σμimplied vol · KKmax(S−K,0)udβρᵢⱼfat tailsPnL · drawdownyield · TdSₜ = μSₜ dt + σSₜ dWₜC = S·N(d₁) − K·e⁻ʳᵀ·N(d₂)σₚ² = wᵀΣwd₁ = [ln(S/K) + (r + σ²⁄2)T] ⁄ σ√T∂V/∂t + ½σ²S²·∂²V/∂S² + rS·∂V/∂S − rV = 0Sharpe = (Rₚ − R_f) ⁄ σₚVaR₉₉ = μ − 2.33σρ = Cov(X,Y) ⁄ σₓσᵧE[Rₚ] = Σ wᵢ·E[Rᵢ]φ(x) = e^(−x²/2) ⁄ √2πE[Rᵢ] = R_f + βᵢ(E[Rₘ] − R_f)dXₜ = θ(μ − Xₜ)dt + σ dWₜσ²ₜ = ω + α·ε²ₜ₋₁ + β·σ²ₜ₋₁C − P = S − K·e⁻ʳᵀΔ = ∂V/∂S Γ = ∂²V/∂S²p = (e^(rΔt) − d) ⁄ (u − d)rₜ = ln(Pₜ ⁄ Pₜ₋₁)E[Xₜ₊₁ | ℱₜ] = Xₜf* = (bp − q) ⁄ bz = (x − μ) ⁄ σP(N=k) = λᵏe^(−λ) ⁄ k!Θ = −∂V/∂Tν = S·φ(d₁)·√Tmin wᵀΣw s.t. wᵀμ = μ*∫Σ∂∞λ
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σμimplied vol · KKmax(S−K,0)udβρᵢⱼfat tailsPnL · drawdownyield · TdSₜ = μSₜ dt + σSₜ dWₜC = S·N(d₁) − K·e⁻ʳᵀ·N(d₂)σₚ² = wᵀΣwd₁ = [ln(S/K) + (r + σ²⁄2)T] ⁄ σ√T∂V/∂t + ½σ²S²·∂²V/∂S² + rS·∂V/∂S − rV = 0Sharpe = (Rₚ − R_f) ⁄ σₚVaR₉₉ = μ − 2.33σρ = Cov(X,Y) ⁄ σₓσᵧE[Rₚ] = Σ wᵢ·E[Rᵢ]φ(x) = e^(−x²/2) ⁄ √2πE[Rᵢ] = R_f + βᵢ(E[Rₘ] − R_f)dXₜ = θ(μ − Xₜ)dt + σ dWₜσ²ₜ = ω + α·ε²ₜ₋₁ + β·σ²ₜ₋₁C − P = S − K·e⁻ʳᵀΔ = ∂V/∂S Γ = ∂²V/∂S²p = (e^(rΔt) − d) ⁄ (u − d)rₜ = ln(Pₜ ⁄ Pₜ₋₁)E[Xₜ₊₁ | ℱₜ] = Xₜf* = (bp − q) ⁄ bz = (x − μ) ⁄ σP(N=k) = λᵏe^(−λ) ⁄ k!Θ = −∂V/∂Tν = S·φ(d₁)·√Tmin wᵀΣw s.t. wᵀμ = μ*∫Σ∂∞λ
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Statistics skill›Module 1 · Describing and estimating›Topic 1.2·~45 minBeginner

Sampling distributions

Skill positionModule 1 of 4 · Topic 2 of 3
1.2 Sampling distributions
—Lab score— / 100Total XP—Streak—

A statistic is a random variable

Draw another sample and the mean changes. Any number computed from a sample, a statistic, is itself random, and its distribution over repeated samples is called its sampling distribution. This is the idea underneath every error bar and p-value: they all describe how a statistic would vary if the data collection were repeated, not how the data vary.

