3.4 Bivariate Distributions

![The support $S$ of $(X, Y)$ in @exm-3-4-8.](ch03/images/fig-3-12.svg){#fig-3-12}

![The subset $S_0$ of the support $S$ where $x \geq y$ in @exm-3-4-8.](ch03/images/fig-3-13.svg){#fig-3-13}

::: {.callout-tip title="Example 3.4.9"}
::: {#exm-3-4-9}

# Example 3.4.9: Determining a Joint pdf by Geometric Methods.

Suppose that a point $(X, Y)$ is selected at random from inside the circle $x^2 + y^2 \leq 9$. We shall determine the joint pdf of $X$ and $Y$.

The support of $(X, Y)$ is the set $S$ of points on and inside the circle $x^2 + y^2 \leq 9$. The statement that the point $(X, Y)$ is selected at random from inside the circle is interpreted to mean that the joint pdf of $X$ and $Y$ is constant over $S$ and is 0 outside $S$. Thus,

$$
f(x, y) = \begin{cases}
c &\text{for }(x, y) \in S, \\
0 &\text{otherwise.}
\end{cases}
$$

We must have

$$
\int_{S}\int f(x, y) \, dx \, dy = c \times \text{(area of }S\text{)} = 1.
$$

Since the area of the circle $S$ is $9\pi$, the value of the constant $c$ must be $1/(9\pi)$.

(sec-3-4-3)=
# Mixed Bivariate Distributions

::: {.callout-tip title="Example 3.4.10"}
::: {#exm-3-4-10}

# Example 3.4.10: A Clinical Trial

Consider a clinical trial (such as the one described in @exm-2-1-12) in which each patient with depression receives a treatment and is followed to see whether they have a relapse into depression. Let $X$ be the indicator of whether or not the first patient is a "success" (no relapse). That is $X = 1$ if the patient does not relapse and $X = 0$ if the patient relapses. Also, let $P$ be the proportion of patients who have no replapse among all patients who might receive the treatment. It is clear that $X$ must have a discrete distribution, but it might be sensible to think of $P$ as a continuous random variable taking its value anywhere in the interval $[0, 1]$. Even though $X$ and $P$ can have neither a joint discrete distribution nor a joint continuous distribution, we can still be interested in the joint distribution of $X$ and $P$.  

Prior to @exm-3-4-10, we have discussed bivariate distributions that were either discrete or continuous. Occasionally, one must consider a mixed bivariate distribution in which one of the random variables is discrete and the other is continuous. We shall use a function $f(x, y)$ to characterize such a joint distribution in much the
same way that we use a joint pmf to characterize a discrete joint distribution or a joint pdf to characterize a continuous joint distribution.

::: {.callout-note title="Definition 3.4.5"}
::: {#def-3-4-5}

# Definition 3.4.5: Joint pmf/pdf

Let $X$ and $Y$ be random variables such that $X$ is discrete and $Y$ is continuous. Suppose that there is a function $f(x, y)$ defined on the $xy$-plane such that, for every pair $A$ and $B$ of subsets of the real numbers,

$$
\Pr(X \in A \text{ and } Y \in B) = \int_{B}\sum_{x \in A}f(x,y) \, dy,
$$ {#eq-3-4-4}

if the integral exists. Then the function $f$ is called the *joint pmf/pdf* of $X$ and $Y$.

Clearly, @def-3-4-5 can be modified in an obvious way if $Y$ is discrete and $X$ is continuous. Every joint pmf/pdf must satisfy two conditions. If $X$ is the discrete random variable with possible values $x_1, x_2, \ldots$ and $Y$ is the continuous random variable, then $f(x, y) \geq 0$ for all $x$, $y$ and

$$
\int_{-\infty}^{\infty}\sum_{i=1}^{\infty} f(x_i, y) \, dy = 1.
$$ {#eq-3-4-5}

Because $f$ is nonnegative, the sum and integral in Eqs. [-@eq-3-4-4] and [-@eq-3-4-5] can be done in whichever order is more convenient.

