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Hit-Rate Inference for Directional Signals

How to properly test whether a signal's win rate on directional bets (up versus down) is genuinely better than chance, using the binomial distribution — and why a hit rate that looks impressive can still be statistically indistinguishable from a coin flip.

Prerequisites: One-Sample and Two-Sample t-Tests

A directional signal says "up" or "down" each day, and over a year of predictions it's right 55% of the time. Fifty-five percent sounds meaningfully better than a coin flip's fifty. But "better than chance" is a statistical question, not a vibe — it depends on exactly how many predictions were made, because a hit rate calculated from a small number of trials is a noisy estimate that can easily land at 55% by pure luck even if the true underlying skill is zero. Hit-rate inference is the standard machinery — built on the binomial distribution — for deciding whether an observed win rate is actually distinguishable from a fair coin.

An analogy: is this coin actually biased?

If you flip a coin 20 times and get 11 heads (55%), nobody would claim the coin is biased — that's an utterly unremarkable outcome for a fair coin, easily explained by ordinary randomness. If you flip the same coin 20,000 times and get 11,000 heads (still exactly 55%), that's an entirely different story — a fair coin landing 55% heads over 20,000 flips is astronomically unlikely, so you'd be quite confident the coin is genuinely biased. The same hit rate, 55%, is either meaningless or overwhelming evidence, purely depending on the sample size. A trading signal's hit rate needs exactly this same scrutiny before anyone calls it skill.

The idea, one symbol at a time

Model each prediction as a Bernoulli trial: correct with true probability pp, incorrect with probability 1p1-p. Under the null hypothesis of no skill, p=0.5p = 0.5. Given nn independent predictions with kk correct, the observed hit rate is p^=k/n\hat{p} = k/n. Its standard error, under the null, is:

SE=0.5×(10.5)n=0.5n.\text{SE} = \sqrt{\frac{0.5 \times (1-0.5)}{n}} = \frac{0.5}{\sqrt{n}} .

In plain English: the uncertainty in a hit-rate estimate shrinks proportionally to 1/n1/\sqrt{n} — quadrupling the number of predictions only halves the uncertainty, which is why small samples give such shaky hit-rate estimates. The test statistic for whether p^\hat{p} is significantly different from 0.5 is:

z=p^0.5SE=p^0.50.5/n.z = \frac{\hat{p} - 0.5}{\text{SE}} = \frac{\hat{p} - 0.5}{0.5/\sqrt{n}} .

In plain English: how many standard errors the observed hit rate sits away from the coin-flip benchmark of 50%. Compare z|z| to the usual normal critical value (about 1.96 for a 95% two-sided test); if z<1.96|z| < 1.96, the observed hit rate, however good it looks, is not statistically distinguishable from pure chance.

Worked example 1: a modest sample

A signal makes n=100n = 100 directional predictions and is correct k=58k = 58 times, p^=0.58\hat{p} = 0.58.

SE=0.5100=0.05,z=0.580.50.05=0.080.05=1.6.\text{SE} = \frac{0.5}{\sqrt{100}} = 0.05, \qquad z = \frac{0.58 - 0.5}{0.05} = \frac{0.08}{0.05} = 1.6 .

z=1.6<1.96|z| = 1.6 < 1.96: not statistically significant. A 58% hit rate over just 100 trials is well within the range a coin flip could plausibly produce by chance alone — roughly an 11% probability of seeing a deviation at least this large under the null, not rare enough to be convincing.

Worked example 2: the same hit rate, far more trials

The same signal is instead run for n=2,500n = 2{,}500 predictions, still landing at p^=0.58\hat{p} = 0.58 (k=1,450k = 1{,}450).

SE=0.52500=0.550=0.01,z=0.080.01=8.0.\text{SE} = \frac{0.5}{\sqrt{2500}} = \frac{0.5}{50} = 0.01, \qquad z = \frac{0.08}{0.01} = 8.0 .

z=8.0|z| = 8.0 is enormously beyond the 1.96 threshold — with this many trials, a 58% hit rate is essentially impossible to attribute to a fair coin. Identical hit rate, twenty-five times the sample size, completely different conclusion: the first case was inconclusive noise, the second is overwhelming evidence of real predictive skill (or at least, of p0.5p \ne 0.5; it says nothing yet about whether that edge is large enough to be profitable after costs).

observed hit rate p = 0.50 (null) n=100 (wide) n=2500 (narrow) observed 0.58
At n=100, a hit rate of 0.58 sits well inside the null distribution's bulk; at n=2500, the null distribution has narrowed so much that 0.58 sits far out in its tail.
z=1.96 z=1.6 n=100 z=8.0 n=2500
The identical 58% hit rate produces a z-statistic below the significance threshold at n=100 but far above it at n=2500 — sample size, not the hit rate itself, decides significance.

What this means in practice

Hit-rate inference is the quick sanity check for any signal reported as a win percentage — a discretionary trader's "I'm right 60% of the time," a classifier's directional accuracy, or a signal vendor's marketing claim — before treating it as evidence of skill rather than noise. It's also a reminder that a hit rate alone never tells you about profitability: a 51% hit rate on trades with a 3:1 win/loss ratio can be far more profitable than a 65% hit rate on trades with a poor payoff ratio, so hit-rate significance and economic significance are separate questions that both need to be checked.

A hit rate is a Bernoulli sample proportion, and its standard error under the no-skill null shrinks only as 1/√n. The exact same hit rate can be statistically indistinguishable from a coin flip at a small sample size and overwhelming evidence of skill at a large one — always compute the z-statistic (or an exact binomial test) against n before treating any reported hit rate as meaningful.

A hit-rate test alone says nothing about whether a strategy makes money — it only tests directional accuracy, ignoring the size of wins versus losses. A signal can have a statistically rock-solid hit rate above 50% and still lose money if its small, frequent wins are dwarfed by occasional large losses (or vice versa: a hit rate indistinguishable from 50% can still belong to a profitable strategy with an asymmetric payoff). Never substitute a hit-rate test for an actual test of the strategy's return distribution.

Related concepts

Practice in interviews

Further reading

  • Wasserman, All of Statistics, ch. 10
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