SE⁡(Xˉ)=sn,SE⁡(p^)=p^(1−p^)n,t=xˉ−μ0s/n\operatorname{SE}(\bar X)=\frac{s}{\sqrt{n}},\qquad \operatorname{SE}(\hat p)=\sqrt{\frac{\hat p(1-\hat p)}{n}},\qquad t=\frac{\bar x-\mu_0}{s/\sqrt{n}}SE(Xˉ)=n​s​,SE(p^​)=np^​(1−p^​)​​,t=s/n​xˉ−μ0​​(1)

The standard deviation of a sampling distribution is the standard error. For a mean it is σ/n\sigma/\sqrt{n}σ/n​, estimated by s/ns/\sqrt{n}s/n​; for a proportion p^\hat pp^​ it is p(1−p)/n\sqrt{p(1-p)/n}p(1−p)/n​. When σ\sigmaσ must be estimated from the data, the standardised mean follows a Student-ttt distribution with n−1n-1n−1 degrees of freedom instead of a normal, which has fatter tails for small nnn and approaches the normal as nnn grows. The ttt-statistic measures how many standard errors the sample mean is from a hypothesised value.

Three standard errorsA sample of 64 returns has s=4%s=4\%s=4%: the standard error of the mean is 4/64=0.5%4/\sqrt{64}=0.5\%4/64​=0.5%. In a sample of 100 trades, 30% won: the standard error of that proportion is 0.3⋅0.7/100≈0.046\sqrt{0.3\cdot0.7/100}\approx0.0460.3⋅0.7/100​≈0.046, about 4.6 percentage points. A mean of 0.8%0.8\%0.8% from n=36n=36n=36 with s=3%s=3\%s=3% has t=0.8/(3/6)=1.6t=0.8/(3/6)=1.6t=0.8/(3/6)=1.6: only 1.6 standard errors from zero, not convincing evidence of a nonzero mean.
  • Sampling distribution: how a statistic varies over repeated samples.
  • Standard error: the standard deviation of the sampling distribution; it shrinks like 1/n1/\sqrt{n}1/n​.
  • ttt-distribution: use it for a standardised mean when σ\sigmaσ is estimated, with n−1n-1n−1 degrees of freedom.
Standard deviation is not standard errorA strategy with 1% daily volatility is not uncertain by 1% in its average return: after 400 days the mean return is known to 1%/400=0.05%1\%/\sqrt{400}=0.05\%1%/400​=0.05%. Quoting the wrong one either overstates your uncertainty or hides it. Say which you mean: spread of the data, or precision of an estimate.

Lab · what's graded

  • The standard error of a mean10 pts
  • The standard error of a proportion15 pts
  • A t-statistic15 pts
  • Which distribution15 pts
  • Facts about sampling distributions15 pts
  • How many trades15 pts
  • Standard deviation versus standard error15 pts
0/ 7 answered
Tasks
1Numeric answer10 ptsNot answered

A sample of n = 64 daily returns has sample standard deviation 4%. What is the standard error of the mean, in percent?

SE=SE =SE=accepted: ± 0.005
2Numeric answer15 ptsNot answered

In 100 trades, 30% were winners. What is the standard error of that proportion? Give it to four decimals.

SE=SE =SE=accepted: ± 0.0005
3Numeric answer15 ptsNot answered

A strategy has a mean daily return of 0.8% over n = 36 days with sample standard deviation 3%. What is the t-statistic for the hypothesis that the true mean is 0?

t=t =t=accepted: ± 0.005
4Choice15 ptsNot answered

Returns are roughly normal, n = 12 and the standard deviation is estimated from the data. Which distribution does the standardised sample mean follow?

5Choice15 ptsNot answered

Select every true statement.

6Numeric answer15 ptsNot answered

You want the standard error of a win rate near 50% to be 0.01 (one percentage point). How many trades n do you need? Use p = 0.5.

n=n =n=accepted: ± 0.1%
7Explain in words15 ptsNot answered

In two sentences, say what the standard deviation and the standard error each describe.

Math modeKeyboard or LaTeX inputTask 1 of 7 · Ctrl ⏎ to check

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