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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?

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

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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.

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  • $\begingroup$ 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 $$\frac{\omega_s}{\omega_t}=e^{k(s-t)}$$ $\endgroup$ Sep 21, 2022 at 11:48
  • $\begingroup$ @OliverQueen Yes, exactly! $\langle \rangle$ is my notation for expected value. $\endgroup$
    – stochastic
    Sep 21, 2022 at 11:57
  • $\begingroup$ Oh. I had no idea that the expected value could be written as $<\omega_t>$ instead of $\mathbb{E}[\omega_t]$...thanks $\endgroup$ Sep 21, 2022 at 12:03

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