Mean and SD of Random Variables
Introduced| English |
|---|
| expected value/ekˈspektɪd ˈvæljuː/ |
| variance/ˈveərɪəns/ |
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
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$$\mu_X = \sum x_i \, P(x_i)$$
- 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.
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 $\mu_X$.
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).
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.
Variance = 3.2 + 0.8 = 4, so SD = √4 = 2.
The expected value of a random variable must be one of its possible values.
It's a long-run average — e.g. 2.5 children — need not be attainable.
The standard deviation of a random variable measures...
SD = typical spread around the mean.
The mean of a random variable is also called its ___ value (one word).
Mean = expected value = long-run average.