# 5.1: Joint Distributions of Discrete Random Variables

• • Contributed by Kristin Kuter
• Associate Professor (Mathematics Computer Science) at Saint Mary's College

In this chapter we consider two or more random variables defined on the same sample space and discuss how to model the probability distribution of the random variables jointly. We will begin with the discrete case by looking at the joint probability mass function for two discrete random variables. In the following section, we will consider continuous random variables.

### Definition $$\PageIndex{1}$$

If discrete random variables $$X$$ and $$Y$$ are defined on the same sample space $$S$$, then their joint probability mass function (joint pmf) is given by
$$p(x,y) = P(X=x\ \ \text{and}\ \ Y=y),\notag$$
where $$(x,y)$$ is a pair of possible values for the pair of random variables $$(X,Y)$$, and $$p(x,y)$$ satisfies the following conditions:

• $$0 \leq p(x,y) \leq 1$$
• $$\displaystyle{\mathop{\sum\sum}_{(x,y)}p(x,y) = 1}$$
• $$\displaystyle{P\left((X,Y)\in A\right)) = \mathop{\sum\sum}_{(x,y)\in A} p(x,y)}$$

Note that conditions #1 and #2 in Definition 5.1.1 are required for $$p(x,y)$$ to be a valid joint pmf, while the third condition tells us how to use the joint pmf to find probabilities for the pair of random variables $$(X,Y)$$.

In the discrete case, we can obtain the joint cumulative distribution function (joint cdf) of $$X$$ and $$Y$$ by summing the joint pmf:
$$F(x,y) = P(X\leq x\ \text{and}\ Y\leq y) = \sum_{x_i \leq x} \sum_{y_j \leq y} p(x_i, y_j),\notag$$
where $$x_i$$ denotes possible values of $$X$$ and $$y_j$$ denotes possible values of $$Y$$. From the joint pmf, we can also obtain the individual probability distributions of $$X$$ and $$Y$$ separately as shown in the next definition.

### Definition $$\PageIndex{2}$$

Suppose that discrete random variables $$X$$ and $$Y$$ have joint pmf $$p(x,y)$$. Let $$x_1, x_2, \ldots, x_i, \ldots$$ denote the possible values of $$X$$, and let $$y_1, y_2, \ldots, y_j, \ldots$$ denote the possible values of $$Y$$. The marginal probability mass functions (marginal pmf's) of $$X$$ and $$Y$$ are respectively given by the following:
\begin{align*}
p_X(x) &= \sum_j p(x, y_j) \quad(\text{fix a value of}\ X\ \text{and sum over possible values of}\ Y) \\
p_Y(y) &= \sum_i p(x_i, y) \quad(\text{fix a value of}\ Y\ \text{and sum over possible values of}\ X)
\end{align*}

### Example $$\PageIndex{1}$$

Consider again the probability experiment of Example 3.3.2, where we toss a fair coin three times and record the sequence of heads $$(h)$$ and tails $$(t)$$. Again, we let random variable $$X$$ denote the number of heads obtained. We also let random variable $$Y$$ denote the winnings earned in a single play of a game with the following rules, based on the outcomes of the probability experiment (this is the same as Example 3.6.2):

• player wins $1 if first $$h$$ occurs on the first toss • player wins$2 if first $$h$$ occurs on the second toss
• player wins $3 if first $$h$$ occurs on the third toss • player loses$1 if no $$h$$ occur

Note that the possible values of $$X$$ are $$x=0,1,2,3$$, and the possible values of $$Y$$ are $$y=-1,1,2,3$$. We represent the joint pmf using a table:

### Table 1: joint pmf of $$X$$ and $$Y$$

$$p(x,y)$$ $$X$$
$$Y$$ 0 1 2 3
-1 1/8 0 0 0
1 0 1/8 2/8 1/8
2 0 1/8 1/8 0
3 0 1/8 0 0

The values in Table 1 give the values of $$p(x,y)$$. For example, consider $$p(0,-1)$$:
$$p(0,-1) = P(X=0\ \text{and}\ Y=-1) = P(ttt) = \frac{1}{8}.\notag$$
Since the outcomes are equally likely, the values of $$p(x,y)$$ are found by counting the number of outcomes in the sample space $$S$$ that result in the specified values of the random variables, and then dividing by $$8$$, the total number of outcomes in $$S$$. The sample space is given below, color coded to help explain the values of $$p(x,y)$$:
$$S = \{{\color{green}ttt}, {\color{orange}htt}, {\color{orange}tht}, {\color{orange}tth}, {\color{blue}hht}, {\color{blue}hth}, {\color{blue}thh}, {\color{purple} hhh}\}\notag$$

Given the joint pmf, we can now find the marginal pmf's. Note that the marginal pmf for $$X$$ is found by computing sums of the columns in Table 1, and the marginal pmf for $$Y$$ corresponds to the row sums. (Note that we found the pmf for $$X$$ in Example 3.3.2 as well, it is a binomial random variable. We also found the pmf for $$Y$$ in Example 3.6.2.)

