Let $\{x_t: t\geq 0\}$ be a Ornstein-Uhlenbeck (OU) process given by
$$dx_t = \theta ( \mu – x_t) dt + \sigma dB_t$$
where $\mu$ is a constant.
Also, let $\{e_t: t\geq 0 \}$ follows a Cox–Ingersoll–Ross (CIR) process: For a given initial value $e_0>0$,
$$
de_t = a(1-e_t) dt + \sigma_e \sqrt{e_t} dW_t,
$$
where $a$ is some constant and $B_t, W_t$ are standard Brownian motions with $dB_t dW_t = 0$.
I wonder is it possible to calculate the following conditional expectation: for $t\geq 0$,
$$
\mathbb{E}[e_t|x_t ] =?
$$
My attempt:
Rewrite the CIR process as an integral representation:
$$
e_t = e_0 + \int_0^t a(1-e_s)ds + \sigma_e \int_0^t \sqrt{e_s} dW_s
$$
Then I try to substitute this into $\mathbb{E}[e_t| x_t]$ but get stuck quickly.
I guess that if $e_t$ and $x_t$ are independent for all $t$, then $\mathbb{E}[e_t|x_t] = \mathbb{E}[e_t]$, but still get stuck on how to proceed further.
Any help is appreciated. Thank you.
Best Answer
By Levy's characterisation of Brownian motion it should follow that $B$ and $W$ are independent Brownian motions. Therefore, by the definition of the conditional expectation we have $\Bbb E [e_t \vert x_t] = \Bbb E [e_t ]$. Now observe that $$\Bbb E[e_t] = e_0 + \int_0^t a (1 - \Bbb E [e_s] ) ds$$ since the last part vanishes as martingale. It follows that $\varphi (t) = \Bbb E [e_t] $ satisfies the differential equation $\frac{\partial}{\partial t} \varphi (t) = a (1 - \varphi (t))$ with initial condition $\varphi (0) = e_0$. By variation of constants it follows that $$\Bbb E [e_t] = 1 + (e_0 -1)e^{-at}.$$