Unit 5, lesson 37

Ito lemma and the stochastic chain rule

How Itô's lemma extends the chain rule to Itô processes, where its second-order correction comes from, and how the multivariate Itô formula, the Itô product rule and the generator of a diffusion follow from it.

  • Video 37
  • Itô integrals and stochastic differential equations

The domestic value of a foreign stock is the product of its price in foreign currency and the exchange rate. In a constructed model both follow geometric Brownian motions, with drifts 0.080.08 and 0.060.06 per year, volatilities 0.20.2 and 0.30.3 per square-root year and correlation 0.50.5 between their Brownian drivers. The ordinary product rule suggests that the expected domestic value grows at 0.140.14 per year, but the correct rate is 0.170.17. This lesson states Itô’s lemma, explains the second-order term that the ordinary chain rule misses, and derives from it the multivariate formula, the product rule and the generator of a diffusion, which together account for the extra 0.030.03.

Key terms

Itô’s lemma

Definition Let WW be a standard Brownian motion with respect to a filtration (Ft)(\mathcal{F}_t) under P\mathbb{P}, and let XX be an Itô process on [0,T][0, T],

Xt=X0+∫0tbs ds+∫0tσs dWs,\begin{aligned} X_t &= X_0 + \int_0^t b_s \, ds \\[0.4em] &\quad + \int_0^t \sigma_s \, dW_s , \end{aligned}

with X0X_0 an F0\mathcal{F}_0-measurable random variable, bb and σ\sigma predictable, and ∫0T∣bs∣ ds<∞\int_0^T \lvert b_s \rvert \, ds < \infty and ∫0Tσs2 ds<∞\int_0^T \sigma_s^2 \, ds < \infty almost surely, so that both integrals are defined. In differential form dXt=bt dt+σt dWtdX_t = b_t \, dt + \sigma_t \, dW_t, and the quadratic variation of XX is [X]t=∫0tσs2 ds[X]_t = \int_0^t \sigma_s^2 \, ds. Let f(t,x)f(t, x) have a continuous first derivative in tt and continuous first and second derivatives in xx on [0,T]×R[0, T] \times \mathbb{R}, which is written f∈C1,2f \in C^{1,2}. Itô’s lemma, also called the stochastic chain rule, says that

df(t,Xt)=σt ∂xf dWt+(∂tf+bt ∂xf+12σt2 ∂xxf) dt,\begin{aligned} &df(t, X_t) = \sigma_t \, \partial_x f \, dW_t \\[0.4em] &\quad + \bigl(\partial_t f + b_t \, \partial_x f \\[0.4em] &\qquad + \tfrac12 \sigma_t^2 \, \partial_{xx} f\bigr)\, dt , \end{aligned}

with every derivative evaluated at (t,Xt)(t, X_t). Equivalently df=∂tf dt+∂xf dXt+12∂xxf d[X]tdf = \partial_t f \, dt + \partial_x f \, dX_t + \tfrac12 \partial_{xx} f \, d[X]_t. The differential notation is shorthand for the identity obtained by integrating both sides from 00 to tt, which holds almost surely for all t∈[0,T]t \in [0, T] at once. The stochastic integral in it is defined because ∂xf(s,Xs)\partial_x f(s, X_s) is continuous in ss and so bounded on each path. It is in general a local martingale, and a martingale when E∫0Tσs2(∂xf)2 ds<∞\mathbb{E}\int_0^T \sigma_s^2 (\partial_x f)^2 \, ds < \infty. If ff is C1,2C^{1,2} only on [0,T]×D[0, T] \times D for an open interval DD that contains every path of XX, the formula still holds, and if XX can leave DD it holds for the times before XX first leaves DD. The theorem is stated here with the idea of its proof, which the next card describes.

In plain words To find how a smooth function of a noisy state changes, take the ordinary first-order terms in time and in the state, and add half the second state derivative times the variance rate of the state. That added term is the only difference from the ordinary chain rule, and it disappears when the state has no Brownian part.

Example For f(x)=x2f(x) = x^2 and X=WX = W, so that b=0b = 0 and σ=1\sigma = 1, the derivatives are ∂xf=2x\partial_x f = 2x and ∂xxf=2\partial_{xx} f = 2, and

d(Wt2)=2Wt dWt+dt.d(W_t^2) = 2 W_t \, dW_t + dt .