**Note: Probabilities of More General Sets.** For a general set $C$ of pairs of real numbers, we can compute $\Pr((X, Y) \in C)$ using the joint pmf/pdf of $X$ and $Y$. For each $x$, let $C_x = \{y \mid (x, y) \in C\}$. Then

$$
\Pr((X, Y) \in C) = \sum_{\text{All }x}\int_{C_x}f(x, y)\, dy,
$$

if all of the integrals exist. Alternatively, for each $y$, define $C^y = \{x \mid (x, y) \in C\}$, and
then

$$
\Pr((X, Y) \in C) = \int_{-\infty}^{\infty}\left[ \sum_{x \in C^y}f(x, y) \right]dy,
$$

if the integral exists.

::: {.callout-tip title="Example 3.4.11"}
::: {#exm-3-4-11}

# Example 3.4.11: A Joint pmf/pdf

Suppose that the joint pmf/pdf of $X$ and $Y$ is

$$
f(x, y) = \frac{xy^{x-1}}{3}, \; \text{ for }x = 1, 2, 3\text{ and }0 < y < 1.
$$

We should check to make sure that this function satisfies @eq-3-4-5. It is easier to integrate over the $y$ values first, so we compute

$$
\sum_{x=1}^{3}\int_{0}^{1}\frac{xy^{x-1}}{3} \, dy = \sum_{x=1}^{3}\frac{1}{3} = 1.
$$

Suppose that we wish to compute the probability that $Y \geq 1/2$ and $X \geq 2$. That is, we want $\Pr(X \in A \text{ and } Y \in B)$ with $A = [2, \infty)$ and $B = [1/2, \infty)$. So, we apply @eq-3-4-4 to get the probability

$$
\sum_{x=2}^{3}\int_{1/2}^{1}\frac{xy^{x-1}}{3} \, dy = \sum_{x=2}^{3}\left(\frac{1 - (1/2)^x}{3}\right) = 0.5417.
$$

For illustration, we shall compute the sum and integral in the other order also. For each $y \in [1/2, 1)$, $\sum_{x=2}^{3}f(x,y) = 2y/3 + y^2$. For $y \geq 1/2$, the sum is 0. So, the probability is

$$
\int_{1/2}^{1}\left[\frac{2}{3}y + y^2\right]dy = \frac{1}{3}\left[1 - \left(\frac{1}{2}\right)^2\right] + \frac{1}{3}\left[1 - \left(\frac{1}{2}\right)^3\right] = 0.5417.
$$

::: {.callout-tip title="Example 3.4.12"}
::: {#exm-3-4-12}

# Example 3.4.12: A Clinical Trial.

A possible joint pmf/pdf for $X$ and $P$ in @exm-3-4-10 is

$$
f(x, p) = p^x(1-p)^{1-x}, \; \text{ for }x = 0, 1\text{ and }0 < p < 1.
$$

Here, $X$ is discrete and $P$ is continuous. The function $f$ is nonnegative, and the reader should be able to demonstrate that it satisfies @eq-3-4-5. Suppose that we wish to compute $\Pr(X \leq 0 \text{ and } P \leq 1/2)$. This can be computed as

$$
\int_{0}^{1/2}(1 - p) \, dp = -\frac{1}{2}\left[ (1 - 1/2)^2 - (1 - 0)^2 \right] = \frac{3}{8}.
$$

Suppose that we also wish to compute $\Pr(X = 1)$. This time, we apply @eq-3-4-4 with $A = \{1\}$ and $B = (0, 1)$. In this case,

$$
\Pr(X = 1) = \int_{0}^{1}p \, dp = \frac{1}{2}.
$$

A more complicated type of joint distribution can also arise in a practical problem.

::: {.callout-tip title="Example 3.4.13"}
::: {#exm-3-4-13}

# Example 3.4.13: A Complicated Joint Distribution

Suppose that $X$ and $Y$ are the times at which two specific components in an electronic system fail. There might be a certain probability $p$ ($0 < p < 1$) that the two components will fail at the same time and a certain probability $1 − p$ that they will fail at different times. Furthermore, if they fail at the same time, then their common failure time might be distributed according to a certain pdf $f(x)$; if they fail at different times, then these times might be distributed according to a certain joint pdf $g(x, y)$.

