Chapter 4 of 5 · 11 min

How sure is the average? The standard error

A sample mean is a random number with its own volatility: the daily standard deviation over the square root of the days. Halving the uncertainty takes four times the data, which is why strategy returns take years to tell apart from zero.

By the end of this chapter you can
  • Compute the standard error of a mean daily return
  • Build a 95% confidence interval and read it against zero
  • Compute a t-statistic and say whether it clears 1.96
  • Solve for the days a target precision needs, and the standard deviation an interval implies
1

The intuition

A strategy averaged 5 basis points a day over 250 days, with a daily standard deviation of 100 basis points. Run it over a different 250 days and the average would be different: its standard error is 100 ÷ √250 = 6.3 basis points, larger than the mean itself. The 95% interval is 5 ± 1.96 × 6.3, from −7.4 to +17.4 basis points. It contains zero. A year of data cannot tell this strategy from one with no edge.

The square root is the painful part. To get the interval down to ±2 basis points you need (1.96 × 100 ÷ 2)² = 9,604 days, thirty-eight years. Precision on a mean costs data at the square of the precision you want, and that is why track records are so hard to judge: a few basis points of edge buried under a hundred of noise takes longer to prove than most strategies survive.

The key idea

SE = s ÷ √n. 95% interval = mean ± 1.96 × SE. t = mean ÷ SE. Days for a margin E: n = (1.96 × s ÷ E)², rounded up. From an interval's half-width: s = half-width × √n ÷ 1.96. Four times the data halves the standard error.

2

Why it works

  • The conventions here: independent, identically distributed daily returns in basis points. The sample is large, so the normal 1.96 is used for a two-sided 95% interval instead of a t value. Required sample sizes round up to a whole day.
  • Why s ÷ √n. The variance of an average of n independent draws is the variance of one draw over n. Take the square root: the standard deviation of the mean is s ÷ √n.
  • The interval and the test are the same thing. The interval excludes zero exactly when |mean ÷ SE| exceeds 1.96. 'Significant at 5%' and 'the 95% interval misses zero' are one statement.
  • More data shifts the bounds, not the center. Four times the days halves the half-width; the mean stays where the data put it.
  • Solving for n. Set 1.96 s ÷ √n = E and square: n = (1.96 s ÷ E)². Halve the margin, quadruple the days.
  • What significance does not rule out. Overfitting, costs, a regime change. A mean that is statistically different from zero under these assumptions is still a mean measured on the past.
Mean 5 bps a day, standard deviation 100 bps, 250 days
Standard error: 100 ÷ √2506.32 bps
Half-width: 1.96 × 6.3212.40 bps
95% interval−7.40 to +17.40 bps: contains zero
t-statistic: 5 ÷ 6.320.79, below 1.96
Days for ±2 bps: (1.96 × 100 ÷ 2)²9,604 days, about 38 years
With 1,000 days: half-width 12.40 ÷ 26.20 bps

The same 5 bps a day over 1,000 days has t = 5 ÷ 3.16 = 1.58: still short of 1.96. It would take about 1,537 days, six years, for this edge to clear the bar.

3

The formulas

SE = s ÷ √n

The standard deviation of the mean.

95% interval = mean ± 1.96 × SE; t = mean ÷ SE

Two views of one test: the interval misses zero when t exceeds 1.96.

n = (1.96 × s ÷ E)², rounded up

Days for a margin of E either side.

s = half-width × √n ÷ 1.96

The daily standard deviation an interval implies.

4

Worked example

Standard error, then half-width, then the two bounds. The follow-up says what the interval lets you conclude, which depends on whether it contains zero.

Drawing the numbers…
5

See it move

Same strategy. Change the days of data, the daily standard deviation, the mean, and the margin you want.

Drawing the numbers…
Try this
  • Add days. The standard error and the half-width fall, the interval narrows around the same mean, and the t-statistic rises.
  • Raise the daily standard deviation. The standard error, the half-width and the days needed rise; the t-statistic falls.
  • Raise the mean. The whole interval shifts up and the t-statistic rises; the width does not change.
  • Widen the margin you will accept. The days needed fall as its square.
6

Run it backwards

The daily standard deviation is known. How many days of data does a target margin need?

Drawing the numbers…

Set 1.96 × s ÷ √n equal to the margin and square: n = (1.96 s ÷ E)². Round up to a whole day.

The follow-up divides by 252 to turn the days into years, which is usually the point of the question.

7

Traps

Using the standard deviation where the standard error belongs.
The spread of one day is s; the spread of the average is s ÷ √n.
Believing twice the data halves the uncertainty.
Four times the data halves it. Precision costs data at the square.
Reading a positive mean as a positive strategy.
Check the interval. If it contains zero, the data cannot tell the strategy from noise.
Reading significance as proof of an edge.
It rules out 'no edge' under the assumptions. Not overfitting, costs, or a regime change.
Forgetting to round n up.
A fraction of a day is not enough data for the margin. Round up.
8

Say it in the interview

The interviewer asks

A strategy made 5 basis points a day over a year with 100 basis points of daily volatility. Is it real?

Say yours out loud first, then compare.
9

Check yourself

5 fresh questions, with new numbers. Answer each one correctly to finish the chapter. Get one wrong and you will see the full working, then you can try it again with new numbers.

Answers within 1% are marked right. Type the number; $, %, x and M are fine. First tries count toward Learned: the topic is Learned once every chapter is done and 75% of first tries were right.

0 of 5
Drawing your questions…
Remember
  • SE = s ÷ √n; the interval is mean ± 1.96 SE; t = mean ÷ SE.
  • Four times the data halves the uncertainty.
  • n = (1.96 s ÷ E)² for a margin E, rounded up.
  • Significance rules out 'no edge' under the assumptions, and nothing more.