Finance-Theory · Unit 14 · Video 6 · Interactive Practice

Size, January, Value, Momentum: Anomalies Out of Sample

IKey Formulas

Formula	Name	What it says
$\text{spread} = \bar R_{\text{top decile}} - \bar R_{\text{bottom decile}}$ , $1 \text{ point} = 100 \text{ bp}$	Decile spread	Sort stocks into ten equal-count bins on one characteristic each month; compare the average returns of the two ends
$12\,\bar R_{\text{all}} = R_{\text{Jan}} + 11\,\bar R_{\text{non-Jan}}$	Removing January	Every year has one January and eleven other months, so $\bar R_{\text{non-Jan}} = (12\,\bar R_{\text{all}} - R_{\text{Jan}})/11$ (all monthly averages)
$P/B = \dfrac{\text{market value of equity}}{\text{book value of equity}}$ , $B/M = \dfrac{1}{P/B}$	Value sort	Low $P/B$ (high $B/M$ ) = value stocks; high $P/B$ = growth stocks
$E[\text{false discoveries}] = N\alpha$ , $P(\text{at least one}) = 1 - (1-\alpha)^N$	Data mining	$N$ characteristics with no true effect, each tested at level $\alpha$ (independent tests)

Key Insight: An anomaly is a factoid: a fact about one sample, 1964–2004, that no theory guarantees. Out of sample (2005–2025; for value, 2008–2025) the size, January and value factors turned slightly negative and momentum kept only a fraction of its old average.

IIDecile Spreads, Then and Later

Each sort's end-to-end spread was large in 1964–2004; its Fama-French factor was milder, and shrank or reversed afterwards.

Sort on

💡 Small stocks are also more volatile than large ones, so part of the size spread is paid for with risk; value and growth stocks carry roughly comparable risk, which makes the value spread the bigger puzzle.

IIITaking January Out

Removing January from the all-months average leaves small and large stocks earning about the same.

Step all months

💡 The January pattern faded too: the Fama-French size factor averaged about $+2.3\%$ in the Januarys of 1964–2004 and about $-0.3\%$ in the Januarys of 2005–2025.

IVData Mining: Significance by Chance

Test enough characteristics that truly do nothing, and some will pass a significance test anyway.

Characteristics N Level α

VQuiz Questions

Problem 1 · Book-to-Market and Where a Stock Sits

Given (illustrative): a fictional firm, Tern Logistics, has a market value of equity of $2.4 billion and a book value of equity of $0.6 billion. Find its book-to-market ratio and the side of a price-to-book sort it falls on.

✅ Correct!

P/B = 2.4/0.6 = 4

, so

B/M = 1/4 = 0.25

: the market prices the firm at four times its book value, the growth end of the sort.

❌ That is $P/B$ , not $B/M$ . Price-to-book puts market value on top:

2.4/0.6 = 4

. Book-to-market is its reciprocal.

❌ Right ratio, wrong side. Value stocks have LOW

P/B

(HIGH

B/M

): plenty of book value the market does not seem to appreciate. A

B/M

0.25

is low.

❌ Not quite. Compute

P/B

first, then take its reciprocal.

Show solution

P/B = \frac{\text{market value of equity}}{\text{book value of equity}} = \frac{2.4}{0.6} = 4, \qquad B/M = \frac{1}{P/B} = \frac{1}{4} = 0.25

The price is four times the accounting value of the equity, so the firm sits toward the high- $P/B$ (low- $B/M$ ) end of the sort, where growth and technology stocks sit. In 1964–2004 that end earned about $10.4\%$ a year against about $16.7\%$ at the low- $P/B$ value end.

Problem 2 · The Non-January Average

Given (illustrative): a fictional smallest-size decile averages $1.5\%$ per month across all months and $6.6\%$ in January. Find its average return in the other eleven months.

✅ Correct!

(12 \times 1.5\% - 6.6\%)/11 = 11.4\%/11 \approx 1.04\%

: January alone carries a large share of the decile's return.