The joint distribution of $X$ and $Y$ in this example is not continuous, because there is positive probability $p$ that $(X, Y)$ will lie on the line $x = y$. Nor does the joint distribution have a joint pmf/pdf or any other simple function to describe it. There are ways to deal with such joint distributions, but we shall not discuss them in this
text.

(sec-3-4-4)=
# Bivariate Cumulative Distribution Functions

The first calculation in @exm-3-4-12, namely, $\Pr(X \leq 0 \text{ and } Y \leq 1/2)$, is a generalization of the calculation of a CDF to a bivariate distribution. We formalize the generalization as follows.

```{figure} images/fig-3-14.svg
:label: fig-3-14
:enumerator: 3.14
:align: center
:width: 50%

The probability of a rectangle
```

::: {.callout-note title="Definition 3.4.6"}
::: {#def-3-4-6}

# Definition 3.4.6: Joint (Cumulative) Distribution Function/CDF

The *joint Cumulative Distribution Function* (*joint CDF*) of two random variables $X$ and $Y$ is defined as the function $F$ such that for all values of $x$ and $y$ ($-\infty < x < \infty$ and $-\infty < y < \infty$),

$$
F(x, y) = \Pr(X \leq x \text{ and }Y \leq y).
$$

It is clear from @def-3-4-6 that $F(x, y)$ is monotone increasing in $x$ for each fixed $y$ and is monotone increasing in $y$ for each fixed $x$.

If the joint CDF of two arbitrary random variables $X$ and $Y$ is $F$, then the probability that the pair $(X, Y)$ will lie in a specified rectangle in the $xy$-plane can be found from $F$ as follows: For given numbers $a < b$ and $c < d$,

$$
\begin{align*}
\Pr&(a < X \leq b \text{ and }c < Y \leq d) \\
&= \Pr(a < X \leq b \text{ and }Y \leq d) - \Pr(a < X \leq b \text{ and }Y \leq c) \\
&= \left[\Pr(X \leq b \text{ and } Y \leq d) - \Pr(X \leq a \text{ and }Y \leq d)\right] \\
&\phantom{\Pr} - \left[\Pr(X \leq b \text{ and }Y \leq c) - \Pr(X \leq a \text{ and }Y \leq c)\right] \\
&= F(b, d) - F(a, d) - F(b, c) + F(a, c).
\end{align*}
$$

Hence, the probability of the rectangle $C$ sketched in @fig-3-14 is given by the combination of values of $F$ just derived. It should be noted that two sides of the rectangle are included in the set $C$ and the other two sides are excluded. Thus, if there are points or line segments on the boundary of $C$ that have positive probability, it is important to distinguish between the weak inequalities and the strict inequalities in @eq-3-4-6.

::: {.callout-caution title="Theorem 3.4.5"}
::: {#thm-3-4-5}

# Theorem 3.4.5

Let $X$ and $Y$ have a joint CDF $F$. The CDF $F_1$ of just the single random variable $X$ can be derived from the joint CDF $F$ as $F_1(x) = \lim_{y \rightarrow \infty} F(x, y)$. Similarly, the CDF $F_2$ of $Y$ equals $F_2(y) = \lim_{x \rightarrow \infty} F(x, y)$, for $0 < y < \infty$.

::: {.proof}

We prove the claim about $F_1$ as the claim about $F_2$ is similar. Let $-\infty < x < \infty$. Define

$$
\begin{align*}
B_0 &= \{ X \leq x \text{ and } Y \leq 0 \}, \\
B_n &= \{ X \leq x \text{ and } n-1 < Y \leq n \}, \; \text{ for }n = 1, 2, \ldots, \\
A_m &= \bigcup_{n=0}^{m}B_n, \; \text{ for }m = 1, 2, \ldots.
\end{align*}
$$

