Use the standard error and t-statistic to decide whether a performance measure is statistically significant.
The idea
A performance estimate such as alpha or an average excess return is a noisy sample figure, so we test whether it differs from zero. The t-statistic is the estimate divided by its standard error, as follows. t = estimate ÷ SE. For a mean of n returns with standard deviation σ, the standard error is found as follows. SE = σ ÷ √n. For a regression alpha, the standard error comes from the regression output. With enough observations to treat t as roughly normal, an absolute value above about 2 rejects zero, and 1.96 is the exact 5% two-sided level. The sample size needed grows very quickly as the noise-to-signal ratio rises, as follows. n = (t × σ ÷ estimate)². Small alphas therefore need decades of data. The usual mistake is ranking managers by the size of alpha rather than by its t-statistic.
Formulas
t-statistic of a performance estimate
t = estimate ÷ SE
- estimate
- the alpha or mean excess return
- SE
- its standard error
Reject zero if |t| exceeds about 2, or 1.96 at 5% two-sided, in large samples. Inverting the ratio or using the standard deviation in place of SE are the traps.
Standard error of a mean return
SE = σ ÷ √n
- σ
- the population standard deviation
- n
- the sample size
Use the same periodicity for σ and n. Dividing by n instead of √n badly overstates t.
Observations needed for alpha to reach a target t-statistic
n = (t × σ ÷ α)²
- t
- the target t-statistic, about 2
- σ
- the residual standard deviation per period
- α
- the expected alpha per period
The whole ratio is squared, including t. Measure σ and α over the same period; n is in those periods.
A practice question
The standard error of an estimated mean excess return is sigma/sqrt(n). If a manager's track record is extended from 4 years to 16 years of data with the same volatility, the standard error:
- is cut by three-quarters.Why not: Treats the standard error as proportional to 1/n instead of 1/sqrt(n).
- is unchanged, because volatility did not change.Why not: Ignores that the standard error shrinks as the sample grows.
- is cut in half. - correct
- is cut by a quarter.Why not: Takes the standard error as falling in proportion to the sample size multiple (1/4) instead of its square root.
The number of observations rises by a factor of 4, so the standard error is multiplied by 1/sqrt(4) = 1/2.
A harder one
A manager's regression shows a monthly alpha of 0.20% with a standard error of 0.12%. At a 5% two-sided significance level, which conclusion is correct?
- t = 1.67, which exceeds 1.0, so the alpha is statistically significant.Why not: Uses a critical value of 1 rather than about 1.96.
- t = 1.67, below about 1.96, so zero alpha cannot be rejected. - correct
- t = 16.7, so the alpha is highly significant.Why not: Mixes units: treats 0.20% as 0.2 but 0.12% as 0.012.
- t = 0.60, so the alpha is not significant.Why not: Inverts the ratio (standard error divided by alpha).
t = 0.20/0.12 = 1.67. The 5% two-sided critical value is 1.96, so the alpha is not significantly different from zero.
The hardest tier
A manager's expected alpha is 0.30% per month and the monthly residual standard deviation is 2.5%. Assuming these are the true values, about how many months of data are needed for the expected t-statistic to reach 2?
- 17Why not: Computes 2 x 2.5/0.30 and forgets to square it.
- 69Why not: Computes (2.5/0.30)² and omits the critical value of 2.
- 278 - correct
- 139Why not: Squares only the residual part: 2 x 2.5²/0.30², so the multiplier is not squared with the rest.
t = alpha/(sigma/sqrt(n)), so n = (t x sigma/alpha)² = (2 x 2.5/0.30)² = 278 months, roughly 23 years.
In the app: 6 practice questions on this outcome, spaced review that brings it back before you forget it.
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