Finance-Theory · Unit 23 · Video 2 · Interactive Practice

Cycles of Efficiency: 130 Years of S&P Serial Correlation

IKey Formulas

Formula	Name	What you need
$\rho_1 = \dfrac{\mathrm{Cov}(R_t,\, R_{t-1})}{\mathrm{Var}(R_t)}$	First-order autocorrelation	This month's return and last month's
$\mathrm{Cov}(R_t, R_{t-1}) = \mathsf{E}[(R_t - \mu)(R_{t-1} - \mu)]$ , $\ \mathrm{Var}(R_t) = \mathsf{E}[(R_t - \mu)^2]$	Built from deviations	The mean return $\mu = \mathsf{E}[R_t]$
$(R_t + c) - (\mu + c) = R_t - \mu$	A constant mean drops out	The same shift $c$ in every month
$\hat\rho_1 = \dfrac{\sum_{t=2}^{n} (R_t - \bar R)(R_{t-1} - \bar R)}{\sum_{t=1}^{n} (R_t - \bar R)^2}$	One window's estimate	$n$ months of returns, e.g. $n = 60$ for five years

Key Insight: A random walk predicts $\rho_1 \approx 0$ in every window. The rolling five-year serial correlation of the S&P Composite (1871–2003) instead swings, from 40–50% in the late 1800s to a surge in the 1990s and a low level through 2003: efficiency comes in cycles, set by who is trading.

IIVisualization 1 — One Window, One Number

Sliding a fixed window along the returns turns one autocorrelation into a history of them.

Window Window ends at month 720

💡 The four eras' built-in values, 0.30, −0.05, 0.30 and 0.15, are ours; the video's bars come from Shiller's S&P data, which shows the same kind of rise and fall with no steady drift toward zero.

IIIVisualization 2 — Does a Positive Average Return Create Correlation?

Adding the same constant to every month moves the mean and steepens the price drift, but leaves every deviation unchanged.

Constant added to every month (c, %) 0

IVQuiz Questions

Problem 1 · From Correlation to Covariance

Given: Over a 60-month window, monthly returns have standard deviation $\sigma = 5\%$ and first-order autocorrelation $\rho_1 = -0.05$ . Find $\mathrm{Cov}(R_t, R_{t-1})$ in squared percentage points.

✅ Correct!

\mathrm{Cov} = \rho_1 \cdot \mathrm{Var} = -0.05 \times 25 = -1.25

: a small negative covariance, close to the random walk's zero.

❌ Close, but…

\rho_1

divides the covariance by the variance

\sigma^2 = 25

, not by

\sigma = 5

; multiply

-0.05

25

❌ Check the sign. The covariance has the sign of

\rho_1

, because the variance in the denominator is positive.

❌ Not quite. That is

\mathrm{Var}/\rho_1 = 25 / (-0.05)

; the definition gives

\mathrm{Cov} = \rho_1 \times \mathrm{Var}

Show solution

The variance is the square of the standard deviation, in squared percentage points:

\mathrm{Var}(R_t) = \sigma^2 = 5^2 = 25

Rearrange the definition $\rho_1 = \mathrm{Cov}(R_t, R_{t-1}) / \mathrm{Var}(R_t)$ :

\mathrm{Cov}(R_t, R_{t-1}) = \rho_1 \cdot \mathrm{Var}(R_t) = -0.05 \times 25 = -1.25

The covariance is $-1.25$ squared percentage points.

Problem 2 · Leverage and Serial Correlation

Given: Over a 120-month window the index's monthly returns $R_t$ have first-order autocorrelation $\rho_1$ . A leveraged fund earns $R'_t = 2R_t - 0.4\%$ every month (twice the index return, less a fixed 0.4% financing cost). What is the fund's first-order autocorrelation $\rho'_1$ ?

✅ Correct! The fixed cost cancels in every deviation, and the factor 2 scales numerator and denominator by

2^2 = 4

alike.

❌ Not quite. Each deviation doubles, so the covariance gains a factor of

2 \times 2 = 4

, and so does the variance; the ratio does not change.

