Itô’s lemma and the computations you will be asked for

SC · Chapter 315 min readAsked at Two Sigma, Citadel, DE Shaw, Jane Street

After this lesson you should be able to

  • State the properties of Brownian motion and why it has quadratic variation tt.
  • Apply Itô’s lemma, including the second-order term.
  • Compute the standard moments and decide whether a process is a martingale.

Stochastic calculus differs from ordinary calculus in exactly one place: (dW)2=dt(dW)^2 = dt rather than zero. Every formula that follows — Itô’s lemma, the geometric Brownian motion solution, the Black–Scholes PDE — is ordinary calculus plus that one extra term, and interviews test whether you know where it comes from.

Why there is a second term at all. An ordinary chain rule keeps only the first-order term because (dt)2(\mathrm{d}t)^2 vanishes faster than dt\mathrm{d}t. Brownian motion moves like dt\sqrt{\mathrm{d}t}, so its square is of order dt\mathrm{d}t and refuses to vanish. That single fact is all of Itô calculus: the second-order term survives, and it survives as a drift rather than as noise, which is why a convex function of a martingale is not a martingale.

Definition 3.1

Brownian motion

Standard Brownian motion, W0=0,Wt−Ws∼N(0,t−s),independent incrementsW_0 = 0,\quad W_t - W_s \sim \mathcal{N}(0, t-s),\quad \text{independent increments} — Those three properties plus continuity define it. Everything else — the covariance Cov(Ws,Wt)=min⁡(s,t)\mathrm{Cov}(W_s, W_t) = \min(s,t), the non-differentiability, the scaling Wct=dc WtW_{ct} \overset{d}{=} \sqrt{c}\,W_t — is derived from them, and being able to derive rather than recite is what is being tested.

Derivation 3.2

Quadratic variation

Why (dW)2=dt(dW)^2 = dt, which is the whole of the difference from ordinary calculus.

  1. E[(Wt+Δ−Wt)2]=Δ\mathbb{E}\big[(W_{t+\Delta} - W_t)^2\big] = \Delta

    The increment has variance Δ\Delta.

  2. Var[(Wt+Δ−Wt)2]=2Δ2\mathrm{Var}\big[(W_{t+\Delta} - W_t)^2\big] = 2\Delta^2

    A chi-squared with one degree of freedom, scaled.

  3. ∑i(ΔWi)2→t  as the mesh→0\sum_{i} (\Delta W_i)^2 \to t \ \text{ as the mesh} \to 0

    The mean sums to tt; the variance sums to O(Δ)→0O(\Delta) \to 0, so the limit is deterministic.

[W]t=t,written (dW)2=dt[W]_t = t, \quad \text{written } (dW)^2 = dt

Equation 3.3

Itô’s lemma

For dX=μ dt+σ dWdX = \mu\,dt + \sigma\,dW. It is a second-order Taylor expansion in which (dX)2(dX)^2 survives because it equals σ2dt\sigma^2 dt.

df=(∂f∂t+μ∂f∂x+12σ2∂2f∂x2)dt+σ∂f∂x dWdf = \left(\frac{\partial f}{\partial t} + \mu\frac{\partial f}{\partial x} + \frac{1}{2}\sigma^2\frac{\partial^2 f}{\partial x^2}\right)dt + \sigma\frac{\partial f}{\partial x}\,dW
12σ2fxx\tfrac12 \sigma^2 f_{xx}
The Itô term. Ordinary calculus has no analogue.

Example 3.4

Geometric Brownian motion

Given dS=μS dt+σS dWdS = \mu S\,dt + \sigma S\,dW, find d(ln⁡S)d(\ln S) and solve for StS_t.

Show the worked solutionHide the worked solution

Worked solution

  1. Formula
    f=ln⁡S:fS=1S,fSS=−1S2f = \ln S: \quad f_S = \frac{1}{S}, \quad f_{SS} = -\frac{1}{S^2}
  2. Substitute
    d(ln⁡S)=(μ−12σ2)dt+σ dWd(\ln S) = \left(\mu - \tfrac12\sigma^2\right)dt + \sigma\,dW
  3. Solve
    ln⁡St=ln⁡S0+(μ−12σ2)t+σWt\ln S_t = \ln S_0 + \left(\mu - \tfrac12\sigma^2\right)t + \sigma W_t
  4. St=S0exp⁡ ⁣[(μ−12σ2)t+σWt]S_t = S_0 \exp\!\left[\left(\mu - \tfrac12\sigma^2\right)t + \sigma W_t\right]
  5. Answer
    St=S0e(μ−σ2/2)t+σWtS_t = S_0 e^{(\mu - \sigma^2/2)t + \sigma W_t}

Sanity check. The −σ2/2-\sigma^2/2 is the Itô correction, and it is the reason the *median* return is below the mean. It also explains why E[St]=S0eμt\mathbb{E}[S_t] = S_0 e^{\mu t} exactly, with the correction cancelling against the lognormal expectation.

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