)\) is defined so that \[ \Pr(X = 1) = 0.2;\quad \Pr(X = 2) = 0.3;\quad \Pr(X = 3) = 0.5. \] Then, \[\begin{align*} f_{Y\mid X = 1}(y) &= 1\quad\text{for $0 < y < 1$};\\ f_{Y\mid X = 2}(y) &= 0.5\quad\text{for $0 < y < 2$};\\ f_{Y\mid X = 3}(y) &= 1\quad\text{for $1 < y < 2$}. \end{align*}\]

1. Determine the joint density–mass function.
2. Find \(\Pr(Y > 1)\).

Exercise 5.6 A random vector \((X, Y)\) is defined so that \[ \Pr(X = -1) = 0.3;\quad \Pr(X = 1) = 0.7. \] Then, \[\begin{align*} f_{Y\mid X = -1}(y) &= 0.5\quad\text{for $-1 < y < 1$};\\ f_{Y\mid X = 1}(y) &= 0.25\quad\text{for $-2 < y < 2$}. \end{align*}\]

1. Determine the joint density–mass function.
2. Find \(\Pr(Y > 1)\).

Exercise 5.7 The pair of random variables \((X, Y)\) have the joint probability function given by \[ \Pr(X = x, Y = y) = k\,|x - y| \] for \(x = 0, 1, 2\) and \(y = 1, 2, 3\).

1. Find the value \(k\).
2. Construct a table of probabilities for this distribution.
3. Find \(\Pr(X \le 1, Y = 3)\).
4. Find \(\Pr(X + Y \ge 3)\).

Exercise 5.8 For what value of \(k\) is \(f(x,y) = kxy\) (for \(0 \le x \le 1\); \(0 \le y \le 1\), a valid joint PDF?

1. Then, find \(\Pr(X \le x_0, Y\le y_0)\).
2. Hence evaluate \(\Pr\left(X \le (3/8), Y \le (5/8) \right)\).

Exercise 5.9 For the random vector \((X, Y)\), the conditional PDF of \(Y\) given \(X = x\) is \[ f_{Y \mid X = x}(y) = \frac{2(x + y)}{2x + 1}, \] for \(0 < y <1\). The marginal PDF of \(X\) is given by \[ g_X(x) = x + \frac{1}{2} \] for \(0 <x < 1\).

1. Find \(F_{Y\mid X=x}(y)\).
2. Hence evaluate \(\Pr(Y < 3/4 \mid X = 1/3)\).
3. Find the joint PDF, \(f_{X, Y}(x, y)\), of \(X\) and \(Y\).
4. Find \(\Pr(Y < X)\).

Exercise 5.10 Consider a random process where a fair coin is tossed twice. Let \(X\) be the number of heads observed in the two tosses, and \(Y\) be the number of heads on the first toss of the coin.

1. Construct the table of the joint probability function for \(X\) and \(Y\).
2. Determine the marginal probability function for \(X\).
3. Determine the conditional distribution of \(X\) given one head appeared on the first toss.
4. Determine if the variables \(X\) and \(Y\) are independent or not, justifying your answer with necessary calculation or argument.

Exercise 5.11 Two fair, six-sided dice are rolled, and the numbers on the top faces observed. Event \(A\) is the maximum of the two numbers, and Event \(B\) is the minimum of the two numbers. Then, define \(C\) as \(0\) if the maximum is odd, and as \(1\) otherwise; and define \(D\) as \(0\) if the minimum is divisible by three, and as \(1\) otherwise.

Construct the joint probability function for \(C\) and \(D\).

Exercise 5.12 Consider the joint PDF \[ f_{X, Y}(x, y) = \begin{cases} c x(y + 1) & \text{where $x + y < 2$ with $x > 0$ and $y > 0$};\\ 0 & \text{elsewhere}. \end{cases} \]

1. Draw the region over which the joint PDF is defined.
2. Compute the value of \(c\).
3. Compute \(\Pr(Y < 1 \mid X > 1)\).
4. Compute \(\Pr(Y < 1 \mid X > 0.25)\).
5. Compute \(\Pr(Y < 1)\)

Exercise 5.13 Consider the joint PDF \[ f_{X, Y}(x, y) = \begin{cases} k ( 1 - x) y & \text{for the region $R$ below};\\ 0 & \text{elsewhere}, \end{cases} \] where the region \(R\) is shown in Fig. 5.8 (left panel).

1. Determine the value of \(k\).
2. Compute \(\Pr(X > Y)\).
3. Compute \(\Pr(X > 0.5)\).

Exercise 5.14 Consider the joint PDF \[ f_{X, Y}(x, y) = \begin{cases} k (x + 2y) y & \text{for the region $A$ below};\\ 0 & \text{elsewhere}, \end{cases} \] where the region \(A\) is shown in Fig. 5.8 (right panel).

1. Determine the value of \(k\).
2. Compute \(\Pr(X > Y)\).
3. Compute \(\Pr(X > 0.5)\).

FIGURE 5.8: The region \(R\) (left) and the region \(A\) (right).

Exercise 5.15 Consider two random variables \(X\) and \(Y\) that have uniform distributions, with probability density functions \[\begin{align*} f_X(x) &= 1\quad\text{for $0 < x < 1$; and}\\ f_Y(y) &= 1\quad\text{for $0 < y < 1$} \end{align*}\]

1. Using simulation, show that the distribution of \(X + Y\) has a triangular distribution.
2. What is the mean and variance of the resulting distribution, based on the simulation?
3. Explain how your answers change if the distribution of  \(Y\) changes to \[ f_Y(y) = \frac{1}{2}\quad\text{for $-1 < y < 1$}. \]

Exercise 5.16 Consider the joint probability density function \[ f_{X, Y} = xy \] defined over the shaded region in Fig. 5.9 (left panel).

1. Find the two marginal distributions.
2. Find the conditional distribution \(X\mid Y=y\).
3. Find the conditional distribution \(X\mid Y=1/2\).

Exercise 5.17 Consider the joint probability density function \[ f_{X, Y} = x + y \] defined over the shaded region in Fig. 5.9 (right panel).

1. Find the two marginal distributions.
2. Find the conditional distribution \(Y\mid X=x\).
3. Find the conditional distribution \(Y\mid X=1\).

FIGURE 5.9: Two regions over which joint probability density functions are defined

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Peter K. Dunn, Taryn Axelsen 2026: CC BY-NC-SA 4.0