Unit 5, lesson 38

Stochastic differential equations and solution concepts

What it means to solve a stochastic differential equation, strong and weak solutions, the global Lipschitz theorem on existence and pathwise uniqueness, the Ornstein-Uhlenbeck process and the Euler step.

  • Video 38
  • Itô integrals and stochastic differential equations

A constructed short-rate model starts at 0.020.02, pulls the rate toward 0.040.04 with speed 0.50.5 per year and shocks it with volatility 0.010.01 per square-root year. What is the distribution of the rate in two years, and how close is a simulation that moves forward in steps of a tenth of a year? Neither question has an answer until the equation has a meaning, which needs care because Brownian paths have no derivative, and until a solution is known to exist and to be unique. This lesson defines strong and weak solutions, states the global Lipschitz theorem on existence and uniqueness, solves the Ornstein-Uhlenbeck equation exactly and compares its exact transition with an Euler step.

Key terms

Stochastic differential equation

Definition Let WW be a standard Brownian motion with respect to a filtration (Ft)(\mathcal{F}_t) under P\mathbb{P}, let bb and σ\sigma be measurable functions on [0,T]×R[0, T] \times \mathbb{R}, and let ξ\xi be an F0\mathcal{F}_0-measurable initial condition. The stochastic differential equation

dXt=b(t,Xt) dt+σ(t,Xt) dWt\begin{aligned} dX_t &= b(t, X_t)\, dt \\[0.4em] &\quad + \sigma(t, X_t)\, dW_t \end{aligned}

with X0=ξX_0 = \xi is shorthand for the integral equation

Xt=ξ+∫0tb(s,Xs) ds+∫0tσ(s,Xs) dWs,\begin{aligned} X_t &= \xi + \int_0^t b(s, X_s)\, ds \\[0.4em] &\quad + \int_0^t \sigma(s, X_s)\, dW_s , \end{aligned}

required to hold almost surely for all t∈[0,T]t \in [0, T] at once. A solution is a process XX with continuous paths, adapted to (Ft)(\mathcal{F}_t), such that ∫0T∣b(s,Xs)∣ ds<∞\int_0^T \lvert b(s, X_s) \rvert \, ds < \infty and ∫0Tσ(s,Xs)2 ds<∞\int_0^T \sigma(s, X_s)^2 \, ds < \infty almost surely, so that both integrals are defined, and such that the integral equation holds. Brownian paths are nowhere differentiable, so dWtdW_t is not the derivative of WW times dtdt, and the equation is not an ordinary differential equation along each path. Equations in several dimensions are defined in the same way, with a vector drift and a matrix diffusion coefficient.

In plain words The equation prescribes the drift and the noise loading as functions of the current time and state, and a solution is a process whose accumulated drift and accumulated Itô integral rebuild it at every time. Writing an equation down does not by itself produce a process that solves it.

Example Geometric Brownian motion dSt=μSt dt+σSt dWtdS_t = \mu S_t \, dt + \sigma S_t \, dW_t with a constant S0>0S_0 > 0 has the candidate solution St=g(t,Wt)S_t = g(t, W_t) with g(t,w)=S0exp⁡((μ−12σ2)t+σw)g(t, w) = S_0 \exp\bigl((\mu - \tfrac12 \sigma^2) t + \sigma w\bigr). Itô’s lemma with ∂tg=(μ−12σ2)g\partial_t g = (\mu - \tfrac12 \sigma^2) g, ∂wg=σg\partial_w g = \sigma g and ∂wwg=σ2g\partial_{ww} g = \sigma^2 g gives

dSt=(μ−12σ2)St dt+12σ2St dt+σSt dWt,\begin{aligned} &dS_t = \bigl(\mu - \tfrac12 \sigma^2\bigr) S_t \, dt \\[0.4em] &\quad + \tfrac12 \sigma^2 S_t \, dt + \sigma S_t \, dW_t , \end{aligned}

which is the equation, so the integral identity holds for all tt on almost every path. The second-order term of Itô’s lemma restores exactly the 12σ2\tfrac12 \sigma^2 that the exponent subtracts.