### Table 2: marginal pmf's for $$X$$ and $$Y$$

$$x$$ $$p_X(x)$$ $$y$$ $$p_Y(y)$$
0 1/8 -1 1/8
1 3/8 1 1/2
2 3/8 2 1/4
3 1/8 3 1/8

Finally, we can find the joint cdf for $$X$$ and $$Y$$ by summing over values of the joint frequency function. For example, consider $$F(1,1)$$:
$$F(1,1) = P(X\leq1\ \text{and}\ Y\leq1) = \sum_{x\leq1}\sum_{y\leq1} p(x,y) = p(0,-1) + p(0,1) + p(-1,1) + p(1,1) = \frac{1}{4}\notag$$
Again, we can represent the joint cdf using a table:

### Table 3: joint cdf of $$X$$ and $$Y$$

$$F(x,y)$$ $$X$$
$$Y$$ 0 1 2 3
-1 1/8 1/8 1/8 1/8
1 1/8 1/4 1/2 5/8
2 1/8 3/8 3/4 7/8
3 1/8 1/2 7/8 1

## Expectations of Functions of Jointly Distributed Discrete Random Variables

We now look at taking the expectation of jointly distributed discrete random variables. Because expected values are defined for a single quantity, we will actually define the expected value of a combination of the pair of random variables, i.e., we look at the expected value of a function applied to $$(X,Y)$$.

### Theorem $$\PageIndex{1}$$

Suppose that $$X$$ and $$Y$$ are jointly distributed discrete random variables with joint pmf $$p(x,y)$$.

If $$g(X,Y)$$ is a function of these two random variables, then its expected value is given by the following:
$$\text{E}[g(X,Y)] = \mathop{\sum\sum}_{(x,y)}g(x,y)p(x,y).\notag$$

### Example $$\PageIndex{2}$$

Consider again the discrete random variables we defined in Example 5.1.1 with joint pmf given in Table 1. We will find the expected value of three different functions applied to $$(X,Y)$$.

1. First, we define $$g(x,y) = xy$$, and compute the expected value of $$XY$$:
\begin{align*}
\text{E}[XY] = \mathop{\sum\sum}_{(x,y)}xy\cdot p(x,y) &= (0)(-1)\left(\frac{1}{8}\right) \\
&\ + (1)(1)\left(\frac{1}{8}\right) + (2)(1)\left(\frac{2}{8}\right) + (3)(1)\left(\frac{1}{8}\right) \\
&\ + (1)(2)\left(\frac{1}{8}\right) + (2)(2)\left(\frac{1}{8}\right) \\
&\ + (1)(3)\left(\frac{1}{8}\right) \\
&= \frac{17}{8} = 2.125
\end{align*}

2. Next, we define $$g(x) = x$$, and compute the expected value of $$X$$:
\begin{align*}
\text{E}[X] = \mathop{\sum\sum}_{(x,y)}x\cdot p(x,y) &= (0)\left(\frac{1}{8}\right) \\
&\ + (1)\left(\frac{1}{8}\right) + (2)\left(\frac{2}{8}\right) + (3)\left(\frac{1}{8}\right) \\
&\ + (1)\left(\frac{1}{8}\right) + (2)\left(\frac{1}{8}\right) \\
&\ + (1)\left(\frac{1}{8}\right)\\
&= \frac{12}{8} = 1.5
\end{align*}
Recall that $$X\sim\text{binomial}(n = 3, p = 0.5)$$, and that the expected value of a binomial random variable is given by $$np$$. Thus, we can verify the expected value of $$X$$ that we calculated above using Theorem 5.1.1 using this fact for binomial distributions: $$\text{E}[X] = np = 3(0.5) = 1.5$$.

3. Lastly, we define $$g(x,y) = y$$, and calculate the expected value of $$Y$$:
\begin{align*}
\text{E}[Y] = \mathop{\sum\sum}_{(x,y)}y\cdot p(x,y) &= (-1)\left(\frac{1}{8}\right) \\
&\ + (1)\left(\frac{1}{8}\right) + (1)\left(\frac{2}{8}\right) + (1)\left(\frac{1}{8}\right) \\
&\ + (2)\left(\frac{1}{8}\right) + (2)\left(\frac{1}{8}\right) \\
&\ + (3)\left(\frac{1}{8}\right) \\
&= \frac{10}{8} = 1.25
\end{align*}
Again, we can verify this result by reviewing the calculations done in Example 3.6.2.