Integrating gives Wt2=2∫0tWs dWs+tW_t^2 = 2 \int_0^t W_s \, dW_s + t, the identity that left-endpoint sums produce in the Itô integral lesson. For Yt=exp⁡(0.2Wt−0.02t)Y_t = \exp(0.2 W_t - 0.02 t) take f(t,x)=exp⁡(0.2x−0.02t)f(t, x) = \exp(0.2 x - 0.02 t), with ∂tf=−0.02f\partial_t f = -0.02 f, ∂xf=0.2f\partial_x f = 0.2 f and ∂xxf=0.04f\partial_{xx} f = 0.04 f. The drift is −0.02Yt+12⋅0.04 Yt=0-0.02 Y_t + \tfrac12 \cdot 0.04 \, Y_t = 0, so dYt=0.2Yt dWtdY_t = 0.2 Y_t \, dW_t, and the second-order term exactly cancels the explicit time trend.

Second-order correction

Definition The second-order correction in Itô’s lemma is the term 12σt2 ∂xxf(t,Xt) dt\tfrac12 \sigma_t^2 \, \partial_{xx} f(t, X_t) \, dt, which equals 12∂xxf d[X]t\tfrac12 \partial_{xx} f \, d[X]_t and has no counterpart in the ordinary chain rule. It comes from the Taylor expansion over a partition 0=t0<⋯<tn=t0 = t_0 < \dots < t_n = t,

Δfk≈∂tf Δtk+∂xf ΔXk+12 ∂xxf (ΔXk)2,\begin{aligned} \Delta f_k &\approx \partial_t f \, \Delta t_k + \partial_x f \, \Delta X_k \\[0.4em] &\quad + \tfrac12 \, \partial_{xx} f \, (\Delta X_k)^2 , \end{aligned}

with the derivatives at (tk,Xtk)(t_k, X_{t_k}). As the mesh tends to zero, the first-order sums converge to ∫0t∂tf ds\int_0^t \partial_t f \, ds and to the left-endpoint integral ∫0t∂xf dXs\int_0^t \partial_x f \, dX_s. In the square (ΔXk)2(\Delta X_k)^2, the parts of the form b2(Δtk)2b^2 (\Delta t_k)^2 and 2bσ Δtk ΔWk2 b \sigma \, \Delta t_k \, \Delta W_k add up to zero in the limit, while σ2(ΔWk)2\sigma^2 (\Delta W_k)^2 adds up to ∫0tσs2 ds\int_0^t \sigma_s^2 \, ds because the quadratic variation of WW is tt. These limits are often recorded as the multiplication rules (dt)2=0(dt)^2 = 0, dt dWt=0dt \, dW_t = 0 and (dWt)2=dt(dW_t)^2 = dt, which are bookkeeping for quadratic covariations and not identities between numbers. Controlling the Taylor remainder and the coefficients requires localization estimates that are stated, not carried out, here. Under the hypotheses of the lemma the correction is exact, not a truncation. Its sign follows the curvature of ff in xx. It is positive where ff is convex, negative where ff is concave, and zero when ff is linear in xx or σ=0\sigma = 0, in which case Itô’s lemma reduces to the ordinary chain rule.

In plain words A Brownian increment over a step of length hh has size about h\sqrt{h}, so its square has size about hh, the same order as the time step. The squared shocks therefore add up to a nonzero amount, and the curvature of ff turns that amount into a drift.

Example Let a positive price satisfy dSt=μSt dt+σSt dWtdS_t = \mu S_t \, dt + \sigma S_t \, dW_t with constants μ\mu and σ>0\sigma > 0, and apply the lemma to log⁡x\log x on the open interval (0,∞)(0, \infty), with derivatives 1/x1/x and −1/x2-1/x^2. Then

dlog⁡St=(μ−12σ2) dt+σ dWt.\begin{aligned} d \log S_t &= \bigl(\mu - \tfrac12 \sigma^2\bigr)\, dt \\[0.4em] &\quad + \sigma \, dW_t . \end{aligned}