Strong solution

Definition Fix in advance the filtered probability space, the Brownian motion WW and the initial condition ξ\xi, which is F0\mathcal{F}_0-measurable and hence independent of WW. A strong solution is a solution XX on this space, driven by this WW, that is adapted to the filtration generated by ξ\xi and WW and completed with the null sets. Its value at time tt is then determined, up to a null set, by ξ\xi and the Brownian path up to time tt, so the given inputs determine the output. The word strong describes how the solution depends on its inputs. It says nothing about the accuracy of a numerical method.

In plain words A strong solution is a recipe that turns a given noise path into a state path. Anyone who feeds the same initial value and the same Brownian path into it gets the same state path.

Example The geometric Brownian motion formula is a strong solution, because StS_t is an explicit function of S0S_0 and WtW_t. With the constructed values μ=0.08\mu = 0.08, σ=0.2\sigma = 0.2 and S0=100S_0 = 100 currency units, every path with W1=0.5W_1 = 0.5 has

S1=100 e0.06+0.2⋅0.5=100 e0.16≈117.35\begin{aligned} S_1 &= 100 \, e^{0.06 + 0.2 \cdot 0.5} \\[0.4em] &= 100 \, e^{0.16} \approx 117.35 \end{aligned}

currency units. The Ornstein-Uhlenbeck solution below is also a strong solution, since it is built from the initial value and an integral against the given WW.

Existence and uniqueness theorem

Definition Suppose that bb and σ\sigma are measurable on [0,T]×R[0, T] \times \mathbb{R} and that there are constants LL and KK such that for all t∈[0,T]t \in [0, T] and all real xx and yy

∣b(t,x)−b(t,y)∣+∣σ(t,x)−σ(t,y)∣≤L∣x−y∣,\begin{aligned} &\lvert b(t, x) - b(t, y) \rvert \\[0.4em] &\quad + \lvert \sigma(t, x) - \sigma(t, y) \rvert \\[0.4em] &\quad \le L \lvert x - y \rvert , \end{aligned}

which is the global Lipschitz condition, and

∣b(t,x)∣+∣σ(t,x)∣≤K(1+∣x∣),\begin{aligned} &\lvert b(t, x) \rvert + \lvert \sigma(t, x) \rvert \\[0.4em] &\quad \le K (1 + \lvert x \rvert) , \end{aligned}

which is the linear growth condition. Let WW be a Brownian motion for (Ft)(\mathcal{F}_t) and let ξ\xi be F0\mathcal{F}_0-measurable with E[ξ2]<∞\mathbb{E}[\xi^2] < \infty. Then the equation has a strong solution XX on [0,T][0, T] with E[sup⁡t≤TXt2]<∞\mathbb{E}[\sup_{t \le T} X_t^2] < \infty, and pathwise uniqueness holds. Pathwise uniqueness means that any two solutions XX and YY on the same space, with the same Brownian motion and the same initial condition, satisfy Xt=YtX_t = Y_t for all t∈[0,T]t \in [0, T] almost surely. The theorem gives sufficient conditions, not a definition of a solution and not the only route to existence. For coefficients that do not depend on tt, the Lipschitz condition implies the growth condition. The proof is sketched here. Picard iteration starts from Xt(0)=ξX^{(0)}_t = \xi and sets

Xt(k+1)=ξ+∫0tb(s,Xs(k)) ds+∫0tσ(s,Xs(k)) dWs.\begin{aligned} &X^{(k+1)}_t = \xi \\[0.4em] &\quad + \int_0^t b(s, X^{(k)}_s)\, ds \\[0.4em] &\quad + \int_0^t \sigma(s, X^{(k)}_s)\, dW_s . \end{aligned}

The Cauchy-Schwarz inequality bounds the drift integrals, the Itô isometry with Doob’s maximal inequality bounds the stochastic integrals, and the Lipschitz condition turns differences of coefficients into differences of states. The resulting bound on E[sup⁡s≤t∣Xs(k+1)−Xs(k)∣2]\mathbb{E}[\sup_{s \le t} \lvert X^{(k+1)}_s - X^{(k)}_s \rvert^2] is of order (Ct)k/k!(Ct)^k / k! for a constant CC, which is summable in kk, so the iterates converge to a solution. For two solutions, the same estimates applied to u(t)=E[sup⁡s≤t∣Xs−Ys∣2]u(t) = \mathbb{E}[\sup_{s \le t} \lvert X_s - Y_s \rvert^2], after stopping both processes so that uu is finite, give u(t)≤C∫0tu(s) dsu(t) \le C \int_0^t u(s) \, ds, and Gronwall’s inequality forces u=0u = 0.

