Mean and SD of Random Variables
Introduced| English | Meaning |
|---|---|
| expected value/ekˈspektɪd ˈvæljuː/ | The probability-weighted mean of the possible values of a random variable. |
| variance/ˈveərɪəns/ | A measure of spread based on the squared deviations of values from their mean. |
The long-run average
- The mean (or expected value 期望值) of a random variable is its long-run average outcome.
- Written $\mu_X$ or $E(X)$ — the balance point of the distribution.
- It's a weighted average: each value weighted by its probability.
- Over many repetitions, the average result closes in on $\mu_X$.
Computing the mean
-
$$\mu_X = \sum x_i \, P(x_i)$$
- Multiply each value by its probability, then add them all up.
- The mean need not be a possible value (a "family of $2.5$ kids").
- It's an average, not a prediction of any single trial.
A discrete probability distribution with its mean and standard deviation
Variance and standard deviation
- The variance 方差 $\sigma_X^2 = \sum (x_i - \mu_X)^2 \, P(x_i)$ — the probability-weighted average squared distance from the mean.
- The standard deviation $\sigma_X = \sqrt{\sigma_X^2}$ brings it back to the original units.
- Bigger $\sigma_X$ = outcomes are more spread out around the mean.
- Square the deviations, weight by probability, sum, then square-root.
Interpreting the SD
- $\sigma_X$ is the typical distance of an outcome from the mean $\mu_X$.
- Small $\sigma_X$: results cluster tightly near the expected value.
- Large $\sigma_X$: results swing widely from trial to trial.
- Together, $\mu_X$ (center) and $\sigma_X$ (spread) summarize the whole distribution.
The expected value is a long-run average, not a guaranteed or even possible outcome. $E(X)=2.5$ children means the average over many families, not that any family has $2.5$ kids. And compute the SD from variance: square the deviations first, weight, sum, then take the square root — don't average the raw distances.
A game: win $5$ dollars with probability $0.2$, else win $0$. Let $X$ be the winnings.
- Mean: $\mu_X = 5(0.2) + 0(0.8) = 1$ dollar — the long-run average payout.
- Variance: $(5-1)^2(0.2) + (0-1)^2(0.8) = 3.2 + 0.8 = 4$.
- SD: $\sigma_X = \sqrt{4} = 2$ dollars — typical distance from the $1$ average.
The mean (expected value) $\mu_X = \sum x_i P(x_i)$ is the long-run average — a probability-weighted average that needn't be a possible value. The variance $\sigma_X^2 = \sum (x_i-\mu_X)^2 P(x_i)$ and standard deviation $\sigma_X = \sqrt{\sigma_X^2}$ measure the typical distance of an outcome from $\mu_X$.
Write from an expected value of 3
- A random variable pays 0 with probability 0.25 and 4 with probability 0.75.
- The lesson expected value 2 is another distribution.
What is the expected value?
Multiply each value by its probability and add.
What is the expected value?
Difficulty 1/5 · 100 coins + a small bonus on first completion today
Three quarters of 4 is 3. Other options — 2 (the other distribution); 4 (the larger payout).
Weighting values by probability
The mean is the balance point of this weighted distribution.
X wins 5 with prob 0.2, else 0. Find the mean E(X) = 5(0.2)+0(0.8).
Difficulty 2/5 · 140 coins + a small bonus on first completion today
5(0.2) + 0(0.8) = 1.
With mean 1: variance = (5−1)²(0.2)+(0−1)²(0.8). Then find the standard deviation.
Difficulty 2/5 · 140 coins + a small bonus on first completion today
Variance = 3.2 + 0.8 = 4, so SD = √4 = 2.
The expected value of a random variable must be one of its possible values.
Difficulty 1/5 · 100 coins + a small bonus on first completion today
It's a long-run average — e.g. 2.5 children — need not be attainable.
The standard deviation of a random variable measures...
Difficulty 1/5 · 100 coins + a small bonus on first completion today
SD = typical spread around the mean.
The mean of a random variable is also called its ___ value (one word).
Difficulty 1/5 · 100 coins + a small bonus on first completion today
Mean = expected value = long-run average.