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.

  • 2
    $\begingroup$ If I'm not severely miscalculating $\langle X \rangle_t = 1$ so that $\langle M \rangle_t = \langle X \rangle_t = 1$ for all $t \geq 0$. In particular, since if $W$ is a Brownian motion $\langle W \rangle_t = t$, $M$ is not a Brownian motion. $\endgroup$ – Rhys Steele Oct 31 '17 at 18:25
  • $\begingroup$ Ok, that might be the reason I could not verify $E[MtMs]=t \wedge s$. I assume that $< . > _t $ is the sharp bracket, but we did not cover this in the lecture yet hence I don't want to use this. Could it be true that $M_t$ is a martingale? So maybe just the task is just wrong. $\endgroup$ – user397268 Nov 1 '17 at 8:13
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    $\begingroup$ $M_t$ isn't a martingale. (It's in fact not even a local martingale since its semimartingale decomposition has non-zero finite variation part. If you know about semimartingale decompositions then this is another way to see that $M_t$ is not a Brownian motion.) $\endgroup$ – Rhys Steele Nov 1 '17 at 8:33
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    $\begingroup$ It suffices to show that $\mathbb{E}(M_t^2) \neq t$ ... $\endgroup$ – saz Nov 1 '17 at 16:28
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    $\begingroup$ @Diamir The result is correct, see my answer below for one possible approach. $\endgroup$ – saz Nov 2 '17 at 16:34

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 $$\mathbb{E}(M_t^2) \neq t.$$


$$\mathbb{E}(X_t X_s) = e^{-(t+s)} \mathbb{E}(B_{e^{2t}} B_{e^{2s}}) = e^{-(t+s)} \min\{e^{2s},e^{2t}\} \tag{1}$$


$$\mathbb{E}((M_t+X_0)^2) = \mathbb{E}(X_t^2) + 2 \mathbb{E} \left( \int_0^t X_t X_s \, ds \right) + \mathbb{E} \left( \int_0^t \int_0^t X_s X_r \, ds \, dr \right). \tag{2}$$

We have already calculated the first term on the right-hand side. Using $(1)$ we find

$$\begin{align*} \mathbb{E} \left( \int_0^t X_s X_t \, ds \right) &= e^{-t} \int_0^t e^s \, ds = e^{-t} (e^t-1) = 1-e^{-t}. \end{align*}$$


$$\begin{align*} \mathbb{E} \left( \int_0^t \int_0^t X_r X_s \, ds \, dr \right) &= 2 \int_0^t \int_0^r \mathbb{E}(X_s X_r) \, ds \, dr \\ &= 2 \int_0^t e^{-r} \int_0^r e^{s} \, ds \, dr \\ &= 2 \int_0^t e^{-r} (e^r-1) \, dr \\ &= 2t+2(e^{-t}-1). \end{align*}$$

Combining all the calculations, we conclude

$$\mathbb{E}((M_t+X_0)^2) = 1 + 2 (1-e^{-t}) + 2t+2(e^{-t}-1) = 1+2t$$


$$\mathbb{E}(M_t^2) = 2t.$$


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