In plain words The Lipschitz condition stops the coefficients from reacting too sharply to a small difference in state, which keeps two candidate solutions driven by the same noise together. The growth condition stops the state from running off to infinity in finite time.

Example The Ornstein-Uhlenbeck coefficients b(x)=κ(θ−x)b(x) = \kappa(\theta - x) and σ(x)=σ\sigma(x) = \sigma satisfy both conditions with L=κL = \kappa, and for the constructed κ=0.5\kappa = 0.5, θ=0.04\theta = 0.04 and σ=0.01\sigma = 0.01 the constant K=0.5K = 0.5 works, since 0.02+0.01≤0.50.02 + 0.01 \le 0.5. Geometric Brownian motion satisfies them with L=K=∣μ∣+σL = K = \lvert \mu \rvert + \sigma. The square-root coefficient σx\sigma \sqrt{x} of the Cox-Ingersoll-Ross model is not Lipschitz at 00, since x/x→∞\sqrt{x}/x \to \infty as x↓0x \downarrow 0, so the theorem does not apply to it and specialized results are needed. Without the growth condition, the deterministic equation dXt=Xt2 dtdX_t = X_t^2 \, dt with X0=2X_0 = 2 has the solution Xt=2/(1−2t)X_t = 2/(1 - 2t), which explodes at t=0.5t = 0.5, so no solution exists on [0,1][0, 1]. Without the Lipschitz condition, dXt=3Xt2/3 dtdX_t = 3 X_t^{2/3} \, dt with X0=0X_0 = 0 has the two solutions Xt=0X_t = 0 and Xt=t3X_t = t^3.

Weak solution

Definition A weak solution with initial law ν\nu consists of a filtered probability space (Ω,F,(Ft),P)(\Omega, \mathcal{F}, (\mathcal{F}_t), \mathbb{P}), a Brownian motion WW for (Ft)(\mathcal{F}_t) and a continuous adapted process XX, all constructed together, such that X0X_0 has law ν\nu and the integral equation holds almost surely on that space. Every strong solution is a weak solution. A weak solution still solves the integral equation with its own driver, so it is more than a process with the right distribution. The equation has uniqueness in law when any two weak solutions with the same initial law give processes XX with the same law on the space of paths. Pathwise uniqueness implies uniqueness in law, and together with the existence of a weak solution it implies the existence of a strong solution, the Yamada-Watanabe theorem, stated here without proof.

In plain words For a weak solution the noise may be built together with the state, so only the law of the inputs is prescribed, not a particular Brownian path. Uniqueness in law says that all such constructions produce the same distribution of state paths, which is all that an expectation such as a model price E[g(XT)]\mathbb{E}[g(X_T)] depends on.

Example Tanaka’s equation is dXt=sgn⁡(Xt) dWtdX_t = \operatorname{sgn}(X_t) \, dW_t with X0=0X_0 = 0, where sgn⁡(x)=1\operatorname{sgn}(x) = 1 for x>0x > 0 and sgn⁡(x)=−1\operatorname{sgn}(x) = -1 for x≤0x \le 0. Its coefficient is not continuous, so the theorem above does not apply. A weak solution is built from any Brownian motion BB by setting X=BX = B and Wt=∫0tsgn⁡(Bs) dBsW_t = \int_0^t \operatorname{sgn}(B_s) \, dB_s. This WW is a Brownian motion by Lévy’s characterization, which a later lesson of this unit covers, and

∫0tsgn⁡(Xs) dWs=∫0tsgn⁡(Bs)2 dBs=Bt=Xt.\begin{aligned} &\int_0^t \operatorname{sgn}(X_s) \, dW_s \\[0.4em] &\quad = \int_0^t \operatorname{sgn}(B_s)^2 \, dB_s \\[0.4em] &\quad = B_t = X_t . \end{aligned}

Every solution is a Brownian motion, so uniqueness in law holds. Pathwise uniqueness fails, because −X-X solves the equation with the same WW, as the path spends zero time at the level 00. It is also a known result, stated here without proof, that no strong solution exists. A model price depends only on the law and is the same for every weak solution, while a simulation that feeds given shocks into the equation needs a strong solution.

