Let $W=(W_t)_{t \geq 0}$ be a Browniwn motion. Do the processes $$X_t = W_{e^t} \quad \text{and} \quad Y_t = \exp \left(- \frac{t^2}{2} \right) W_{e^t}$$ have the no-memory property, i.e. are the sets $\sigma(X_t; t \leq a)$ and $\sigma(X_{t+a}-X_a; t \geq 0)$, $a>0$, respectively $\sigma(Y_t; t \leq a)$ and $\sigma(Y_{t+a}-Y_a; t \geq 0)$, $a>0$, independent?
$X$ has this property because of the independent increment of Brownian Motion. But $Y$ is a little bit tricky. I try to form a function such that the increment of $Y$ is a function of the increment of $X$ and the result follows but I fail to find the function. So does $Y$ have this property and why?
No, $Y$ does not have the no-memory property. Since $$Y_2 -Y_1 = \exp(-2) W_{\exp(2)} - \exp(-1/2) W_{\exp(1)}$$ we have \begin{align*} \mathbb{E}((Y_2-Y_1) Y_1) &= \exp(-5/2) \mathbb{E}(W_{\exp(2)} W_{\exp(1)}) - \exp(-1) \mathbb{E}(W_{\exp(1)}^2). \end{align*} Using that $$\mathbb{E}(W_s W_t) = \min\{s,t\}, \qquad s,t \geq 0,$$ we get
$$\mathbb{E}((Y_2-Y_1) Y_1) =\exp(-5/2) \exp(1) - \exp(-1) \exp(1) = \exp(-3/2) -1 \neq 0.$$
As $\mathbb{E}(Y_1)=0$, this shows that $$\mathbb{E}((Y_2-Y_1) Y_1) \neq \mathbb{E}(Y_2-Y_1) \mathbb{E}(Y_1)$$
and therefore $Y_2-Y_1$ is not independent from $Y_1$. Thus, $\sigma(Y_t; t \leq 1)$ and $\sigma(Y_{t+1}-Y_1; t \geq 0)$ are not independent.