In this question,
Ornstein-Uhlenbeck process: Markov, but not martingale?
, the process
$$
X_t = x e^{-\lambda t} + \sigma \int_0^t e^{-\lambda(t-s)} d W_s \,
$$
where $x$ is a constant and
$ E[X_{t'} | \mathcal{F}_t]$, $0 \leq t < t'$
is computed as
$$
E[X_{t'} | \mathcal{F}_t] = X_t e^{-\lambda {t'}} (\neq X_t) , \qquad 0 \leq t < t'
$$,
where $\mathcal{F}_t$ is generated by $W_t$.
Why is this?
I thought
\begin{align}
E[X_{t'} | \mathcal{F}_t]
&=x e^{-\lambda t'}
+
E[\sigma \int_t^{t'} e^{-\lambda(t-s)} d W_s|\mathcal{F}_t]
+
E[\sigma \int_{0}^{t} e^{-\lambda(t-s)} d W_s|\mathcal{F}_t]\\
&=x e^{-\lambda t'}
+
0
+
\sigma \int_{0}^{t} e^{-\lambda(t-s)} d W_s\neq X_t,
\end{align}
where the second term is $0$ because of the independence, and the third term is due to the $\mathcal{F}_t$-measurability.
I understand $X$ is not martingale (well, if my calculation is correct), but don't understand why $ E[X_{t'} | \mathcal{F}_t]= X_t e^{-\lambda {t'}}$.
Best Answer
When changing $X_t$ to $X_{t'}$ you need to change $t$ to $t'$ everywhere. Thus
\begin{align} E[X_{t'} | \mathcal{F}_t] &=x e^{-\lambda t'} + E[\sigma \int_t^{t'} e^{-\lambda(t'-s)} d W_s|\mathcal{F}_t] + E[\sigma \int_{0}^{t} e^{-\lambda(t'-s)} d W_s|\mathcal{F}_t]\\ &=x e^{-\lambda t'} + 0 + \sigma \int_{0}^{t} e^{-\lambda(t'-s)} d W_s\neq X_t, \\ &=e^{-\lambda (t'-t)}X_t \end{align} It seems the original computation was not correct or not correctly reported.