A distribution contains more information than its mean. Expectations, dispersion, and higher moments describe different features and supply the algebra used in sampling theory.
1. Expectation is a probability-weighted average
These expectations require the relevant integrability conditions; \(\mathbb{E}|Y|<\infty\) ensures a finite mean. The expectation need not be a possible observation. In the crash example\(, \mathbb{E}[M]=.35\) even though no computer crashes .35 times. A repayment of 110 with probability .99 and zero otherwise has expected repayment 108.90; the expected net payoff after lending 100 is 8.90.
Linearity: \(\mathbb{E}[a+\)bX+cY\(]=a+\)bE\([X]+\)cE[Y]. Independence is unnecessary. However\(, \mathbb{E}[g(Y)]\) generally differs from \(g(\mathbb{E}[Y])\). For example\(, \mathbb{E}[Y^{2}]\) differs from \((\mathbb{E}[Y])^{2}\) whenever \(Y\) has positive variance.
2. Variance and standard deviation
Variance is nonnegative and is zero only when \(Y\) is constant with probability one. It has squared units; standard deviation has the units of \(Y\). Expanding \((Y-\mu )^{2}\) gives \(\mathbb{E}[Y^{2}]-2\mu \mathbb{E}[Y]+\mu ^{2}=\mathbb{E}[Y^{2}]-\mu ^{2}\).
Worked example. For M in Module \(1, \mathbb{E}[M^{2}]=.10+4(.06)+9(.03)+16(.01)=.77\). Therefore \(\operatorname{Var}(M)=.77-.35^{2}=.6475\) and \(\operatorname{SD}(M)\approx .8047\).
Bernoulli. Since \(Y^{2}=Y, \mathbb{E}[Y]=p, \mathbb{E}[Y^{2}]=p\), and \(\operatorname{Var}(Y)=p(1-p)\). Variance is largest at \(p=.5\) and vanishes at \(p=0\) or 1. This identity later supplies the sampling uncertainty of a sample proportion.
3. Affine transformations and units
Derivation. Subtracting the new mean leaves \(a+\)bY\(-(a+b\mu )=b(Y-\mu )\). Squaring and taking expectation produces \(b^{2}\sigma ^{2}\). Taking the nonnegative square root gives \(|b|\sigma\), including when b is negative.
If after-tax income is \(2000+.8X\), the mean becomes \(2000+.8\mu _{X}\) and the standard deviation becomes \(.8\sigma _{X}. A\) fixed transfer changes location but not spread. Converting dollars to cents multiplies means and standard deviations by 100 and variances by 10,000.
4. Moments, skewness, kurtosis, and outliers
Standardized moments require \(\sigma >0\) and existence of the relevant moment. Positive skewness commonly signals a long right tail; negative skewness commonly signals a long left tail. Symmetry implies zero skewness when the third moment exists. The converse is not a general theorem: an asymmetric distribution can have zero third central moment.
Normal kurtosis is 3; excess kurtosis subtracts 3. High kurtosis indicates a large standardized fourth moment and sensitivity to extreme observations. It does not rank every tail probability at every cutoff. Two distributions can share a mean and variance while differing greatly in shape and outlier frequency. Positive changes of units preserve skewness and kurtosis; multiplying by a negative number reverses skewness.
These expansions explain why statements about finite fourth moments appear later: applying a finite-variance LLN to \(Y^{2}\) requires \(\operatorname{Var}(Y^{2})<\infty\). The moment restriction belongs to a theorem'\(s\) justification, not merely to a descriptive statistic.
5. Earnings and model adequacy
SW Table 2.4 compares earnings conditional on education and gender. For women with a high-school diploma, the reported mean is 15.49 and median 14.03; for women with a four-year degree, they are 25.42 and 22.44. The comparison includes dispersion and upper percentiles, not just a mean shift. These are historical textbook summaries of 2012 full-time workers, not current estimates or causal education effects.
The stock-return box in §2.4 supplies a second lesson: matching typical variation with a normal distribution can badly understate extreme outcomes. This concerns the distribution of individual observations; a CLT for averages is a separate claim. Time-varying volatility and heavy tails are two distinct ways to reconsider the model.
Check your understanding
Question. If \(\operatorname{Var}(Y)=9\), what is \(\operatorname{SD}(10-2Y)\)? Is \(\mathbb{E}[Y^{2}]=9\)?
Solution. \(\operatorname{SD}(10-2Y)=6. \mathbb{E}[Y^{2}]=9+\mu ^{2}\), which equals 9 only if \(\mu =0\).
Sources: SW §2.2, pp. 65–71; Table 2.4, pp. 79–80; stock-return box, pp. 85–86; Exercises 2.21 and Appendix 2.1. General sign and moment caveats are explicit mathematical clarifications.