Let $W$ be a one dimensional Brownian motion and define, $$ X_t=W_{(\text{exp}(\beta t)-1)}\\ \hat{W}_t=\frac{1}{\sqrt{\beta}}\int_0^te^{-\frac{\beta s}{2}}dX_s $$

Show that $\hat{W}_t$ is a local martingale in its natural filtration and compute its quadratic variation.

To show that it's a local martingale, is it just straight forward differentiation on $\hat{W}_t$ like below,

$$ d\hat{W}_t=\frac{1}{\sqrt{\beta}}e^{-\frac{\beta t}{2}}dW_{(\text{exp}(\beta t)-1)} $$

  • 2
    $\begingroup$ No, you have to show it satisfies the definition. All you have done there is rewrite your definition of $\hat{W_t}$ in the shorthand using "d"s. $\endgroup$
    – muaddib
    Jun 12, 2015 at 0:23

3 Answers 3


As @muaddib pointed out, you have simply rewritten the definition of $\hat{W}_t$ - but this doesn't show that $(\hat{W}_t)_{t \geq 0}$ is a martingale.


  1. Show that $(X_t)_{t \geq 0}$ is a martingale with respect to its canonical filtration $$\mathcal{F}_t := \sigma(X_s; s \leq t) = \sigma(W_s; s \leq e^{\beta t}-1).$$
  2. Conclude that $(\hat{W}_t)_{t \geq 0}$ is a martingale with respect to $(\mathcal{F}_t)_{t \geq 0}$.
  3. Using the tower property show that $(\hat{W}_t)_{t \geq 0}$ is also a martingale with respect to its natural filtration.

I came up with a solution involving properties of time change as follows...

Let $\theta(t)=\int_0^t\xi^2ds$, which we assume is finite.

Then the stochastic integral $Y_t=\int_0^t\xi dW_s$ exists, and $Y_t$ is a Gaussian process with independent increments.

The variance of $Y_t-Y_s$ is given by,

$$ \begin{align} Var\left[Y_t-Y_s\right]&=\mathbb{E}\left[\left(Y_t-Y_s\right)^2\right]-\mathbb{E}\left[Y_t-Y_s\right]^2\\ &=\mathbb{E}\left[\left(\int_0^t \xi dW_s - \int_0^s\xi dW_u \right)^2\right]-0\\ &=\mathbb{E}\left[\left(\int_s^t \xi dW_u \right)^2\right]\\ &=\mathbb{E}\left[\left(\int_s^t \xi^2du \right)^2\right]\\ &=\theta(t)-\theta(s)\\ &=\mathbb{E}\left[\left( W_{\theta(t)}-W_{\theta(s)} \right)^2\right]\\ \end{align} $$

So $Y_t$ has the same distribution as time changed Brownian Motion $W_{\theta(t)}$. From the definition of $Y_t$ we have,

$$ Y_t=\int_0^t\xi dW_s=\int_0^t\sqrt{\theta'(s)}dB_s \sim W_{\theta(t)} $$

For our problem, $ \theta(t) = e^{\beta t}-1$ , and $\sqrt{\theta'(t)}=\sqrt{\beta}e^{\frac{\beta t}{2}}$

From this we finally have,

$$ \hat{W}_t=\frac{1}{\sqrt{\beta}}\int_0^t e^{-\frac{bs}{2}}\sqrt{\beta} e^{\frac{bs}{2}}dW_s=\int_0^t dW_s $$

Which is a martingale.

For quadratic variation, we know that $\langle \hat{W}\rangle_t$ is the process such that $ \hat{W}_t^2 - \langle W\rangle_t$ is a martingale. Therefore,

$$ \begin{align} \mathbb{E}\left[ \hat{W}_t^2 - \langle \hat{W}\rangle_t \right] &= 0\\ \Rightarrow\mathbb{E}\left[ \langle \hat{W}\rangle_t \right] &= \mathbb{E}\left[ \hat{W}_t^2 \right]\\ &= \mathbb{E}\left[ \left(\int_0^t dW_s \right)^2\right]\\ &= \mathbb{E}\left[ \left(\int_0^t 1^2 ds \right)^2\right]\\ \Rightarrow \langle \hat{W}\rangle_t = t \end{align} $$

This shows that $\langle \hat{W}\rangle_t $ is a Brownian Motion in its natural filtration.

  • $\begingroup$ So you actually claim that $W_t = \hat{W}_t$ almost surely? Doesn't this sound this somewhat odd to you? Do you really think that we can expect this equality? More specific: Why does $$\int_0^t \sqrt{\theta'(t)} \, dB_s = W_{\theta(t)}$$ hold true? (There are so many typos in this equality that one has to guess what you mean.) $\endgroup$
    – saz
    Jun 12, 2015 at 16:57
  • $\begingroup$ sorry for the typos, I've fixed up. $\endgroup$
    – Danny
    Jun 13, 2015 at 1:03
  • $\begingroup$ i think it's some thing like we know that $W_{ct}$ is the same as $\sqrt{c}W_t$. So if we let $Z_t=\frac{1}{\sqrt{c}}\int_0^t dW_{cs}$, then $Z_t$ has mean 0 and variance t as well. The above is just a more complicated form of this same argument involving arbitrary functions. $\endgroup$
    – Danny
    Jun 13, 2015 at 1:40
  • $\begingroup$ btw, how do i add line breaks in comments. i tried the help and it says two blank spaces, but it's not working for me $\endgroup$
    – Danny
    Jun 13, 2015 at 1:44
  • $\begingroup$ You have to be more careful; the equality $W_{ct} = \sqrt{c} W_t$ holds in distribution and not almost surely. That's why I asked whether you think that we can really expect $W_t = \hat{W}_t$ almost surely - it might well be that they are equal in distribution, but not almost surely. (Since you want to investigate martingale properties, we need "almost surely"!) $\endgroup$
    – saz
    Jun 13, 2015 at 5:23

@saz i tried out the following according to your hints,

Let $\mathcal{F}_t:= \sigma(X_s;s\le t)$, then we have,

$$ \begin{align} \mathbb{E}[X_t|\mathcal{F}_s]&=\mathbb{E}[X_t-X_s+X_s|\mathcal{F}_s]\\ &=\mathbb{E}[X_t-X_s\mathcal|{F}_s] + \mathbb{E}[X_s|\mathcal{F}_s]\\ &=X_s \end{align} $$

So $X_t$ is a martingale under its natural filtration, and $\hat{W}_t $ is a martingale with respect to $\mathcal{F}_t$.

But I'm not sure how to what is the tower property part you have in mind. Can you kindly elaborate?


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