With the constructed values μ=0.08\mu = 0.08 and σ=0.2\sigma = 0.2, the log drift is 0.08−0.02=0.060.08 - 0.02 = 0.06 per year. Since log⁡S1\log S_1 is normal with mean log⁡S0+0.06\log S_0 + 0.06, the median of S1S_1 is S0e0.06≈1.0618 S0S_0 e^{0.06} \approx 1.0618\, S_0, while E[S1]=S0e0.08≈1.0833 S0\mathbb{E}[S_1] = S_0 e^{0.08} \approx 1.0833\, S_0. Over one trading day, h=1/252h = 1/252, the drift μh≈0.000317\mu h \approx 0.000317 and the correction 12σ2h≈0.0000794\tfrac12 \sigma^2 h \approx 0.0000794 are of the same order, while the shock σh≈0.0126\sigma \sqrt{h} \approx 0.0126 is much larger. No change of probability measure and no risk premium is involved, since the 0.020.02 comes only from the curvature of the logarithm.

Multivariate Itô formula

Definition Let W=(W1,…,Wm)⊤W = (W^1, \dots, W^m)^{\top} be a standard mm-dimensional Brownian motion with independent coordinates, and let XX be an nn-dimensional Itô process dXt=bt dt+Σt dWtdX_t = b_t \, dt + \Sigma_t \, dW_t, with predictable bt∈Rnb_t \in \mathbb{R}^n and Σt∈Rn×m\Sigma_t \in \mathbb{R}^{n \times m} whose entries satisfy the integrability conditions of the one-dimensional case. The matrix at=ΣtΣt⊤a_t = \Sigma_t \Sigma_t^{\top} is symmetric and positive semidefinite, and d[Xi,Xj]t=atij dtd[X^i, X^j]_t = a^{ij}_t \, dt. For f∈C1,2([0,T]×Rn)f \in C^{1,2}([0, T] \times \mathbb{R}^n), with gradient ∇f\nabla f and Hessian ∇2f\nabla^2 f in the state variables,

df(t,Xt)=∇f⊤Σt dWt+(∂tf+bt⊤∇f) dt+12tr⁡(at∇2f) dt,\begin{aligned} &df(t, X_t) = \nabla f^{\top} \Sigma_t \, dW_t \\[0.4em] &\quad + \bigl(\partial_t f + b_t^{\top} \nabla f\bigr)\, dt \\[0.4em] &\quad + \tfrac12 \operatorname{tr}\bigl(a_t \nabla^2 f\bigr)\, dt , \end{aligned}

with the derivatives at (t,Xt)(t, X_t). Written out, the correction is 12∑i,jatij ∂ijf dt\tfrac12 \sum_{i,j} a^{ij}_t \, \partial_{ij} f \, dt. With two coordinates each mixed term appears twice, for (i,j)=(1,2)(i, j) = (1, 2) and for (2,1)(2, 1), and since a12=a21a^{12} = a^{21} and ∂12f=∂21f\partial_{12} f = \partial_{21} f the two halves combine into at12 ∂12f dta^{12}_t \, \partial_{12} f \, dt. Independent Brownian coordinates do not make the coordinates of XX independent, because shared loadings in Σ\Sigma correlate them.

In plain words Every pair of state coordinates contributes its covariance rate times the curvature of ff in that pair of directions. Knowing each coordinate’s own equation is not enough, because the cross terms need the covariance between coordinates.

Example Take two stock prices in one currency with the constructed parameters of the opening, dXt=0.08Xt dt+0.2Xt dWt1dX_t = 0.08 X_t \, dt + 0.2 X_t \, dW^1_t and dYt=0.06Yt dt+0.3Yt dBtdY_t = 0.06 Y_t \, dt + 0.3 Y_t \, dB_t, where B=0.5W1+0.75 W2B = 0.5 W^1 + \sqrt{0.75}\, W^2 is a Brownian motion with correlation 0.50.5 with W1W^1. The covariance rates are a11=0.04X2a^{11} = 0.04 X^2, a22=0.09Y2a^{22} = 0.09 Y^2 and a12=0.5⋅0.2⋅0.3 XY=0.03 XYa^{12} = 0.5 \cdot 0.2 \cdot 0.3 \, XY = 0.03 \, XY. The exchange ratio R=X/YR = X/Y has ∂xxf=0\partial_{xx} f = 0, ∂xyf=−1/y2\partial_{xy} f = -1/y^2 and ∂yyf=2x/y3\partial_{yy} f = 2x/y^3 for f(x,y)=x/yf(x, y) = x/y, so its correction is

a12∂xyf+12a22∂yyf=(−0.03+0.09) Rt.\begin{aligned} &a^{12} \partial_{xy} f + \tfrac12 a^{22} \partial_{yy} f \\[0.4em] &\quad = (-0.03 + 0.09)\, R_t . \end{aligned}

