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
uantCraft AcademyRoadmapsSkills
Ctrl K
AO
+1.2k Learners
LoginJoin ↗
RoadmapsSkills
AO
+1.2k Learners
Login
  • English
  • Français
  • Español
  • العربية
  • Português
  • Deutsch
  • 日本語
  • 中文
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
uantCraft AcademyRoadmapsSkills
Ctrl K
AO
+1.2k Learners
LoginJoin ↗
RoadmapsSkills
AO
+1.2k Learners
Login
  • English
  • Français
  • Español
  • العربية
  • Português
  • Deutsch
  • 日本語
  • 中文
Stochastic Calculus skill›Module 1 · Random paths and integrals›Topic 1.3·~1 hrAdvanced

The Itô integral

Skill positionModule 1 of 5 · Topic 3 of 3
1.3 The Itô integral
—Lab score— / 100Total XP—Streak—

Adding up the gains of a trader

Think of a trader who holds fif_ifi​ shares during the iii-th small time step. The number is chosen at the start of the step. The gain over that step is fi ΔWif_i\,\Delta W_ifi​ΔWi​, the holding times the random move. Adding all the gains and letting the steps shrink gives the Itô integral. The key rule is that fff is read at the start of each step, never at the end, because nobody knows the next move in advance. This is what makes the integral a fair game: its mean is 000.

∫0Tf dW=lim⁡∑if(ti) (Wti+1−Wti),E[∫0Tf dW]=0,E[(∫0Tf dW)2]=E∫0Tf2 dt\int_0^T f\,dW=\lim\sum_i f(t_i)\,(W_{t_{i+1}}-W_{t_i}),\qquad E\Big[\int_0^T f\,dW\Big]=0,\qquad E\Big[\Big(\int_0^T f\,dW\Big)^{2}\Big]=E\int_0^T f^{2}\,dt∫0T​fdW=limi∑​f(ti​)(Wti+1​​−Wti​​),E[∫0T​fdW]=0,E[(∫0T​fdW)2]=E∫0T​f2dt(1)

The last formula is the Itô isometry. It turns the variance of a random gain into an ordinary integral of f2f^{2}f2, which you can compute. Reading fff at the start of the step also changes some answers. For example ∫0TW dW=(WT2−T)/2\int_0^T W\,dW=(W_T^{2}-T)/2∫0T​WdW=(WT2​−T)/2, not WT2/2W_T^{2}/2WT2​/2 as ordinary calculus would say. The missing −T/2-T/2−T/2 comes from the quadratic variation of the last topic.

Three shares for four yearsHold f=3f=3f=3 shares all the time, so ∫043 dW=3W4\int_0^4 3\,dW=3W_4∫04​3dW=3W4​. The variance of W4W_4W4​ is 444, so the gain has variance 9×4=369\times4=369×4=36. The isometry gives the same number without any random variable: ∫0432 dt=36\int_0^4 3^{2}\,dt=36∫04​32dt=36.
  • Itô integral: the sum of holding times move, with the holding read at the start of each step.
  • Mean and isometry: the mean is 000 and the variance is E∫f2 dtE\int f^{2}\,dtE∫f2dt.
  • Extra term: ∫W dW=(WT2−T)/2\int W\,dW=(W_T^{2}-T)/2∫WdW=(WT2​−T)/2, with the −T/2-T/2−T/2 coming from quadratic variation.

Lab · what's graded

  • A variance from the isometry15 pts
  • The integral of W against W15 pts
  • A discrete trader's gain15 pts
  • A holding that grows with time15 pts
  • Why read the holding at the start15 pts
  • Match the formulas10 pts
  • Why the mean is zero15 pts
0/ 7 answered
Tasks
1Numeric answer15 ptsNot answered

A trader always holds f=3f=3f=3 shares for T=4T=4T=4 years. What is the variance of the gain ∫043 dW\int_0^4 3\,dW∫04​3dW?

v=v =v=accepted: ± 0.01
2Numeric answer15 ptsNot answered

At T=1T=1T=1 the path ends at W1=1.2W_1=1.2W1​=1.2. What is ∫01W dW=(W12−1)/2\int_0^1 W\,dW=(W_1^{2}-1)/2∫01​WdW=(W12​−1)/2?

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

A trader holds 2, 1, 32,\,1,\,32,1,3 shares in three steps, chosen before each step. The price moves are 0.5, −1, 0.40.5,\,-1,\,0.40.5,−1,0.4. What is the total gain ∑fiΔWi\sum f_i\Delta W_i∑fi​ΔWi​?

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

A trader holds f(t)=tf(t)=tf(t)=t shares at time ttt from 000 to 333. What is the variance of the gain ∫03t dW\int_0^3 t\,dW∫03​tdW?

v=v =v=accepted: ± 0.01
5Choice15 ptsNot answered

Why does the Itô integral use the holding at the start of each step?

6Match10 ptsNot answered

Match each item with its value.

Pick the partner of each item on the left.

The mean of an Itô integral
The variance of an Itô integral
∫0TW dW\int_0^T W\,dW∫0T​WdW
7Explain in words15 ptsNot answered

In one or two sentences, explain why an Itô integral has mean zero.

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

Log in to run your code.

Every lecture is free to read. Create an account or log in to run code, run tests and submit this lab.

  • In-browser IDE with a bash terminal — nothing to install
  • Instant grading, XP and your place on the skill
Create an accountLog in
OR
Continue with GitHub