A constructed short-rate model starts at , pulls the rate toward with speed per year and shocks it with volatility 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 be a standard Brownian motion with respect to a filtration under , let and be measurable functions on , and let be an -measurable initial condition. The stochastic differential equation
with is shorthand for the integral equation
required to hold almost surely for all at once. A solution is a process with continuous paths, adapted to , such that and almost surely, so that both integrals are defined, and such that the integral equation holds. Brownian paths are nowhere differentiable, so is not the derivative of times , 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 with a constant has the candidate solution with . Itô’s lemma with , and gives
which is the equation, so the integral identity holds for all on almost every path. The second-order term of Itô’s lemma restores exactly the that the exponent subtracts.
Strong solution
Definition Fix in advance the filtered probability space, the Brownian motion and the initial condition , which is -measurable and hence independent of . A strong solution is a solution on this space, driven by this , that is adapted to the filtration generated by and and completed with the null sets. Its value at time is then determined, up to a null set, by and the Brownian path up to time , 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 is an explicit function of and . With the constructed values , and currency units, every path with has
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 .
Existence and uniqueness theorem
Definition Suppose that and are measurable on and that there are constants and such that for all and all real and
which is the global Lipschitz condition, and
which is the linear growth condition. Let be a Brownian motion for and let be -measurable with . Then the equation has a strong solution on with , and pathwise uniqueness holds. Pathwise uniqueness means that any two solutions and on the same space, with the same Brownian motion and the same initial condition, satisfy for all 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 , the Lipschitz condition implies the growth condition. The proof is sketched here. Picard iteration starts from and sets
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 is of order for a constant , which is summable in , so the iterates converge to a solution. For two solutions, the same estimates applied to , after stopping both processes so that is finite, give , and Gronwall’s inequality forces .
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 and satisfy both conditions with , and for the constructed , and the constant works, since . Geometric Brownian motion satisfies them with . The square-root coefficient of the Cox-Ingersoll-Ross model is not Lipschitz at , since as , so the theorem does not apply to it and specialized results are needed. Without the growth condition, the deterministic equation with has the solution , which explodes at , so no solution exists on . Without the Lipschitz condition, with has the two solutions and .
Weak solution
Definition A weak solution with initial law consists of a filtered probability space , a Brownian motion for and a continuous adapted process , all constructed together, such that has law 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 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 depends on.
Example Tanaka’s equation is with , where for and for . Its coefficient is not continuous, so the theorem above does not apply. A weak solution is built from any Brownian motion by setting and . This is a Brownian motion by Lévy’s characterization, which a later lesson of this unit covers, and
Every solution is a Brownian motion, so uniqueness in law holds. Pathwise uniqueness fails, because solves the equation with the same , as the path spends zero time at the level . 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 , and and a deterministic initial value , the Ornstein-Uhlenbeck equation is
with . Its coefficients are globally Lipschitz, so it has a unique strong solution. To find it, apply the product rule to . The factor has no Brownian part, so there is no covariation term, and
because cancels the in the drift. Integrating from to and dividing by gives
The stochastic integral has a deterministic integrand, so it is normal with mean , and the Itô isometry gives its variance . Hence is normal with
As the law tends to the normal law with mean and variance . The same formulas with in place of and in place of give the conditional law of given , which is the exact transition. The solution formula also holds for , where the variance is when , but for deviations from grow instead of decaying.
In plain words The state is pulled toward at a rate proportional to its distance from , 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 . Mean reversion is a property of the drift, and a single path still moves away from much of the time.
Example With the constructed values , , per year and per square-root year, the decay factor at is , so
about percent, and
in squared decimal-rate units, a standard deviation of about . The law is normal, so . This is small but positive, and the model allows negative rates. The long-run standard deviation is .
Euler step
Definition For the equation and a step with grid times , the Euler step, also called the Euler-Maruyama step, freezes both coefficients at the start of the step,
with independent standard normal variables , so that 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 and satisfy the global Lipschitz condition, with and with taken to be the increment of the driving Brownian motion, the root-mean-square error at the grid times is of order , 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 with , the Euler step adds the drift and, for the constructed draw , the shock . The new state is , about percent. The exact transition over the same step has
compared with the Euler mean , and standard deviation , compared with . An exact sampler fed the same standard normal number gives .
Check yourself
- In the constructed Ornstein-Uhlenbeck model with , , and , find the mean, variance and standard deviation of , and the probability that .
Show the answer
The decay factor is , so
The variance is , so the standard deviation is about . Then
since .
- Decide whether each pair of coefficients satisfies the hypotheses of the existence and uniqueness theorem on , and for each pair that fails say what goes wrong. The pairs are and for geometric Brownian motion, then and for , then and with .
Show the answer
The first pair satisfies both conditions, since and , so there is a unique strong solution, the explicit exponential formula. The second pair fails the Lipschitz condition at , because is unbounded as , 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 explodes at , so there is no solution on .
- A classmate says that because Euler approximations converge, a single Euler step of length gives the right mean of in the constructed Ornstein-Uhlenbeck model. Compute the mean of the Euler approximation at with one step and with twenty steps , and compare both with the exact mean.
Show the answer
Taking expectations in the Euler step gives the recursion for the means
With the factor is , so the Euler mean jumps to exactly , while the exact mean is . With the factor is , and after twenty steps in place of the exact , so the Euler mean is
within about of the exact mean. The exact one-step factor is , and Euler replaces it by its first-order Taylor polynomial , which is accurate only when is small. Convergence is a statement about . For the factor is negative and the Euler means overshoot and oscillate, and for they diverge.