Adding the first-order drift (0.08−0.06)Rt(0.08 - 0.06) R_t gives the drift 0.08Rt0.08 R_t, of which the mixed term contributes −0.03Rt-0.03 R_t. The diffusion part is Rt(0.2 dWt1−0.3 dBt)R_t (0.2 \, dW^1_t - 0.3 \, dB_t), with variance rate (0.04+0.09−0.06)Rt2=0.07Rt2(0.04 + 0.09 - 0.06) R_t^2 = 0.07 R_t^2. So RR is again a geometric Brownian motion, and from R0=2R_0 = 2 its mean after one year is 2e0.08≈2.16662 e^{0.08} \approx 2.1666, not the ratio of means 2e0.02≈2.04042 e^{0.02} \approx 2.0404.

Itô product rule

Definition Let XX and YY be Itô processes driven by the same mm-dimensional Brownian motion, dXt=btX dt+ΣtX dWtdX_t = b^X_t \, dt + \Sigma^X_t \, dW_t and dYt=btY dt+ΣtY dWtdY_t = b^Y_t \, dt + \Sigma^Y_t \, dW_t, with row vectors ΣtX,ΣtY∈R1×m\Sigma^X_t, \Sigma^Y_t \in \mathbb{R}^{1 \times m} and the integrability conditions above. Then

d(XtYt)=Xt dYt+Yt dXt+d[X,Y]t,\begin{aligned} &d(X_t Y_t) = X_t \, dY_t \\[0.4em] &\quad + Y_t \, dX_t + d[X, Y]_t , \end{aligned}

where d[X,Y]t=ΣtX(ΣtY)⊤ dtd[X, Y]_t = \Sigma^X_t (\Sigma^Y_t)^{\top} \, dt is the quadratic covariation. This is the multivariate formula for f(x,y)=xyf(x, y) = xy, whose Hessian has zeros on the diagonal and ones off it, so the correction is 12⋅2 at12 dt=at12 dt\tfrac12 \cdot 2\, a^{12}_t \, dt = a^{12}_t \, dt. When one factor has no Brownian part the covariation vanishes and the ordinary product rule holds. For a constant rate rr, for example, d(e−rtSt)=e−rt(dSt−rSt dt)d(e^{-rt} S_t) = e^{-rt} (dS_t - r S_t \, dt), the rule behind discounted prices.

In plain words The change in a product is each factor times the change in the other, plus the joint movement of the two. With two noisy factors the joint movement adds to the drift, and positive correlation raises the growth of the product.

Example In the constructed model of the opening, the stock price XX starts at X0=100X_0 = 100 foreign currency units and the exchange rate YY at Y0=50Y_0 = 50 domestic units per foreign unit, with the drifts, volatilities and correlation of the previous card. The covariation rate is 0.5⋅0.2⋅0.3 XtYt=0.03 XtYt0.5 \cdot 0.2 \cdot 0.3 \, X_t Y_t = 0.03 \, X_t Y_t, so the domestic value Z=XYZ = XY satisfies

dZtZt=0.17 dt+0.2 dWt1+0.3 dBt,\begin{aligned} \frac{dZ_t}{Z_t} &= 0.17 \, dt + 0.2 \, dW^1_t \\[0.4em] &\quad + 0.3 \, dB_t , \end{aligned}

with drift 0.08+0.06+0.03=0.170.08 + 0.06 + 0.03 = 0.17. Constant coefficients give lognormal moments of every order, so the stochastic integral is a martingale and m(t)=E[Zt]m(t) = \mathbb{E}[Z_t] solves m′(t)=0.17 m(t)m'(t) = 0.17\, m(t). From Z0=5000Z_0 = 5000 domestic currency units,

E[Z1]=5000 e0.17≈5926.52,\begin{aligned} \mathbb{E}[Z_1] &= 5000 \, e^{0.17} \\[0.4em] &\approx 5926.52 , \end{aligned}

while the product of the means is E[X1] E[Y1]=5000 e0.14≈5751.37\mathbb{E}[X_1] \, \mathbb{E}[Y_1] = 5000 \, e^{0.14} \approx 5751.37. With correlation −0.5-0.5 and the same marginal equations, the expected domestic value is 5000 e0.11≈5581.395000 \, e^{0.11} \approx 5581.39. The two marginal laws do not determine the mean of the product.

