4.9 Discrete Distributions
1. Discrete vs. Continuous Variables
A random variable $X$ takes values in a domain. It can be:
- Discrete: e.g., $X \in \{10, 20, 30\}$ or $X \in \{0, 1, 2, 3, \dots\}$. A discrete variable takes values in a finite or countable set.
- Continuous: e.g., $X \in [10, 20]$ or $X \in \mathbb{R}$. A continuous variable takes values in an interval or intervals.
This section covers discrete random variables.
Discrete Random Variable
Let $X$ be a variable that takes the values 10, 20, and 30 with probabilities 0.2, 0.3, and 0.5. This can be shown in a table:
| $X$ | 10 | 20 | 30 |
|---|---|---|---|
| $P(X=x)$ | 0.2 | 0.3 | 0.5 |
To express that the probability of $X=10$ is 0.2, we write $P(X=10) = 0.2$. Similarly, $P(X=20) = 0.3$ and $P(X=30) = 0.5$.
For any discrete random variable $X$ with values $x_1, x_2, x_3, \dots$ and probabilities $p_1, p_2, p_3, \dots$:
For any discrete random variable $X$ with values $x_1, x_2, x_3, \dots$ and probabilities $p_1, p_2, p_3, \dots$:
- (i) $p_i \ge 0$ for all $i$ (probabilities are non-negative)
- (ii) $\sum p_i = 1$, meaning $p_1 + p_2 + p_3 + \dots = 1$ (the sum of probabilities is 1)
2. The Expected Value $\mu = E(X)$
The mean $\mu$, or expected value $E(X)$, is defined as:
$E(X) = \sum x_i p_i = x_1 p_1 + x_2 p_2 + x_3 p_3 + \dots$
For the previous example, the expected value is:
$E(X) = 10 \times 0.2 + 20 \times 0.3 + 30 \times 0.5 = 23$
Explanation for $\mu = E(X)$
This mean is identical to the standard statistical mean. Consider these ten numbers:
This mean is identical to the standard statistical mean. Consider these ten numbers:
$10, 10, 20, 20, 20, 30, 30, 30, 30, 30$
The probabilities of selecting 10, 20, or 30 match the table (0.2, 0.3, 0.5). The mean is:
$\mu = \dfrac{10 \times 2 + 20 \times 3 + 30 \times 5}{10} = 10 \times \dfrac{2}{10} + 20 \times \dfrac{3}{10} + 30 \times \dfrac{5}{10} = \mathbf{23}$
EXAMPLE 1
Consider the following probability distribution:
| $X$ | 10 | 20 | 30 |
|---|---|---|---|
| $P(X=x)$ | $a$ | $b$ | 0.5 |
Given that $E(X) = 23$, find $a$ and $b$.
Solution: Use two equations based on the properties of distributions:
Solution: Use two equations based on the properties of distributions:
- Sum of probabilities is 1: $a + b + 0.5 = 1 \implies a + b = 0.5$
- Expected value formula: $10a + 20b + 30(0.5) = 23 \implies 10a + 20b = 8$
Probability distributions apply to betting games:
EXAMPLE 2
Using the previous table, select one number from 10, 20, 30 at random.
- Selecting 10 earns 6 points
- Selecting 20 earns 1 point
- Selecting 30 loses 2 points
What is the expected number of points per game?
Solution: Organize the data in a table:
| $X$ | 10 | 20 | 30 |
|---|---|---|---|
| Profit | 6 points | 1 point | -2 points |
| Prob | 0.2 | 0.3 | 0.5 |
Calculate the expected profit:
$\text{Expected profit} = 6 \times 0.2 + 1 \times 0.3 - 2 \times 0.5 = 1.2 + 0.3 - 1.0 = \mathbf{0.5}$
We earn an average of 0.5 points per game.
$\text{Expected profit} = 6 \times 0.2 + 1 \times 0.3 - 2 \times 0.5 = 1.2 + 0.3 - 1.0 = \mathbf{0.5}$
We earn an average of 0.5 points per game.
Explanation: Playing this game 10 times yields an expected 5 points. In 10 games, we expect to get:
- 2 times the number 10 $\implies 2 \times 6 = 12$ points
- 3 times the number 20 $\implies 3 \times 1 = 3$ points
- 5 times the number 30 $\implies 5 \times (-2) = -10$ points
EXAMPLE 3
We throw two dice.
- TWO SIXES earns 15€
- ONE SIX earns 1€
- NO SIX loses 1€
Find the expected profit per game.
Solution: Organize the data in a table:
| Result | TWO SIXES | ONE SIX | NO SIX |
|---|---|---|---|
| Profit | 15€ | 1€ | -1€ |
| Prob | $\dfrac{1}{36}$ | $\dfrac{10}{36}$ | $\dfrac{25}{36}$ |
The expected profit per game is:
$\text{Expected profit} = 15 \times \dfrac{1}{36} + 1 \times \dfrac{10}{36} - 1 \times \dfrac{25}{36} = \dfrac{15 + 10 - 25}{36} = \mathbf{0}$
This is a FAIR GAME! The expected net profit is zero.
Notice: If the first prize were 14€ instead of 15€, the expected profit would be $-\dfrac{1}{36}$. If we play 36,000 times, we expect to lose 1,000€.
3. Median and Mode
These measures are defined for probability distributions as follows:
- MODE = The value $X=a$ with the highest probability.
- MEDIAN = The value $X=m$ where the cumulative probability splits the distribution (reaches 0.5).
EXAMPLE 4 (Finding Mode and Median)
| $X$ | 10 | 20 | 30 |
|---|---|---|---|
| $P(X=x)$ | 0.4 | 0.3 | 0.3 |
- MODE = 10 (Highest probability)
- MEDIAN = 20 (Cumulative probability reaches 0.5 at 20)
| $X$ | 10 | 20 | 30 |
|---|---|---|---|
| $P(X=x)$ | 0.2 | 0.3 | 0.5 |
- MODE = 30 (Highest probability)
- MEDIAN = 25 (At $X=20$, the cumulative sum is 0.5. The midpoint between 20 and 30 divides the distribution: $\dfrac{20+30}{2} = 25$).
4. VarianceHL ONLY
The variance of a discrete random variable is:
$Var(X) = E(X - \mu)^2$
Expanded, this is:
$Var(X) = (x_1 - \mu)^2 \times p_1 + (x_2 - \mu)^2 \times p_2 + (x_3 - \mu)^2 \times p_3 + \dots$
An alternative formula is:
$Var(X) = E(X^2) - \mu^2$
where
$E(X^2) = x_1^2 \times p_1 + x_2^2 \times p_2 + x_3^2 \times p_3 + \dots$
EXAMPLE 5
Using the initial probability distribution:
| $X$ | 10 | 20 | 30 |
|---|---|---|---|
| $P(X=x)$ | 0.2 | 0.3 | 0.5 |
Since $\mu = E(X) = 23$, the variance using the first formula is:
$Var(X) = (10-23)^2 \times 0.2 + (20-23)^2 \times 0.3 + (30-23)^2 \times 0.5 = \mathbf{61}$
Using the alternative formula with $E(X^2)$:
$$
\begin{aligned}
E(X^2) &= 10^2 \times 0.2 + 20^2 \times 0.3 + 30^2 \times 0.5 \\
&= 100(0.2) + 400(0.3) + 900(0.5) \\
&= 20 + 120 + 450 \\
&= 590 \\
\implies Var(X) &= E(X^2) - \mu^2 = 590 - 23^2 = \mathbf{61}
\end{aligned}
$$