Squared residuals expose a change that the fitted mean misses
The constructed example below has 12 equally spaced observations. Returns y are measured in percentage points and fitted to y = a + b × t, with t = 1 through 12. OLS gives a = 0.2 and b = 0.1. The second six residuals have four times the amplitude of the first six. This is an arithmetic demonstration, not market data or evidence that returns follow a time trend.
- Within each six-observation segment the residual mean is zero. Descriptive sample variance uses Σ(e - segment mean)² / 5: 0.024 pp² in the first segment and 0.384 pp² in the second. Their square roots are 0.154919 pp and 0.619677 pp, a fourfold spread difference.
- These are descriptive standard deviations of fitted residuals. The divisor 5 is not a claim that each segment provides an unbiased estimate of the underlying error variance after the full-sample regression.
- Plot signed residuals against fitted values as well as squared residuals against time. Curvature or a level shift in signed residuals can indicate a deficient mean model. A variance chart alone cannot distinguish that problem from changing error variance.
| t | Return y (pp) | Fitted y (pp) | Residual e (pp) | e² (pp²) |
|---|---|---|---|---|
| 1 | 0.4 | 0.3 | 0.1 | 0.01 |
| 2 | 0.2 | 0.4 | -0.2 | 0.04 |
| 3 | 0.6 | 0.5 | 0.1 | 0.01 |
| 4 | 0.7 | 0.6 | 0.1 | 0.01 |
| 5 | 0.5 | 0.7 | -0.2 | 0.04 |
| 6 | 0.9 | 0.8 | 0.1 | 0.01 |
| 7 | 1.3 | 0.9 | 0.4 | 0.16 |
| 8 | 0.2 | 1.0 | -0.8 | 0.64 |
| 9 | 1.5 | 1.1 | 0.4 | 0.16 |
| 10 | 1.6 | 1.2 | 0.4 | 0.16 |
| 11 | 0.5 | 1.3 | -0.8 | 0.64 |
| 12 | 1.8 | 1.4 | 0.4 | 0.16 |