Generator of a diffusion

Definition Let XX solve dXt=b(Xt) dt+Σ(Xt) dWtdX_t = b(X_t) \, dt + \Sigma(X_t) \, dW_t in Rn\mathbb{R}^n with continuous coefficients bb and Σ\Sigma, let a=ΣΣ⊤a = \Sigma \Sigma^{\top}, and write Ex\mathbb{E}^x for the expectation when X0=xX_0 = x. The generator of the diffusion acts on f∈C2(Rn)f \in C^2(\mathbb{R}^n) by

Lf(x)=b(x)⊤∇f(x)+12tr⁡(a(x)∇2f(x)).\begin{aligned} &\mathcal{L} f(x) = b(x)^{\top} \nabla f(x) \\[0.4em] &\quad + \tfrac12 \operatorname{tr}\bigl(a(x) \nabla^2 f(x)\bigr) . \end{aligned}

Itô’s lemma reads df(Xt)=Lf(Xt) dt+∇f(Xt)⊤Σ(Xt) dWtdf(X_t) = \mathcal{L} f(X_t) \, dt + \nabla f(X_t)^{\top} \Sigma(X_t) \, dW_t, so f(Xt)−f(X0)−∫0tLf(Xs) dsf(X_t) - f(X_0) - \int_0^t \mathcal{L} f(X_s) \, ds is a local martingale. If ff also has compact support, the stochastic integral has a bounded integrand and is a martingale, and

Lf(x)=lim⁡t↓0Ex[f(Xt)]−f(x)t,\begin{aligned} &\mathcal{L} f(x) \\[0.4em] &\quad = \lim_{t \downarrow 0} \frac{\mathbb{E}^x[f(X_t)] - f(x)}{t} , \end{aligned}

so the generator is the instantaneous expected rate of change of ff along the diffusion. When the equation has a solution from each starting point that is unique in law, XX is a Markov process, a result stated here. For a function g∈C1,2g \in C^{1,2} of time and state, the drift of g(t,Xt)g(t, X_t) is (∂tg+Lg)(t,Xt)(\partial_t g + \mathcal{L} g)(t, X_t). If ∂tg+Lg=0\partial_t g + \mathcal{L} g = 0, then g(t,Xt)g(t, X_t) is a local martingale, and when it is a true martingale on [0,T][0, T] it satisfies g(t,Xt)=E[g(T,XT)∣Ft]g(t, X_t) = \mathbb{E}[g(T, X_T) \mid \mathcal{F}_t]. This equation is the starting point of the pricing equations of later lessons, which need the same martingale and integrability conditions.

In plain words The generator says how fast the expected value of f(X)f(X) starts to change from a known state. Its first-order part uses the drift and its second-order part uses the covariance rates, exactly as in Itô’s lemma.

Example For geometric Brownian motion dSt=μSt dt+σSt dWtdS_t = \mu S_t \, dt + \sigma S_t \, dW_t the generator is Lf(x)=μxf′(x)+12σ2x2f′′(x)\mathcal{L} f(x) = \mu x f'(x) + \tfrac12 \sigma^2 x^2 f''(x). For f(x)=x2f(x) = x^2 it gives Lf(x)=(2μ+σ2)x2\mathcal{L} f(x) = (2\mu + \sigma^2) x^2, which is 0.20 x20.20\, x^2 for the constructed μ=0.08\mu = 0.08 and σ=0.2\sigma = 0.2. The function x2x^2 does not have compact support, but the lognormal moments make the stochastic integral ∫2σSt2 dWt\int 2 \sigma S_t^2 \, dW_t a martingale, so E[St2]=S02e0.2t\mathbb{E}[S_t^2] = S_0^2 e^{0.2 t}. From S0=100S_0 = 100 currency units this is 10000 e0.2≈12214.0310000\, e^{0.2} \approx 12214.03 at t=1t = 1, and together with E[S1]=100 e0.08≈108.33\mathbb{E}[S_1] = 100\, e^{0.08} \approx 108.33 it gives a standard deviation of about 21.8821.88 currency units for S1S_1.