❌ Divide by eleven, not twelve. After January is removed,

11.4\%

is shared by the eleven remaining months:

11.4\%/11 \approx 1.04\%

❌ The all-months average includes January. A January far above the average pulls the average up, so the other months must be lower than

1.5\%

❌ The sum uses twelve months.

12 \times 1.5\% = 18\%

is the sum of the twelve monthly averages;

11 \times 1.5\%

leaves one month out before January is subtracted.

❌ Not quite. Use

12\,\bar R_{\text{all}} = R_{\text{Jan}} + 11\,\bar R_{\text{non-Jan}}

Show solution

Each year contributes one January and eleven other months, so the twelve monthly averages add up as

12\,\bar R_{\text{all}} = R_{\text{Jan}} + 11\,\bar R_{\text{non-Jan}}

\bar R_{\text{non-Jan}} = \frac{12 \times 1.5\% - 6.6\%}{11} = \frac{18\% - 6.6\%}{11} = \frac{11.4\%}{11} \approx 1.04\% \text{ per month}

The $18\%$ is a sum of monthly averages used only for this subtraction, not an annual return.

Problem 3 · The Momentum Crash of 2009

Given: from March to May 2009 past losers rose $163\%$ and past winners rose $8\%$ (the video's figures). A position is $1 long the past-winner portfolio and $1 short the past-loser portfolio over those months; ignore interest on the short-sale proceeds. Find the position's profit or loss per $1 long, as a percentage of that $1.

✅ Correct! The long leg gains

8\%

of its $1, the short leg loses

163\%

of its $1: a winners-minus-losers spread of

8 - 163 = -155

percentage points, so the position loses $1.55 per $1 long, more than the whole long stake. A short position takes the full force of the losers' rebound.

❌ A short leg earns minus the asset's return. When the losers rise

163\%

, the short loses

163\%

; it does not add to the gain.

❌ That is a different number.

-34\%

is the Fama-French momentum factor in April 2009 alone, a diversified factor over one month, not this decile-end position over three months.

❌ Check the sign. The short leg is the one that rose most, so the position loses: winners minus losers,

8\% - 163\%

❌ Not quite. Long-short return = return on the long leg minus return on the short leg.

Show solution

The long leg: $1 in past winners grows to $1.08, a gain of 8 cents.

The short leg: the past-loser portfolio sold short for $1 is worth $2.63 when it is bought back, a loss of $1.63.

R_{\text{long-short}} = R_{\text{winners}} - R_{\text{losers}} = 8\% - 163\% = -155\%

The position loses $1.55 per $1 long. Momentum's out-of-sample average of about $+1.4\%$ a year (2005–2025, against about $+10\%$ in 1964–2004) includes this crash.

Problem 4 · Mining 60 Characteristics

Given (illustrative): a researcher tests 60 fictional stock characteristics, none of which truly affects returns, each at the $10\%$ significance level, with independent tests. Find the expected number that test significant and the probability that at least one does.

Expected false discoveries

Probability of at least one

✅ Correct!

60 \times 0.10 = 6

expected false discoveries, and

1 - 0.9^{60} \approx 99.8\%

: a search this wide is almost certain to turn up something that looks significant.

❌ Check the expected count. Each of the 60 has a

10\%

chance of passing by luck:

N\alpha = 60 \times 0.10

❌ That counts the ones that fail.

54 = 60 \times 0.9

is the expected number NOT significant.

❌ Check the probability. Find the chance that all 60 fail,

0.9^{60}

, and subtract it from 1.

❌ That is the chance of no false discovery.

0.9^{60} \approx 0.18\%

; at least one is its complement.

❌ Never certain. All 60 can fail by chance, with probability

0.9^{60} \approx 0.18\% > 0

Show solution

E[\text{false discoveries}] = N\alpha = 60 \times 0.10 = 6

P(\text{none}) = (1 - 0.10)^{60} = 0.9^{60} \approx 0.0018

P(\text{at least one}) = 1 - 0.9^{60} \approx 0.9982 \approx 99.8\%

A characteristic that passes such a search is a candidate, not a finding: it still has to hold up in a sample it was not selected on.

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