Let $(B_t)_{t \ge 0}$ be a Brownian Motion. Define the process $(X_t)_{t \ge 0}$ by $X_t := e^{-t} B_{e^{2t}}$. Now I am supposed to show that
$M_t := X_t – X_0 + \int_0^t X_s \, ds$ defines a Brownian Motion.
I tried to show that $E[M_t M_s] = t \wedge s$ but I failed.
I already know that $(X_t)_{t \ge 0}$ is a stationary process with normal distributed random variables. So is it easier to "just" verify the definition of a Brownian motion in this case? Any help is appreciated.
Best Answer
As was already pointed out in the above comments, $(M_t)_{t \geq 0}$ is not a Brownian motion. To prove it suffices to show that $$\mathbb{E}(M_t^2) \neq t.$$
Clearly,
$$\mathbb{E}(X_t X_s) = e^{-(t+s)} \mathbb{E}(B_{e^{2t}} B_{e^{2s}}) = e^{-(t+s)} \min\{e^{2s},e^{2t}\} \tag{1}$$
and
$$\mathbb{E}((M_t+X_0)^2) = \mathbb{E}(X_t^2) + 2 \mathbb{E} \left( \int_0^t X_t X_s \, ds \right) + \mathbb{E} \left( \int_0^t \int_0^t X_s X_r \, ds \, dr \right). \tag{2}$$
We have already calculated the first term on the right-hand side. Using $(1)$ we find
$$\begin{align*} \mathbb{E} \left( \int_0^t X_s X_t \, ds \right) &= e^{-t} \int_0^t e^s \, ds = e^{-t} (e^t-1) = 1-e^{-t}. \end{align*}$$
Similarly,
$$\begin{align*} \mathbb{E} \left( \int_0^t \int_0^t X_r X_s \, ds \, dr \right) &= 2 \int_0^t \int_0^r \mathbb{E}(X_s X_r) \, ds \, dr \\ &= 2 \int_0^t e^{-r} \int_0^r e^{s} \, ds \, dr \\ &= 2 \int_0^t e^{-r} (e^r-1) \, dr \\ &= 2t+2(e^{-t}-1). \end{align*}$$
Combining all the calculations, we conclude
$$\mathbb{E}((M_t+X_0)^2) = 1 + 2 (1-e^{-t}) + 2t+2(e^{-t}-1) = 1+2t$$
implying
$$\mathbb{E}(M_t^2) = 2t.$$