Ornstein-Uhlenbeck process

Definition For constants κ>0\kappa > 0, θ\theta and σ>0\sigma > 0 and a deterministic initial value x0x_0, the Ornstein-Uhlenbeck equation is

dXt=κ(θ−Xt) dt+σ dWt\begin{aligned} dX_t &= \kappa(\theta - X_t)\, dt \\[0.4em] &\quad + \sigma \, dW_t \end{aligned}

with X0=x0X_0 = x_0. Its coefficients are globally Lipschitz, so it has a unique strong solution. To find it, apply the product rule to eκtXte^{\kappa t} X_t. The factor eκte^{\kappa t} has no Brownian part, so there is no covariation term, and

d(eκtXt)=eκt(κXt dt+dXt)=eκt(κθ dt+σ dWt),\begin{aligned} &d\bigl(e^{\kappa t} X_t\bigr) \\[0.4em] &\quad = e^{\kappa t}\bigl(\kappa X_t \, dt + dX_t\bigr) \\[0.4em] &\quad = e^{\kappa t}\bigl(\kappa\theta \, dt + \sigma \, dW_t\bigr), \end{aligned}

because κXt dt\kappa X_t \, dt cancels the −κXt dt-\kappa X_t \, dt in the drift. Integrating from 00 to tt and dividing by eκte^{\kappa t} gives

Xt=θ+(x0−θ)e−κt+σ∫0te−κ(t−s) dWs.\begin{aligned} X_t &= \theta + (x_0 - \theta) e^{-\kappa t} \\[0.4em] &\quad + \sigma \int_0^t e^{-\kappa (t - s)} \, dW_s . \end{aligned}

The stochastic integral has a deterministic integrand, so it is normal with mean 00, and the Itô isometry gives its variance σ2∫0te−2κ(t−s) ds\sigma^2 \int_0^t e^{-2\kappa(t - s)} \, ds. Hence XtX_t is normal with

E[Xt]=θ+(x0−θ)e−κt,Var⁡(Xt)=σ22κ(1−e−2κt).\begin{aligned} &\mathbb{E}[X_t] = \theta + (x_0 - \theta) e^{-\kappa t} , \\[0.4em] &\operatorname{Var}(X_t) \\[0.4em] &\quad = \frac{\sigma^2}{2\kappa} \bigl(1 - e^{-2\kappa t}\bigr) . \end{aligned}

As t→∞t \to \infty the law tends to the normal law with mean θ\theta and variance σ2/(2κ)\sigma^2/(2\kappa). The same formulas with XsX_s in place of x0x_0 and t−st - s in place of tt give the conditional law of XtX_t given Fs\mathcal{F}_s, which is the exact transition. The solution formula also holds for κ≤0\kappa \le 0, where the variance is σ2t\sigma^2 t when κ=0\kappa = 0, but for κ<0\kappa < 0 deviations from θ\theta grow instead of decaying.

In plain words The state is pulled toward θ\theta at a rate proportional to its distance from θ\theta, while shocks of constant size keep arriving. The initial distance decays exponentially, and older shocks are damped more than recent ones, which caps the variance at σ2/(2κ)\sigma^2/(2\kappa). Mean reversion is a property of the drift, and a single path still moves away from θ\theta much of the time.

