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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Stochastic Calculus skill›Module 4 · Models beyond Brownian motion›Topic 4.2·~1 hrAdvanced

Jumps: Poisson and Merton

Skill positionModule 4 of 5 · Topic 2 of 3
4.2 Jumps: Poisson and Merton
—Lab score— / 100Total XP—Streak—

When prices jump

Brownian motion moves in small steps, but real prices sometimes jump: an earnings surprise, a central bank decision, a crash. A Poisson process counts such events. It has a rate λ\lambdaλ, the average number of jumps per year, and the chance of exactly kkk jumps in time ttt is e−λt(λt)k/k!e^{-\lambda t}(\lambda t)^{k}/k!e−λt(λt)k/k!. In the Merton model the stock follows the usual diffusion and, at each jump time, is multiplied by a random factor JJJ. Here k=E[J−1]k=E[J-1]k=E[J−1] is the average jump size, and the term −λk-\lambda k−λk takes back the drift that jumps add on average, so the expected return stays μ\muμ.

P(Nt=k)=e−λt(λt)kk!,E[Nt]=λt,ln⁡J∼N(m,δ2)⇒k=em+δ2/2−1,Var[ln⁡STS0]=σ2T+λT(m2+δ2)P(N_t=k)=\frac{e^{-\lambda t}(\lambda t)^{k}}{k!},\quad E[N_t]=\lambda t,\quad \ln J\sim\mathcal N(m,\delta^{2})\Rightarrow k=e^{m+\delta^{2}/2}-1,\quad \mathrm{Var}\big[\ln\tfrac{S_T}{S_0}\big]=\sigma^{2}T+\lambda T(m^{2}+\delta^{2})P(Nt​=k)=k!e−λt(λt)k​,E[Nt​]=λt,lnJ∼N(m,δ2)⇒k=em+δ2/2−1,Var[lnS0​ST​​]=σ2T+λT(m2+δ2)(1)

Jumps change what a model can do. Returns get fat tails: most days look like the diffusion, but a few are far bigger. The variance has two sources, the diffusion part σ2T\sigma^{2}Tσ2T and the jump part λT(m2+δ2)\lambda T(m^{2}+\delta^{2})λT(m2+δ2). And hedging becomes harder: you cannot follow a sudden jump with small trades in the stock, so an option's price depends on your view of the jump risk and not only on σ\sigmaσ.

Counting jumpsWith λ=3\lambda=3λ=3 jumps per year, the chance of no jump in half a year is e−1.5≈0.2231e^{-1.5}\approx0.2231e−1.5≈0.2231, and the expected number of jumps is 1.51.51.5. If ln⁡J\ln JlnJ is normal with m=−0.05m=-0.05m=−0.05 and δ=0.1\delta=0.1δ=0.1, then k=e−0.05+0.005−1≈−0.0440k=e^{-0.05+0.005}-1\approx-0.0440k=e−0.05+0.005−1≈−0.0440: the average jump is a drop of 4.4%4.4\%4.4%. With σ=0.2\sigma=0.2σ=0.2, T=1T=1T=1, λ=2\lambda=2λ=2 the log-return variance is 0.04+2×(0.0025+0.01)=0.0650.04+2\times(0.0025+0.01)=0.0650.04+2×(0.0025+0.01)=0.065.
  • Poisson process: counts jumps at rate λ\lambdaλ. The expected count in time ttt is λt\lambda tλt.
  • Merton model: diffusion plus jumps of random size, with a drift correction −λk-\lambda k−λk.
  • Fat tails: jumps add rare big moves, and a jump cannot be fully hedged.

Lab · what's graded

  • No jump at all15 pts
  • The average jump15 pts
  • Variance with jumps15 pts
  • What jumps add15 pts
  • Match the symbols15 pts
  • Simulating a jump path10 pts
  • Why jumps are hard to hedge15 pts
0/ 7 answered
Tasks
1Numeric answer15 ptsNot answered

Jumps arrive at rate λ=3\lambda=3λ=3 per year. What is the probability of no jump in t=0.5t=0.5t=0.5 years, e−λte^{-\lambda t}e−λt? Give four decimals.

p=p =p=accepted: ± 0.0005
2Numeric answer15 ptsNot answered

The jump factor JJJ has ln⁡J\ln JlnJ normal with mean m=−0.05m=-0.05m=−0.05 and standard deviation δ=0.1\delta=0.1δ=0.1. What is the average jump size k=em+δ2/2−1k=e^{m+\delta^{2}/2}-1k=em+δ2/2−1? Give four decimals.

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

With σ=0.2\sigma=0.2σ=0.2, T=1T=1T=1, λ=2\lambda=2λ=2, m=−0.05m=-0.05m=−0.05 and δ=0.1\delta=0.1δ=0.1, what is the log-return variance σ2T+λT(m2+δ2)\sigma^{2}T+\lambda T(m^{2}+\delta^{2})σ2T+λT(m2+δ2)?

v=v =v=accepted: ± 0.001
4Choice15 ptsNot answered

Compared with plain Brownian motion, what do jumps add to a return model?

5Match15 ptsNot answered

Match each symbol with its meaning.

Pick the partner of each item on the left.

λ\lambdaλ
JJJ
k=E[J−1]k=E[J-1]k=E[J−1]
−λk-\lambda k−λk
6Order the steps10 ptsNot answered

Put the steps of one time step in order.

Drag to reorder, or focus an item, press Space to lift it, the arrow keys to move it and Space to drop it.

  1. 1Multiply the price by the diffusion and jump factors
  2. 2Draw the diffusion shock
  3. 3Draw the number of jumps in the step
  4. 4Draw a size for each jump
7Explain in words15 ptsNot answered

In one or two sentences, explain why a delta hedge cannot fully protect against a jump in the stock price.

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

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