Then $\{X \leq x\} = \bigcup_{n=0}^{\infty}B_n$, and $A_m = \{X \leq x \text{ and }Y \leq m\}$ for $m = 1, 2, \ldots$. It follows that $\Pr(A_m) = F(x, m)$ for each $m$. Also,

$$
\begin{align*}
F_1(x) &= \Pr(X \leq x) = \Pr\left( \bigcup_{n=1}^{\infty}B_n \right) \\
&= \sum_{n=0}^{\infty}\Pr(B_n) = \lim_{m \rightarrow \infty} \Pr(A_m) \\
&= \lim_{m \rightarrow \infty}F(x, m) = \lim_{y \rightarrow \infty}F(x, y),
\end{align*}
$$

where the third equality follows from countable additivity and the fact that the $B_n$ events are disjoint, and the last equality follows from the fact that $F(x, y)$ is monotone increasing in $y$ for each fixed $x$.

::: {.callout-tip title="Example 3.4.15"}
::: {#exm-3-4-15}

# Example 3.4.15: Demands for Utilities

We can compute the joint CDF for water and electric demand in @exm-3-4-4 by using the joint pdf that was given in @eq-3-4-2. If either $x \leq 4$ or $y \leq 1$, then $F(x, y) = 0$ because either $X \leq x$ or $Y \leq y$ would be impossible. Similarly, if both $x \geq 200$ and $y \geq 150$, $F(x, y) = 1$ because both $X \leq x$ and $Y \leq y$ would be sure events. For other values of $x$ and $y$, we compute

$$
F(x, y) = \begin{cases}
\int_{4}^{x}\int_{1}^{y}\frac{1}{29204}\, dy \, dx = \frac{xy}{29204} &\text{for }4 \leq x \leq 200, 1 \leq y \leq 150, \\
\int_{4}^{x}\int_{1}^{150}\frac{1}{29204}\, dy \, dx = \frac{x}{196} &\text{for }4 \leq x \leq 200, y > 150, \\
\int_{4}^{200}\int_{1}^{y}\frac{1}{29204}\, dy \, dx = \frac{y}{149} &\text{for }x > 200, 1 \leq y \leq 150.
\end{cases}
$$

The reason that we need three cases in the formula for $F(x, y)$ is that the joint pdf in @eq-3-4-2 drops to 0 when $x$ crosses above 200 or when $y$ crosses above 150; hence, we never want to integrate $1/29204$ beyond $x = 200$ or beyond $y = 150$. If one takes the limit as $y \rightarrow \infty$ of $F(x, y)$ (for fixed $4 \leq x \leq 200$), one gets the second case in the formula above, which then is the CDF of $X$, $F_1(x)$. Similarly, if one takes the $\lim_{x \rightarrow \infty} F(x, y)$ (for fixed $1 \leq y \leq 150$), one gets the third case in the formula, which then is the CDF of $Y$, $F_2(y)$.

### Summary

The joint CDF of two random variables $X$ and $Y$ is $F(x, y) = \Pr(X \leq x \text{ and } Y \leq y)$. The joint pdf of two continuous random variables is a nonnegative function $f$ such that the probability of the pair $(X, Y)$ being in a set $C$ is the integral of $f(x, y)$ over the set $C$, if the integral exists. The joint pdf is also the second mixed partial derivative of the joint CDF with respect to both variables. The joint pmf of two discrete random variables is a nonnegative function $f$ such that the probability of the pair $(X, Y)$ being in a set $C$ is the sum of $f(x, y)$ over all points in $C$. A joint pmf can be strictly positive at countably many pairs $(x, y)$ at most. The joint pmf/pdf of a discrete random variable $X$ and a continuous random variable $Y$ is a nonnegative function $f$ such that the probability of the pair $(X, Y)$ being in a set $C$ is obtained by summing $f(x, y)$ over all $x$ such that $(x, y) \in C$ for each $y$ and then integrating the resulting function of $y$.

### Exercises

::: {#exr-3-4-1}

# Exercise 3.4.1

Suppose that the joint pdf of a pair of random variables $(X, Y)$ is constant on the rectangle where $0 \leq x \leq 2$ and $0 \leq y \leq 1$, and suppose that the pdf is 0 off of this rectangle.

a. Find the constant value of the pdf on the rectangle.
b. Find $\Pr(X \geq Y)$.