❌ Close, but… The covariance does grow by 4, but so does the variance in the denominator:

\mathrm{Var}(2R_t) = 4\,\mathrm{Var}(R_t)

❌ Not quite. A constant subtracted from every month shifts the mean by the same amount, so it never reaches the deviations

R'_t - \mu'

Show solution

The fund's mean is $\mu' = 2\mu - 0.4\%$ , so each deviation is

R'_t - \mu' = (2R_t - 0.4\%) - (2\mu - 0.4\%) = 2(R_t - \mu)

The covariance and the variance both pick up a factor of 4:

\mathrm{Cov}(R'_t, R'_{t-1}) = \mathsf{E}[2(R_t - \mu)\cdot 2(R_{t-1} - \mu)] = 4\,\mathrm{Cov}(R_t, R_{t-1}), \qquad \mathrm{Var}(R'_t) = 4\,\mathrm{Var}(R_t)

\rho'_1 = \frac{4\,\mathrm{Cov}(R_t, R_{t-1})}{4\,\mathrm{Var}(R_t)} = \rho_1

Leverage makes the fund's returns swing twice as far but no more predictable: $\rho'_1 = \rho_1$ .

Problem 3 · Roll the Window Forward

Given: Seven consecutive monthly returns (in %): $1$ , $-1$ , $2$ , $4$ , $3$ , $3$ , $1$ . Using the one-window formula from Key Formulas with $n = 6$ , estimate $\hat\rho_1$ for window A (months 1–6), then roll the window forward one month and estimate it for window B (months 2–7). Find both estimates.

✅ Correct!

6/16 = 0.375

, then

2/16 = 0.125

: moving the window one month cuts the estimate to a third, which is why short windows make such noisy bars.

❌ That is the mean talking.

26/40

and

30/40

use the raw returns; subtract each window's mean (2% in both) from every month first, or the positive average inflates the estimate.

❌ Close, but…

6/15

and

2/7

drop each window's first month from the denominator; the sum of squares runs over all six months.

❌ Not quite. Window B swaps month 1 for month 7, and that changes the adjacent products at both ends; recompute its deviations and products.

Show solution

Window A (months 1–6): $\bar R = (1 - 1 + 2 + 4 + 3 + 3)/6 = 12/6 = 2$ , so the deviations are $-1$ , $-3$ , $0$ , $2$ , $1$ , $1$ .

\text{products: } (-3)(-1) + (0)(-3) + (2)(0) + (1)(2) + (1)(1) = 3 + 0 + 0 + 2 + 1 = 6

\text{squares: } 1 + 9 + 0 + 4 + 1 + 1 = 16, \qquad \hat\rho_1 = \frac{6}{16} = 0.375

Window B (months 2–7): $\bar R = (-1 + 2 + 4 + 3 + 3 + 1)/6 = 12/6 = 2$ , so the deviations are $-3$ , $0$ , $2$ , $1$ , $1$ , $-1$ .

\text{products: } (0)(-3) + (2)(0) + (1)(2) + (1)(1) + (-1)(1) = 0 + 0 + 2 + 1 - 1 = 2

\text{squares: } 9 + 0 + 4 + 1 + 1 + 1 = 16, \qquad \hat\rho_1 = \frac{2}{16} = 0.125

The estimates are $0.375$ for window A and $0.125$ for window B. Five of the six months are shared, yet the estimate falls by two thirds: with so few months each estimate is very noisy, which is why the video's exhibit uses 60-month windows.

Problem 4 · Reading a Cycle

Given: A market's rolling 120-month serial correlation reads $0.30$ in decade A, $-0.05$ in decade B and $0.15$ in decade C, in that order. Which reading fits the video's argument?

✅ Correct! Efficiency is read off how close

\rho_1

is to zero, and

|-0.05|

is the smallest; the rise again in decade C is a cycle, set by the population of traders.

❌ Not quite. The conventional view predicts a drift down to zero that stays there; a rise from

-0.05

back to

0.15

is the cycle the video describes instead.

❌ Not quite. The mean is subtracted from every return before anything is multiplied, so a high average return cannot produce serial correlation.

❌ Check the yardstick. A random walk gives

\rho_1 \approx 0

; distance from zero, not the sign, measures predictability, and

-0.05

is the closest of the three.

Show solution

Under the random walk hypothesis the first-order autocorrelation is statistically indistinguishable from zero, so efficiency is judged by $|\rho_1|$ :

|0.30| > |0.15| > |-0.05|

Decade B is the most efficient and decade A the least. The sequence falls and then rises, so it is not the steady decline toward zero of the conventional view. The video attributes such swings to who is trading: sophisticated investors exploit and compete away correlation (as after the 1990s), while a market crowded with retail investors (as in the 1990s) shows more of it. A constant mean return drops out of $\rho_1$ , so the average return in a decade cannot explain the 0.30.

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