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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Probability skill›Module 3 · Limit theorems and transformations›Topic 3.2·~45 minIntermediate

Central limit theorem

Skill positionModule 3 of 5 · Topic 2 of 3
3.2 Central limit theorem
—Lab score— / 100Total XP—Streak—

Why bell curves are everywhere

The law of large numbers says the average settles near μ\muμ. The central limit theorem says what the leftover error looks like: for independent observations with mean μ\muμ and finite variance σ2\sigma^2σ2, the standardised average Xˉn−μσ/n\dfrac{\bar X_n-\mu}{\sigma/\sqrt{n}}σ/n​Xˉn​−μ​ is approximately a standard normal for large nnn, whatever the shape of the original distribution. Sums and averages of many small independent effects are normal even when the pieces are not.

Xˉn ≈ N ⁣(μ, σ2n),μ∈Xˉn±1.96 σn  (95%)\bar X_n\ \approx\ N\!\left(\mu,\ \frac{\sigma^{2}}{n}\right),\qquad \mu\in\bar X_n\pm1.96\,\frac{\sigma}{\sqrt{n}}\ \ (95\%)Xˉn​ ≈ N(μ, nσ2​),μ∈Xˉn​±1.96n​σ​  (95%)(1)

This is why a 95% confidence interval for a mean is the sample average plus or minus 1.96 standard errors, and why the probability that an average is far from its mean can be read from the normal table. A sum of nnn independent terms each with standard deviation σ\sigmaσ has standard deviation σn\sigma\sqrt{n}σn​: daily volatility scales to annual by multiplying by 252\sqrt{252}252​, which is the same theorem seen from the sum side.

A confidence interval and a probabilityDaily returns have σ=3%\sigma=3\%σ=3%. The average of n=36n=36n=36 days has standard error 3%/6=0.5%3\%/6=0.5\%3%/6=0.5%, so a 95% interval for the true mean is the sample average ±0.98%\pm0.98\%±0.98%. If instead the standard error is 0.25%0.25\%0.25%, the chance that the sample mean exceeds the true mean by 0.5%0.5\%0.5% is P(Z>2)≈0.0228P(Z>2)\approx0.0228P(Z>2)≈0.0228. And 25 independent days of 1% volatility add up to a standard deviation of 1%⋅25=5%1\%\cdot\sqrt{25}=5\%1%⋅25​=5%.
  • CLT: averages and sums of many independent finite-variance terms are approximately normal.
  • Confidence interval: average ±1.96 σ/n\pm1.96\,\sigma/\sqrt{n}±1.96σ/n​ for about 95%.
  • Square-root-of-time: independent daily volatility scales to TTT days by T\sqrt{T}T​.
When the theorem helps lessThe normal approximation is about the middle of the distribution, and it needs independence and finite variance. With heavy-tailed returns it sets in slowly and says little about extremes; with volatility clustering the observations are not independent, so the effective sample is smaller than nnn. Check these assumptions before trusting a 1.96 standard-error interval.

Lab · what's graded

  • A standard error10 pts
  • A tail probability15 pts
  • A confidence interval15 pts
  • What the theorem needs15 pts
  • A sum over days10 pts
  • When the approximation is poor15 pts
  • Fat tails slow the theorem20 pts
0/ 7 answered
Tasks
1Numeric answer10 ptsNot answered

Daily returns have σ=3%\sigma=3\%σ=3%. What is the standard error of the average of n=36n=36n=36 days, in percent?

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

The standard error of a sample mean is 0.25%0.25\%0.25%. By the CLT, what is the probability that the sample mean exceeds the true mean by more than 0.5%0.5\%0.5%? Use P(Z>2)P(Z>2)P(Z>2) and give four decimals.

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

With a standard error of 0.5%0.5\%0.5%, what is the half-width of the 95% confidence interval for the mean, in percent? Use 1.96.

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

Which pair of conditions does the classical CLT need?

5Numeric answer10 ptsNot answered

25 independent daily returns each have standard deviation 1%1\%1%. What is the standard deviation of their sum, in percent?

σsum=\sigma_{\text{sum}} =σsum​=accepted: ± 0.005
6Choice15 ptsNot answered

Select every situation in which the normal approximation to the sample mean is poor.

7Explain in words20 ptsNot answered

In two sentences, explain why averages of heavy-tailed returns need more observations before they look normal.

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

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