## Independent Random Variables

In some cases, the probability distribution of one random variable will not be affected by the distribution of another random variable defined on the same sample space. In those cases, the joint distribution functions have a very simple form, and we refer to the random variables as independent.

### Definition $$\PageIndex{3}$$

Discrete random variables $$X_1, X_2, \ldots, X_n$$ are independent if the joint pmf factors into a product of the marginal pmf's:
$$p(x_1, x_2, \ldots, x_n) = p_{X_1}(x_1)\cdot p_{X_2}(x_2) \cdots p_{X_n}(x_n).\label{indeprvs}$$
It is equivalent to check that this condition holds for the cumulative distribution functions.

Recall the definition of independent events (Definition 2.3.2): $$A$$ and $$B$$ are independent events if $$P(A\cap B) = P(A)\ P(B)$$. This is the basis for the definition of independent random variables because we can write the pmf's in Equation \ref{indeprvs} in terms of events as follows:
$$p(x,y) = P(X=x\ \text{and}\ Y=y) = P(\{X=x\}\cap\{Y=y\}) = P(X=x) P(Y=y) = p_X(x) p_Y(y)\notag$$
In the above, we use the idea that if $$X$$ and $$Y$$ are independent, then the event that $$X$$ takes on a given value $$x$$ is independent of the event that $$Y$$ takes the value $$y$$.

### Example $$\PageIndex{3}$$

Consider yet again the discrete random variables defined in Example 5.1.1. According to the definition, $$X$$ and $$Y$$ are independent if
$$p(x,y) = p_X(x)\cdot p_Y(y),\notag$$
for all pairs $$(x,y)$$. Recall that the joint pmf for $$(X,Y)$$ is given in Table 1 and that the marginal pmf's for $$X$$ and $$Y$$ are given in Table 2. Note that, for $$(x,y) = (0,-1)$$, we have the following
$$p(0,-1) = \frac{1}{8},\ \ p_X(0) = \frac{1}{8},\ \ p_Y(-1) = \frac{1}{8} \quad\Rightarrow\quad p(0,-1) \neq p_X(0)\cdot p_Y(-1).\notag$$
Thus, $$X$$ and $$Y$$ are not independent, or in other words, $$X$$ and $$Y$$ are dependent. This should make sense given the definition of $$X$$ and $$Y$$. The winnings earned depend on the number of heads obtained. So the probabilities assigned to the values of $$Y$$ will be affected by the values of $$X$$.

We also have the following very useful theorem about the expected value of a product of independent random variables, which is simply given by the product of the expected values for the individual random variables.

### Theorem $$\PageIndex{2}$$

If $$X$$ and $$Y$$ are independent random variables, then $$\text{E}[XY] = \text{E}[X]\ \text{E}[Y]$$.

Proof

Assume $$X$$ and $$Y$$ are independent random variables. If we let $$p(x,y)$$ denote the joint pmf of $$(X, Y)$$, then, by Definition 5.1.3, $$p(x,y) = p_X(x)p_Y(y)$$, for all pairs $$(x,y)$$. Using this fact and Theorem 5.1.1, we have
\begin{align*}
\text{E}[XY] &= \mathop{\sum\sum}_{(x,y)}xy\cdot p(x,y) = \mathop{\sum\sum}_{(x,y)}xy\cdot p_X(x)p_Y(y)\\
&= \sum_x\sum_y xyp_(x)P_Y(y) = \sum_x xp_X(x) \left(\sum_y p_Y(y)\right) = \sum_x xp_X(x)\text{E}[Y] \\
&= \text{E}[Y]\sum_x xp_X(x) = \text{E}[Y]\ \text{E}[X].
\end{align*}

Theorem 5.1.2 can be used to show that two random variables are not independent: if $$\text{E}[XY] \neq \text{E}[X]\ \text{E}[Y]$$, then $$X$$ and $$Y$$ cannot be independent. However, beware using Theorem 5.1.2 to show that random variables are independent. Note that Theorem 5.1.2 assumes that $$X$$ and $$Y$$ are independent and then the property about the expected value follows. The other direction does not hold. In other words, if $$\text{E}[XY] = \text{E}[X]\ \text{E}[Y]$$, then $$X$$ and $$Y$$ may or may not be independent.