The Momentum T-Statistic That Isn't
A systematic equity strategy has 36 monthly excess returns (Jan 2022 – Dec 2024): mean 0.90%, standard deviation 2.40%, skewness −0.85, excess kurtosis 5.2, and a lag-1 autocorrelation of 0.35 that is significant at the 5% level. The portfolio rebalances monthly and ranks stocks on 12-month signals, so consecutive monthly holdings overlap heavily. The marketing deck reports 't = 2.25, significant at 5%, capacity $400m.' One reviewer says the fat tails are irrelevant because 36 observations is enough for the central limit theorem; a second says the autocorrelation is a holding-period artefact and can be ignored. The research team also admits it evaluated 47 related variants over this same 36-month window before settling on this one. You must decide whether the evidence supports funding the strategy.
1. Reproduce the reported t-statistic, then state what the correct inference is once the return dynamics are respected. Show your arithmetic.
Read the worked answer
Naive calculation. SE = 2.40/√36 = 0.40; t = 0.90/0.40 = 2.25. That is the number in the deck.
Correction for serial dependence. For an AR(1) process, the variance of the sample mean is inflated relative to σ²/n by the factor (1+ρ)/(1−ρ) = 1.35/0.65 = 2.077. The standard error therefore scales by √2.077 = 1.441, and t_adj = 2.25 / 1.441 ≈ 1.56.
Why even 1.56 is too generous. The autocorrelation is caused by the 12-month overlapping lookback, not by a benign AR(1) process. Overlapping windows imply non-zero autocorrelation out to at least lag 11, so a lag-1 correction captures only part of the dependence. The right tool is a Newey–West / Hansen–Hodrick estimator with lag ≥ 11, or a block bootstrap with blocks of ≥ 12 months. Intuitively, 36 overlapping monthly observations contain closer to 3 non-overlapping annual observations of information than 36 independent ones.
Decision. The reported t of 2.25 is not evidence of skill. The defensible conclusion is 'indistinguishable from zero on this sample', and the honest fix is more calendar time, not a cleverer standard error.
2. Forty-seven variants were tested on the same data. What best-of-k t-statistic would you expect from 47 skill-less variants, and how does that change the deployment decision?
Read the worked answer
Under the null of no skill, the expected maximum of k independent standard normal test statistics is approximately √(2 ln k). For k = 47: √(2 × 3.85) = √7.70 ≈ 2.77.
So a best-of-47 t of 2.25 is below what pure noise typically delivers. The winning variant is less impressive than a single pre-specified test with the same t-statistic.
Family-wise framing. Bonferroni at 5% family-wise requires per-test p < 0.05/47 = 0.00106, i.e. |t| > 3.08. The result fails even the unadjusted 5% threshold once the autocorrelation correction of the previous task is applied, and fails the adjusted threshold by a wide margin.
Sharpe-inflation framing. The deflated-Sharpe logic says the hurdle rises both with the number of trials and with the non-normality of returns (negative skew and fat tails increase the sampling variance of the Sharpe estimate). A −0.85 skew and excess kurtosis of 5.2 make the hurdle higher still.
Decision. Do not deploy. The requirement should be: pre-register a single specification, test it on non-overlapping out-of-sample data of sufficient length, and disclose the trial count (47) in any performance claim. A backtest that survives only because the search was wide is a selection artefact, and presenting it as a track record would be misleading.
3. Is the reviewer who calls kurtosis 'irrelevant because n is large' right? What should be reported instead of a location test as the basis for deployment?
Read the worked answer
Partly right, and beside the point. The central limit theorem concerns the sampling distribution of the mean, but with n = 36 and excess kurtosis of 5.2 the convergence is slow enough that normal-theory p-values are unreliable in the tail - which is exactly the region a deployment decision cares about. More importantly, a t-test is a test of location: it says nothing about whether the return stream is worth owning.
Skew matters more than kurtosis here. A skew of −0.85 is the signature of a strategy that earns small steady gains and takes occasional large losses - precisely the payoff profile where the sample mean over 36 months can look attractive while the left tail is under-sampled. Three years of monthly data is unlikely to contain a representative number of the bad events.
What to report instead:
- Block-bootstrap confidence interval for the mean (handles autocorrelation and fat tails jointly).
- Downside statistics: worst 5% average monthly return, maximum drawdown, Sortino, and the realised loss in the worst rolling 12-month window.
- Sensitivity of the conclusion to removing the best 3 months (a fragility check that often kills overlapping-signal strategies).
- An explicit statement that the sample contains one volatility regime, so tail estimates are lower bounds.
A location test can veto a strategy; it should never be the affirmative case for funding one.