Check yourself

  1. Use Itô’s lemma on Wt4W_t^4 to find E[Wt4]\mathbb{E}[W_t^4].
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With f(x)=x4f(x) = x^4, the derivatives are f′(x)=4x3f'(x) = 4x^3 and f′′(x)=12x2f''(x) = 12x^2, so

d(Wt4)=4Wt3 dWt+6Wt2 dt.\begin{aligned} d(W_t^4) &= 4 W_t^3 \, dW_t \\[0.4em] &\quad + 6 W_t^2 \, dt . \end{aligned}

The integrand 4W34 W^3 is square integrable on [0,t][0, t], because E[Ws6]=15s3\mathbb{E}[W_s^6] = 15 s^3 is integrable in ss, so the stochastic integral has mean zero. Taking expectations,

E[Wt4]=6∫0ts ds=3t2.\mathbb{E}[W_t^4] = 6 \int_0^t s \, ds = 3 t^2 .

At t=1t = 1 this is 33, the fourth moment of a standard normal variable.

  1. A price follows dSt=0.05St dt+0.3St dWtdS_t = 0.05 S_t \, dt + 0.3 S_t \, dW_t with S0>0S_0 > 0. A classmate applies the ordinary chain rule, says that log⁡St\log S_t has drift 0.050.05, and concludes that the median of S2/S0S_2/S_0 is e0.1e^{0.1}. Find the correct log drift, the median and the mean of S2/S0S_2/S_0, and P(S2>S0)\mathbb{P}(S_2 > S_0).
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The second-order correction is −12⋅0.09=−0.045-\tfrac12 \cdot 0.09 = -0.045, so the log drift is 0.05−0.045=0.0050.05 - 0.045 = 0.005. Then log⁡(S2/S0)\log(S_2/S_0) is normal with mean 0.010.01 and variance 0.09⋅2=0.180.09 \cdot 2 = 0.18. The median of S2/S0S_2/S_0 is e0.01≈1.0101e^{0.01} \approx 1.0101, and the mean is e0.05⋅2=e0.1≈1.1052e^{0.05 \cdot 2} = e^{0.1} \approx 1.1052, so the classmate’s value is the mean, not the median. Finally

P(S2>S0)=Φ(0.01/0.18)=Φ(0.0236)≈0.5094.\begin{aligned} &\mathbb{P}(S_2 > S_0) \\[0.4em] &\quad = \Phi\bigl(0.01 / \sqrt{0.18}\bigr) \\[0.4em] &\quad = \Phi(0.0236) \approx 0.5094 . \end{aligned}
  1. In a constructed arithmetic model St=K+WtS_t = K + W_t, so (S1−K)+=(W1)+(S_1 - K)^{+} = (W_1)^{+} is an at-the-money call payoff. Someone applies Itô’s lemma to f(x)=(x)+f(x) = (x)^{+}, using f′(x)=1{x>0}f'(x) = \mathbf{1}_{\{x > 0\}} and f′′(x)=0f''(x) = 0 for x≠0x \ne 0, and concludes that (W1)+=∫011{Ws>0} dWs(W_1)^{+} = \int_0^1 \mathbf{1}_{\{W_s > 0\}} \, dW_s. Show that this is false and name the hypothesis that fails.
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The integrand on the right is bounded and predictable, so the right side has mean zero. The left side has mean

E[(W1)+]=12π≈0.3989>0,\begin{aligned} \mathbb{E}\bigl[(W_1)^{+}\bigr] &= \frac{1}{\sqrt{2\pi}} \\[0.4em] &\approx 0.3989 > 0 , \end{aligned}

so the identity cannot hold. The lemma needs f∈C2f \in C^2, and here f′f' jumps at the strike and f′′f'' is not a function there, so zero curvature away from the kink does not describe the whole function. The missing term is half the local time of WW at zero, from the Itô-Tanaka formula, which is beyond this lesson, and its expectation here must be 0.39890.3989. A smooth pricing function before maturity avoids the problem, with the payoff reached as a limit at maturity.

How this lesson connects