# Can we show that for any $s > t ≥ 0$, we have $\mathbb{E}[\omega_s|\omega_t] =e^{\kappa(s−t)}\omega_t$ and why?

Suppose that the initial state $$\omega_0$$ is normally distributed with mean $$\mu_0$$ and variance $$\sigma_0^2$$. The state then evolves according to the stochastic differential equation

$$\tag{*}d\omega_t=\kappa\omega_tdt+\sigma dZ_t$$

where the driving process $$\{Z_t\}_{t≥0}$$ is a standard Brownian motion, independent of the initial state $$\omega_0$$. The variance rate $$\sigma^2$$ is strictly positive. The percentage drift $$\kappa$$ has unrestricted sign. If $$\kappa < 0$$, then the state follows a mean-reverting Ornstein–Uhlenbeck process. If $$\kappa = 0$$, then $$\omega_t − \omega_0 = \sigma Z_t$$, so the state follows a Brownian motion with zero drift. If $$\kappa > 0$$, then the state process is explosive. The solution of $$(*)$$ is given by $$\omega_t=\omega_0 e^{-\kappa t} +\sigma\int_{0}^te^{\kappa(s-t)}dZ_s$$

The effect of $$\kappa$$ can be seen in the formula for the conditional expectation $$\mathbb{E}[\omega_s|\omega_t]$$.

Can we show that for any $$s > t ≥ 0$$, we have $$\mathbb{E}[\omega_s|\omega_t] =e^{\kappa(s−t)}\omega_t$$ and why?

• Hint $\mathbb E[\omega _s\mid \omega _t]=\mathbb E[\omega _s-\omega _t\mid \omega _t]+\omega _t=\mathbb E[\omega _s-\omega _t]+\omega _t$
– Surb
Sep 21, 2022 at 10:32
• @Surb could please show a small proof? Sep 21, 2022 at 11:44

Take the expected value of both sides of your equation, since the expected value of the Brownian motion is zero, you have $$\frac{d\langle \omega_t\rangle}{dt} = \kappa \langle\omega_t\rangle.$$ which you can integrate from $$t$$ to $$s$$ given the initial condition at $$\omega_t$$ to obtain your result.
• what does the $<>$ bracket means? You mean this $$\frac{d<\omega_t>}{d_t}=\kappa<\omega_t>\Rightarrow \int_t^s\frac{d<\omega_t>}{<\omega_t>}=\int_t^s\kappa dt$$ this gives us $(ln\omega_u)_s^t=k(s-t)$ which gives us $$\frac{\omega_s}{\omega_t}=e^{k(s-t)}$$ Sep 21, 2022 at 11:48
• @OliverQueen Yes, exactly! $\langle \rangle$ is my notation for expected value. Sep 21, 2022 at 11:57
• Oh. I had no idea that the expected value could be written as $<\omega_t>$ instead of $\mathbb{E}[\omega_t]$...thanks Sep 21, 2022 at 12:03