A CLT with unknown \(\sigma\) is not yet an operational procedure. Estimating the variance and justifying its substitution closes the gap between probability theory and usable inference.
1. Sample variance and the \(n-1\) correction
The residuals sum to zero, leaving \(n-1\) independently variable residual coordinates. Estimating the mean draws the residuals toward zero. The divisor \(n-1\) corrects the resulting downward bias under i.i.d. finite-variance sampling.
Thus \(\mathbb{E}[s^{2}]=\sigma ^{2}\). Dividing by \(n\) instead gives expectation \((n-1)\sigma ^{2}/n\). The standard deviation \(s\) itself need not be unbiased for \(\sigma\): the square root is nonlinear.
2. Why sample variance is consistent
SW Appendix 3.3 assumes i.i.d. observations and \(\mathbb{E}[Y^{4}]<\infty\). Under these sufficient conditions, the finite-variance LLN applies to \(Y^{2}\), while the LLN for \(Y\) and continuity give \(\bar{Y} ^{2}\xrightarrow{p} \mu ^{2}\). The prefactor approaches one. A stronger LLN can prove this consistency using only a finite second moment, but that is a weaker-assumption extension, not the particular proof used in the chapter.
3. Three different measures of spread
| Quantity | Meaning | Formula |
|---|---|---|
| Population SD | Spread of individual outcomes | \(\sigma\) |
| Sampling SD of the mean | Spread of \(\bar{Y}\) across repeated samples | \(\sigma /\sqrt{n}\) |
| Estimated standard error | Estimate of that sampling SD | \(\operatorname{SE}(\bar{Y} )=s/\sqrt{n}\) |
4. Continuous mapping and Slutsky'\(s\) theorem
Supporting theorem connection • Aden notes §3.3
Continuous mapping theorem, constant-limit form. If \(T_{n}\xrightarrow{p} \theta\) and g is continuous at \(\theta\), then \(g(T_{n})\xrightarrow{p} g(\theta )\). Thus \(s^{2}\xrightarrow{p} \sigma ^{2}>0\) implies \(s\xrightarrow{p} \sigma\) and \(\sigma /s\xrightarrow{p} 1\).
Slutsky'\(s\) theorem. If \(A_{n}\xrightarrow{d} A\) and \(B_{n}\xrightarrow{p} b\) for a constant b, then \(A_{n}+B_{n}\xrightarrow{d} A+b\) and \(A_{n}B_{n}\xrightarrow{d}\)bA; division is valid when \(b\ne 0\). Independence of \(A_{n}\) and \(B_{n}\) is not required.
This is studentization: replace unknown sampling scale by a consistent estimate. It justifies the approximate standard normal distribution of the t-statistic under a true null. Because both SE and the true sampling SD shrink toward zero, ratio consistency \(\operatorname{SE}/(\sigma /\sqrt{n})\xrightarrow{p} 1\) is the useful property; merely showing both approach zero would not suffice.
5. Sample proportions and a finite-sample convention
For Bernoulli observations\(, \hat{p} =\bar{Y}\) and \(\sum (Y_{i}-\hat{p} )^{2}=n\hat{p} (1-\hat{p} )\). Consequently:
The last expression directly plugs \(\hat{p}\) into the Bernoulli population variance and is the usual large-sample proportion formula used in the text'\(s\) discussion. It is asymptotically equivalent to \(s/\sqrt{n}\), but not exactly equal at finite \(n\). Keep the convention explicit when checking numerical work.
For testing \(H_{0}\):\(p=p_{0}, a\) null-based scale \(\sqrt{p_{0}(1-p_{0})/n}\) yields a score-type statistic. This differs from the plug-in Wald statistic. Near \(p=0\) or 1 or with very few successes, neither a naive Wald interval nor a generic normal rule should be assumed reliable.
Check your understanding
Question. Is \(\operatorname{SE}(\bar{Y} )=s/n\)? Does Slutsky require the sample mean and sample variance to be independent?
Solution. \(\operatorname{SE}=s/\sqrt{n}\). Independence is unnecessary for Slutsky when the scale ratio converges to a constant.
Sources: SW §3.2, pp. 120–123; Appendix 3.3, p. 154; Exercise 3.18; Aden notes pp. 12–14. The Bernoulli divisor distinction explicitly reconciles the sample-variance definition and large-sample plug-in convention.