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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Time Series skill›Module 3 · Volatility›Topic 3.1·~50 minAdvanced

Volatility clustering and ARCH

Skill positionModule 3 of 5 · Topic 1 of 3
3.1 Volatility clustering and ARCH
—Lab score— / 100Total XP—Streak—

When the variance is not constant

Returns are close to uncorrelated, yet they are not independent: the size of today's move tells you about the size of tomorrow's. Quiet weeks follow quiet weeks and turbulent ones follow turbulent ones. The ARCH model of Engle (1982) captures this by letting the variance of the shock depend on recent squared shocks: rt=μ+εtr_t=\mu+\varepsilon_trt​=μ+εt​, εt=σtzt\varepsilon_t=\sigma_tz_tεt​=σt​zt​ with ztz_tzt​ standard normal, and σt2=ω+α1εt−12+⋯+αqεt−q2\sigma_t^{2}=\omega+\alpha_1\varepsilon_{t-1}^{2}+\dots+\alpha_q\varepsilon_{t-q}^{2}σt2​=ω+α1​εt−12​+⋯+αq​εt−q2​. A large shock yesterday raises the variance today, and the shock itself has no memory in its sign. The unconditional variance of an ARCH(1) is ω/(1−α)\omega/(1-\alpha)ω/(1−α) and it needs α<1\alpha<1α<1.

σt2=ω+α εt−12,Var(εt)=ω1−α,LM=T R2∼χq2,σt2=λσt−12+(1−λ) rt−12\sigma_t^{2}=\omega+\alpha\,\varepsilon_{t-1}^{2},\qquad \mathrm{Var}(\varepsilon_t)=\frac{\omega}{1-\alpha},\qquad \mathrm{LM}=T\,R^{2}\sim\chi^{2}_{q},\qquad \sigma_t^{2}=\lambda\sigma_{t-1}^{2}+(1-\lambda)\,r_{t-1}^{2}σt2​=ω+αεt−12​,Var(εt​)=1−αω​,LM=TR2∼χq2​,σt2​=λσt−12​+(1−λ)rt−12​(1)

Even with normal shocks ztz_tzt​, an ARCH model produces returns with fatter tails than a normal distribution, because a calm stretch and a turbulent stretch are mixed in one sample. This is why return histograms show excess kurtosis. To test for ARCH effects, take the residuals ε^t\hat\varepsilon_tε^t​ of a mean model, regress ε^t2\hat\varepsilon_t^{2}ε^t2​ on its own qqq lags and compute Engle's statistic TR2T R^{2}TR2, which is χq2\chi^{2}_qχq2​ under the null of no ARCH. A simple alternative that needs no fitting is the exponentially weighted moving average, σt2=λσt−12+(1−λ)rt−12\sigma_t^{2}=\lambda\sigma_{t-1}^{2}+(1-\lambda)r_{t-1}^{2}σt2​=λσt−12​+(1−λ)rt−12​, with λ=0.94\lambda=0.94λ=0.94 as the RiskMetrics default for daily data. Real returns also show a leverage effect: a fall raises the next variance more than a rise of the same size, which the plain ARCH cannot represent. Matching the long persistence of volatility takes many ARCH lags, and GARCH, covered later in this skill, does it with just a few parameters.

One shock and the next varianceTake an ARCH(1) in percent units with ω=0.2\omega=0.2ω=0.2 and α=0.5\alpha=0.5α=0.5. The long-run variance is 0.2/(1−0.5)=0.40.2/(1-0.5)=0.40.2/(1−0.5)=0.4, a daily volatility of 0.63%0.63\%0.63%. After a quiet day with ε=0.2\varepsilon=0.2ε=0.2 the next variance is 0.2+0.5⋅0.04=0.220.2+0.5\cdot0.04=0.220.2+0.5⋅0.04=0.22. After a shock of ε=−2\varepsilon=-2ε=−2 it is 0.2+0.5⋅4=2.20.2+0.5\cdot4=2.20.2+0.5⋅4=2.2, a volatility of 1.48%1.48\%1.48%, more than twice the long-run level. A model that always used 0.40.40.4 would understate the risk right after the shock.
  • ARCH(qqq): the variance of today's shock is a constant plus a weighted sum of the last qqq squared shocks.
  • Engle's test: regress squared residuals on their lags and use TR2∼χq2T R^{2}\sim\chi^{2}_qTR2∼χq2​.
  • EWMA with λ=0.94\lambda=0.94λ=0.94 is a no-fit alternative; leverage: falls raise volatility more than rises.

Lab · what's graded

  • The long-run variance15 pts
  • Variance after a shock15 pts
  • An EWMA update15 pts
  • Engle's statistic15 pts
  • The tails of an ARCH model10 pts
  • Match the vocabulary15 pts
  • Risk right after a shock15 pts
0/ 7 answered
Tasks
1Numeric answer15 ptsNot answered

An ARCH(1) has ω=0.2\omega=0.2ω=0.2 and α=0.5\alpha=0.5α=0.5 (in percent squared). What is its unconditional variance ω/(1−α)\omega/(1-\alpha)ω/(1−α)?

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

For the same model, yesterday's shock was εt−1=−2\varepsilon_{t-1}=-2εt−1​=−2. What is today's variance σt2=ω+αεt−12\sigma_t^{2}=\omega+\alpha\varepsilon_{t-1}^{2}σt2​=ω+αεt−12​?

s2=s2 =s2=accepted: ± 0.01
3Numeric answer15 ptsNot answered

With λ=0.94\lambda=0.94λ=0.94, yesterday's variance estimate is 1.01.01.0 and yesterday's return was 2%2\%2%. What is today's EWMA variance λσt−12+(1−λ)rt−12\lambda\sigma_{t-1}^{2}+(1-\lambda)r_{t-1}^{2}λσt−12​+(1−λ)rt−12​?

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

Regressing squared residuals on one lag gives R2=0.04R^{2}=0.04R2=0.04 with T=500T=500T=500 observations. What is Engle's statistic TR2T R^{2}TR2 (compare it with the χ12\chi^{2}_1χ12​ 5% value 3.843.843.84)?

lm=lm =lm=accepted: ± 0.1
5Choice10 ptsNot answered

An ARCH(1) with normal shocks ztz_tzt​ generates returns whose unconditional distribution is

6Match15 ptsNot answered

Match each term with its meaning.

Pick the partner of each item on the left.

ARCH model
EWMA variance
Leverage effect
Engle's LM test
7Explain in words15 ptsNot answered

In one or two sentences, explain why a model with constant variance understates risk right after a large shock.

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