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 2 · Random variables and distributions›Topic 2.3·~55 minIntermediate

Common distributions

Skill positionModule 2 of 5 · Topic 3 of 3
2.3 Common distributions
—Lab score— / 100Total XP—Streak—

A small zoo of distributions

The normal distribution N(μ,σ2)N(\mu,\sigma^2)N(μ,σ2) is the bell curve described by its mean and variance. About 68% of the probability lies within one σ\sigmaσ of the mean, 95% within 1.96 σ\sigmaσ and 99.7% within three. Standardising, Z=(X−μ)/σZ=(X-\mu)/\sigmaZ=(X−μ)/σ, turns any normal into N(0,1)N(0,1)N(0,1), whose cumulative distribution Φ\PhiΦ is tabulated. Daily returns are often modelled as normal, which is a useful but imperfect first approximation.

ln⁡S∼N(m,s2) ⇒ E[S]=em+s2/2,P(N=k)=λke−λk!\ln S\sim N(m,s^{2})\ \Rightarrow\ E[S]=e^{m+s^{2}/2},\qquad P(N=k)=\frac{\lambda^{k}e^{-\lambda}}{k!}lnS∼N(m,s2) ⇒ E[S]=em+s2/2,P(N=k)=k!λke−λ​(1)

A price cannot go below zero, so its log is modelled as normal and the price itself is lognormal: skewed to the right, with mean em+s2/2e^{m+s^2/2}em+s2/2 larger than the median eme^{m}em. A Poisson variable counts events in a fixed interval when they arrive independently at a constant average rate λ\lambdaλ: trades in a minute, defaults in a year. The Student-ttt distribution looks like a normal with heavier tails; its degrees of freedom ν\nuν control the heaviness, with small ν\nuν giving far more extreme values and ν→∞\nu\to\inftyν→∞ recovering the normal.

Tail numbers and a lognormal meanA daily return is N(0.05%, 1.2%2)N(0.05\%,\ 1.2\%^2)N(0.05%, 1.2%2). Its 95% one-day value at risk is the loss exceeded only 5% of the time: 1.645⋅1.2%−0.05%=1.924%1.645\cdot1.2\%-0.05\%=1.924\%1.645⋅1.2%−0.05%=1.924%. For a log price N(0.05, 0.22)N(0.05,\ 0.2^2)N(0.05, 0.22) starting from 1, the expected price is e0.05+0.02≈1.0725e^{0.05+0.02}\approx1.0725e0.05+0.02≈1.0725, above the median e0.05≈1.0513e^{0.05}\approx1.0513e0.05≈1.0513. With λ=3\lambda=3λ=3 trades per minute, the chance of a quiet minute with none is e−3≈0.0498e^{-3}\approx0.0498e−3≈0.0498.
  • Normal: symmetric; use zzz-scores. Good for sums, rough for single returns.
  • Lognormal: positive and right-skewed, for prices. Poisson: counts of independent events.
  • Student-ttt: heavy tails; fits returns far better than the normal in the extremes.
Why 'once in 10,000 years' keeps happeningA daily move of five standard deviations has probability about 3×10−73\times10^{-7}3×10−7 under a normal model, once in roughly 13,000 years of trading days. Real equity markets have produced several such days in a few decades. The normal curve is a poor guide to extreme days, which is why risk models use heavier-tailed distributions such as the Student-ttt.

Lab · what's graded

  • Pick the distribution20 pts
  • A normal probability10 pts
  • A value at risk15 pts
  • A Poisson probability15 pts
  • A lognormal mean15 pts
  • Why the Student-t15 pts
  • Normal models and crashes10 pts
0/ 7 answered
Tasks
1Match20 ptsNot answered

Match each quantity to the distribution that models it best.

Pick the partner of each item on the left.

The number of trades arriving in one minute
A stock price at a future date
Daily returns with occasional extreme days
The sum of many small independent effects
2Numeric answer10 ptsNot answered

For a standard normal ZZZ, what is P(∣Z∣<1.96)P(|Z|<1.96)P(∣Z∣<1.96)?

P=P =P=accepted: ± 0.005
3Numeric answer15 ptsNot answered

A daily return is N(0.05%, 1.2%2)N(0.05\%,\ 1.2\%^2)N(0.05%, 1.2%2). Using the 5% quantile z=−1.645z=-1.645z=−1.645, what loss (in percent, positive number) is exceeded only 5% of the time?

L=L =L=accepted: ± 0.01
4Numeric answer15 ptsNot answered

Trades arrive as a Poisson process with λ=3\lambda=3λ=3 per minute. What is the probability of a minute with no trade, P(N=0)P(N=0)P(N=0)? Give it to four decimals.

P(N=0)=P(N=0) =P(N=0)=accepted: ± 0.0005
5Numeric answer15 ptsNot answered

A price starts at 1 and ln⁡S∼N(0.05, 0.22)\ln S\sim N(0.05,\ 0.2^2)lnS∼N(0.05, 0.22) at the horizon. What is E[S]E[S]E[S]? Give it to four decimals.

E[S]=E[S] =E[S]=accepted: ± 0.0005
6Choice15 ptsNot answered

Why is a Student-t often a better model for daily returns than a normal?

7Explain in words10 ptsNot answered

In two sentences, say why a normal model understates the probability of a market crash.

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

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