Do stocks with high momentum or low volatility earn more next month, and by how much? A single regression on all stocks and all months pools 18,000 observations as if they were independent, but stocks in the same month share the same market shock, so their errors are correlated and the t-statistics come out far too big. Fama and MacBeth avoided this with two passes. First, for every month run a cross-sectional regression of next-month returns on today's characteristics: each month gives one slope per characteristic, the price the market paid that month. Second, treat these monthly slopes as a time series: the premium is their average, and its standard error comes from their spread, se=s/T, so the t-statistic is λˉ/(s/T).
One month and four monthsIn one month, four stocks have characteristics x=−3,−1,1,3 (already centred) and next-month returns y=−1,0,3,2 in percent. Then ∑xy=12, Sxx=20, so the slope is 0.6 and the intercept is yˉ=1.0. Suppose four months of such slopes are 0.6,−0.2,0.5,0.1. Their mean is 0.25, the squared deviations add to 0.41, so s2=0.41/3=0.1367, s=0.370, se=0.370/4=0.185 and t=0.25/0.185=1.35. On real data, take 307 months from January 2001 to July 2026 and about 56 stocks per month (17,075 stock-months in all), with three characteristics made comparable by standardising them each month: 12-month momentum skipping the last month, last month's return, and 12-month volatility. The average monthly slopes are +0.178% for momentum (t=1.68), −0.071% for the last month (t=−0.83) and +0.270% for volatility (t=2.00, or 3.2% a year). A single pooled regression gives volatility t=5.47, more than twice the Fama–MacBeth value, and clustering by month gives 2.35. The volatility premium is positive in both halves of the sample (0.27% and 0.27% a month) but the stock list is today's large companies, so survivorship makes high-volatility winners such as NVIDIA over-represented.
The monthly slopes can themselves be autocorrelated, so many studies replace s/T by a Newey–West standard error with a few lags. Here it hardly matters: for volatility t=2.00 becomes 1.99 with four lags. A second point is that the intercept λ^0 is the average return of a stock with average characteristics, 1.00% a month in this sample (t=4.08), which mostly reflects the market and the survivors.
Two passes: a regression per month, then the mean and spread of the monthly slopes.
Standardise the characteristics each month so the slopes are comparable and outliers matter less.
Lab · what's graded
One month's slope15 pts
The premium10 pts
The spread of the slopes10 pts
The standard error10 pts
The t-statistic15 pts
Pooled against Fama–MacBeth15 pts
What survivorship does25 pts
0/ 7 answered
1Numeric answer15 ptsNot answered
Four stocks have centred characteristics x=−3,−1,1,3 and next-month returns y=−1,0,3,2. With ∑xy=12 and Sxx=20, what is the cross-sectional slope? Give one decimal.
b=accepted: ± 0.01
2Numeric answer10 ptsNot answered
Four monthly slopes are 0.6,−0.2,0.5,0.1. What is their mean, the estimated premium? Give two decimals.
m=accepted: ± 0.005
3Numeric answer10 ptsNot answered
The squared deviations from the mean 0.25 add to 0.41. With T=4 months, what is the standard deviation s=0.41/(T−1)? Use 0.1367=0.370 and give three decimals.
s=accepted: ± 0.004
4Numeric answer10 ptsNot answered
With s=0.370 and T=4 months, what is se=s/T? Give three decimals.
se=accepted: ± 0.002
5Numeric answer15 ptsNot answered
The premium is 0.25 and its standard error is 0.185. What is the t-statistic? Give two decimals.
t=accepted: ± 0.03
6Choice15 ptsNot answered
A pooled regression gives the volatility premium t = 5.47, while Fama–MacBeth gives t = 2.00. Why is the pooled value too big?
7Explain in words25 ptsNot answered
The stock list is today's large companies, and the volatility premium came out positive. In one or two sentences, explain how survivorship could inflate the premium.
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