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'}}$.
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.