Ornstein-Uhlenbeck process

An important Ito process in fiance is Ornstein-Uhlenbeck process (or Mean-reverting process). In fiance, It’s called as Hull-White model to describe a stochastic interest rate.

It is given by OU process of the following form.

\displaystyle   d r_t = (a_t - b_t r_t) dt + \sigma_t dW_t, \ \ \ \ \ (1)

where {W} is a 1-dim BM, {b_{t}} and {\sigma_{t}} are positive processes. (1) has an explicit form of

\displaystyle   r_t = r_0 e^{-\int_0^t b(s) ds} + \int_0^t e^{-\int_s^t b(u) du} a(s) ds + \int_0^t \sigma(s) e^{-\int_s^t b(u) du} dW_s. \ \ \ \ \ (2)

In this below, we are going to consider an example of system of SDE with mean field effect.

Example 1

Let {N} be a given natural number, and {N}-dimensional process {X = (X^{i})_{i= 1, 2, \ldots N}} satisfies

\displaystyle d X^{i}_{t} = (\bar X_{t} - X^{i}_{t}) dt + dW^{i}_{t}, \ X_{0}^{i} = x^{i}, \ i = 1, 2, \ldots, N

where {\bar X_{t} = \frac 1 N \sum_{i=1}^{N} X^{i}_{t}. } Existence and uniqueness of the solution is well known with Lipschitz coefficients. We are going to study the asymptotic distribution of {X^{i}} as {N} goes to infinity. By summing up all {N} equations, we observe that

\displaystyle d \bar X_{t} = \frac 1 N \sum_{i=1}^{N} d W^{i}_{t} = \frac 1 {\sqrt N} d B_{t}

for some standard Brownian motion {B}. So we can write {\bar X_{t} = \bar x + \frac 1 {\sqrt N} B_{t} := B^{N, \bar x}_{t}}. We rewrite the equation as

\displaystyle d X^{i}_{t} = (B^{N, \bar x}_{t} - X^{i}_{t}) dt + dW^{i}_{t}, \ X_{0}^{i} = x^{i}, \ i = 1, 2, \ldots, N

and apply the formula , then we have

\displaystyle X_{t}^{N, i} = x^{i} e^{-t} + \int_{0}^{t} e^{-t + s} B^{N, \bar x}_{s}ds + \int_{0}^{t} e^{-t+s} d W_{t}^{i}.

Note that {B_{s}^{N, \bar x} \rightarrow \bar x} both in almost surely and in {L^{2}}. Heuristically, one can replace {B^{N, \bar x}} by constant process {\bar x}, and have asymptotic process {Y^{i} := \lim_{N\rightarrow \infty}X^{N, i}} satisfying

\displaystyle  Y^{i}_{t} = x^{i} e^{-t} + \int_{0}^{t} e^{-t + s} \bar x ds + \int_{0}^{t} e^{-t+s} d W_{t}^{i} = \bar x + e^{-t} (Y_{0}^{i} - \bar x) + \int_{0}^{t} e^{-t+s} d W_{t}^{i}.

This is indeed a solution of McKean-Vlasov equation of the following type, i.e. {(Z, W) = (Y^{i}, W^{i})} satisfies

\displaystyle d Z_{t} = (\mathbb E[Z_{t}] - Z_{t}) dt + dW_{t}, \ \hbox{ s.t. } \mathbb E[Z_{0}] = \bar x.

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