Example With the constructed values x0=0.02x_0 = 0.02, θ=0.04\theta = 0.04, κ=0.5\kappa = 0.5 per year and σ=0.01\sigma = 0.01 per square-root year, the decay factor at t=2t = 2 is e−1≈0.3679e^{-1} \approx 0.3679, so

E[X2]=0.04−0.02 e−1≈0.032642,\begin{aligned} \mathbb{E}[X_2] &= 0.04 - 0.02\, e^{-1} \\[0.4em] &\approx 0.032642 , \end{aligned}

about 3.26423.2642 percent, and

Var⁡(X2)=0.0001(1−e−2)≈0.000086466\begin{aligned} \operatorname{Var}(X_2) &= 0.0001 \bigl(1 - e^{-2}\bigr) \\[0.4em] &\approx 0.000086466 \end{aligned}

in squared decimal-rate units, a standard deviation of about 0.0092990.009299. The law is normal, so P(X2<0)=Φ(−0.032642/0.009299)≈Φ(−3.510)≈0.00022\mathbb{P}(X_2 < 0) = \Phi(-0.032642 / 0.009299) \approx \Phi(-3.510) \approx 0.00022. This is small but positive, and the model allows negative rates. The long-run standard deviation is σ/2κ=0.01\sigma/\sqrt{2\kappa} = 0.01.

Euler step

Definition For the equation dXt=b(t,Xt) dt+σ(t,Xt) dWtdX_t = b(t, X_t) \, dt + \sigma(t, X_t) \, dW_t and a step h>0h > 0 with grid times tn=nht_n = nh, the Euler step, also called the Euler-Maruyama step, freezes both coefficients at the start of the step,

Xn+1E=XnE+b(tn,XnE) h+σ(tn,XnE)h Zn,\begin{aligned} X^{E}_{n+1} &= X^{E}_n + b(t_n, X^{E}_n)\, h \\[0.4em] &\quad + \sigma(t_n, X^{E}_n) \sqrt{h}\, Z_n , \end{aligned}

with independent standard normal variables ZnZ_n, so that h Zn\sqrt{h}\, Z_n has the law of a Brownian increment over one step. It approximates the transition of the solution and in general is not that transition. For coefficients that do not depend on tt and satisfy the global Lipschitz condition, with E[X02]<∞\mathbb{E}[X_0^2] < \infty and with h Zn\sqrt{h}\, Z_n taken to be the increment Wtn+1−WtnW_{t_{n+1}} - W_{t_n} of the driving Brownian motion, the root-mean-square error at the grid times is of order h1/2h^{1/2}, a result stated here and studied with numerical methods later. Pathwise uniqueness, uniqueness in law and convergence of a scheme answer three different questions, and none of them replaces another.

In plain words An Euler step pretends that the drift and the noise loading keep their starting values over the whole step. Over a short step that is close to right, but the exact transition also includes the way the drift changes as the state moves during the step.

Example In the constructed Ornstein-Uhlenbeck model at the state 0.020.02 with h=0.1h = 0.1, the Euler step adds the drift 0.5⋅0.02⋅0.1=0.0010.5 \cdot 0.02 \cdot 0.1 = 0.001 and, for the constructed draw Z=1Z = 1, the shock 0.010.1≈0.0031620.01 \sqrt{0.1} \approx 0.003162. The new state is 0.02+0.001+0.003162=0.0241620.02 + 0.001 + 0.003162 = 0.024162, about 2.41622.4162 percent. The exact transition over the same step has

mean=0.04−0.02 e−0.05≈0.020975,\begin{aligned} \text{mean} &= 0.04 - 0.02 \, e^{-0.05} \\[0.4em] &\approx 0.020975 , \end{aligned}

compared with the Euler mean 0.0210.021, and standard deviation 0.011−e−0.1≈0.0030850.01 \sqrt{1 - e^{-0.1}} \approx 0.003085, compared with 0.0031620.003162. An exact sampler fed the same standard normal number gives 0.020975+0.003085=0.0240600.020975 + 0.003085 = 0.024060.

Check yourself

  1. In the constructed Ornstein-Uhlenbeck model with x0=0.02x_0 = 0.02, θ=0.04\theta = 0.04, κ=0.5\kappa = 0.5 and σ=0.01\sigma = 0.01, find the mean, variance and standard deviation of X1X_1, and the probability that X1<0X_1 < 0.
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The decay factor is e−0.5≈0.6065e^{-0.5} \approx 0.6065, so

E[X1]=0.04−0.02 e−0.5≈0.027869.\begin{aligned} \mathbb{E}[X_1] &= 0.04 - 0.02 \, e^{-0.5} \\[0.4em] &\approx 0.027869 . \end{aligned}

The variance is 0.0001(1−e−1)≈0.0000632120.0001 (1 - e^{-1}) \approx 0.000063212, so the standard deviation is about 0.0079510.007951. Then

P(X1<0)≈Φ(−3.505)≈0.00023,\begin{aligned} \mathbb{P}(X_1 < 0) &\approx \Phi(-3.505) \\[0.4em] &\approx 0.00023 , \end{aligned}

since 0.027869/0.007951≈3.5050.027869 / 0.007951 \approx 3.505.

