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时变均值的 OU：显式解、均值与方差

Determine E and var

专题
Probability / 概率
难度
L4

题目详情

Assume that r(t)r(t) satisfies r(0)=r0r(0) = r_{0} and

dr(t)=(dμ(t)dt+λ(μ(t)−r(t)))t+σ(t)dW(t)dr(t) = \left(\frac{d \mu(t)}{dt} + \lambda (\mu (t) - r(t))\right) t + \sigma (t) dW(t)

Determine r(t)r(t) , its expectation, and variance.

解析

SDE：

dr(t)=(μ′(t)+λ(μ(t)−r(t)))dt+σ(t)dW(t),r(0)=r0.dr(t)=\left(\mu'(t)+\lambda(\mu(t)-r(t))\right)dt+\sigma(t)dW(t),\quad r(0)=r_0.

令 q(t)=eλtr(t)q(t)=e^{\lambda t}r(t)，可得

r(t)=(r0−μ(0))e−λt+μ(t)+e−λt∫0teλsσ(s)dW(s).r(t)=(r_0-\mu(0))e^{-\lambda t}+\mu(t)+e^{-\lambda t}\int_0^t e^{\lambda s}\sigma(s)dW(s).

因此

E[r(t)]=(r0−μ(0))e−λt+μ(t),\boxed{\mathbb{E}[r(t)]=(r_0-\mu(0))e^{-\lambda t}+\mu(t)},

并由 Itô 等距

Var⁡(r(t))=e−2λt∫0te2λsσ(s)2ds.\boxed{\operatorname{Var}(r(t))=e^{-2\lambda t}\int_0^t e^{2\lambda s}\sigma(s)^2ds}.