Expectation and Variance of Stochastic Differential Equations

Question

2. Consider the SDE $$dr_t=\kappa(\theta-r_t)\,dt+\sigma dW_t,\ r_0=x,$$ where $\kappa$, $\theta$ and $\sigma$ are constants. You are given that the solution is $$r_t=\theta+(x-\theta)e^{-\kappa t}+\sigma\int_0^te^{-\kappa(t-s)}\,dW_s.$$ Calculate the mean and variance of $r_t$. You may use the result $$\mathbb E\left[\left(\int_0^tY_s\,dW_s\right)^2\right]=\mathbb E\left[\int_0^tY_s^2\,ds\right],$$ in the calculation of the variance.

Hi, I was wondering if somebody could tell me how to calculate the expectation of an SDE? I believe the expectation of a constant is just equal to the constant.

Answer 1

To compute the expectation and the variance, in addition to the given hint (Ito Isometry), you need to know that if the integrator $W_t$ is an arbitrary martingale, and the integrand $f$ is bounded, then the integral is a martingale, and the expectation of the integral is again zero (proof). Then we can proceed.

For: $$r_t=\theta+(x-\theta)e^{-\kappa t}+\sigma\int_0^te^{-\kappa(t-s)}\,dW_s$$

The Expectaction of $r_t$ $$\begin{align} \mathbb{E}[r_t]&=\mathbb{E}\left[\theta+(x-\theta)e^{-\kappa t}+\sigma\int_0^te^{-\kappa(t-s)}\,dW_s\right]\\ &=\theta+(x-\theta)e^{-\kappa t}+\sigma\mathbb{E}\left[\int_0^te^{-\kappa(t-s)}\,dW_s\right]\\ &=\theta+(x-\theta)e^{-\kappa t} \end{align}$$

The variance of $r_t$

$$\begin{align} Var[r_t]&=\mathbb{E}[r_t^2]-\left(\mathbb{E}[r_t]\right)^2\\ &=\mathbb{E}\left[\left(\theta+(x-\theta)e^{-\kappa t}+\sigma\int_0^te^{-\kappa(t-s)}\,dW_s\right)^2\right]-\left(\theta+(x-\theta)e^{-\kappa t}\right)^2\\ &=(\theta+(x-\theta)e^{-\kappa t})^2+2\sigma(\theta+(x-\theta)e^{-\kappa t})^2\mathbb{E}\left[\int_0^te^{-\kappa(t-s)}\,dW_s\right]+\sigma^2\mathbb{E}\left[\left(\int_0^te^{-\kappa(t-s)}\,dW_s\right)^2\right]-(\theta+(x-\theta)e^{-\kappa t})^2\\ &=\sigma^2\mathbb{E}\left[\int_0^te^{-2\kappa(t-s)}\,ds\right]\\ &=\sigma^2\int_0^te^{-2\kappa(t-s)}\,ds\\ &=\dfrac{\sigma^2}{2\kappa}(1-e^{-2\kappa t}) \end{align}$$

Answer 2

$\newcommand{\IE}{\mathbb{E}}$

As $\IE(dW)=0$ you get the differential equaton for $\IE(r_t)$ $$d\IE(r_t)=\IE(dr_t)=κ(θ−\IE(r_t))dt$$ which has the solution $$ θ−\IE(r_t)=e^{-κt}(θ−\IE(r_0))\implies \IE(r_t)=θ-e^{-κt}(θ−x) $$

Using the Ito formula for the square $$d(r_t)^2=2r_tκ(θ−r_t)dt+2r_tσdW_t+σ^2dt$$ results in $$ d\IE(r^2_t)=(2κθ\IE(r_t)-2κ\IE(r_t^2)+σ^2)dt $$ so that for the variance $Var(r_t)=\IE(r^2_t)-\IE(r_t)^2$ we get the differential equation \begin{align} d[\IE(r^2_t)-\IE(r_t)^2]&=(2κθ\IE(r_t)-2κ\IE(r_t^2)+σ^2)dt-2κ\IE(r_t)(θ−\IE(r_t))dt \\ &=-2κ[\IE(r_t^2)-\IE(r_t)^2]dt+σ^2dt \\ \implies \IE(r^2_t)-\IE(r_t)^2&=\frac{σ^2}{2κ}(1-e^{-2κ}). \end{align}