  1. Decide whether each pair of coefficients satisfies the hypotheses of the existence and uniqueness theorem on [0,1][0, 1], and for each pair that fails say what goes wrong. The pairs are b(x)=μxb(x) = \mu x and σ(x)=σx\sigma(x) = \sigma x for geometric Brownian motion, then b(x)=κ(θ−x)b(x) = \kappa(\theta - x) and σ(x)=σx\sigma(x) = \sigma \sqrt{x} for x≥0x \ge 0, then b(x)=x2b(x) = x^2 and σ(x)=0\sigma(x) = 0 with X0=2X_0 = 2.
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The first pair satisfies both conditions, since ∣μx−μy∣+∣σx−σy∣=(∣μ∣+σ)∣x−y∣\lvert \mu x - \mu y \rvert + \lvert \sigma x - \sigma y \rvert = (\lvert \mu \rvert + \sigma) \lvert x - y \rvert and ∣μx∣+∣σx∣≤(∣μ∣+σ)(1+∣x∣)\lvert \mu x \rvert + \lvert \sigma x \rvert \le (\lvert \mu \rvert + \sigma)(1 + \lvert x \rvert), so there is a unique strong solution, the explicit exponential formula. The second pair fails the Lipschitz condition at 00, because x/x=1/x\sqrt{x}/x = 1/\sqrt{x} is unbounded as x↓0x \downarrow 0, and the coefficient is not even defined for negative states. The theorem then says nothing, which does not mean that the equation is ill posed. Square-root diffusions need specialized existence, uniqueness and boundary results. The third pair is Lipschitz on bounded sets but neither globally Lipschitz nor of linear growth, and the solution Xt=2/(1−2t)X_t = 2/(1 - 2t) explodes at t=0.5t = 0.5, so there is no solution on [0,1][0, 1].

  1. A classmate says that because Euler approximations converge, a single Euler step of length 22 gives the right mean of X2X_2 in the constructed Ornstein-Uhlenbeck model. Compute the mean of the Euler approximation at t=2t = 2 with one step h=2h = 2 and with twenty steps h=0.1h = 0.1, and compare both with the exact mean.
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Taking expectations in the Euler step gives the recursion for the means

mn+1−θ=(1−κh)(mn−θ).\begin{aligned} &m_{n+1} - \theta \\[0.4em] &\quad = (1 - \kappa h)(m_n - \theta) . \end{aligned}

With h=2h = 2 the factor is 1−0.5⋅2=01 - 0.5 \cdot 2 = 0, so the Euler mean jumps to exactly θ=0.04\theta = 0.04, while the exact mean is 0.04−0.02 e−1≈0.0326420.04 - 0.02\, e^{-1} \approx 0.032642. With h=0.1h = 0.1 the factor is 0.950.95, and after twenty steps 0.9520≈0.35850.95^{20} \approx 0.3585 in place of the exact e−1≈0.3679e^{-1} \approx 0.3679, so the Euler mean is

m20=0.04−0.02⋅0.9520≈0.032830,\begin{aligned} m_{20} &= 0.04 - 0.02 \cdot 0.95^{20} \\[0.4em] &\approx 0.032830 , \end{aligned}

within about 0.000190.00019 of the exact mean. The exact one-step factor is e−κhe^{-\kappa h}, and Euler replaces it by its first-order Taylor polynomial 1−κh1 - \kappa h, which is accurate only when κh\kappa h is small. Convergence is a statement about h→0h \to 0. For κh>1\kappa h > 1 the factor is negative and the Euler means overshoot θ\theta and oscillate, and for κh>2\kappa h > 2 they diverge.

How this lesson connects