Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $W_{t}$ be a Wiener Process (a Brownian Motion starting at $W_{0} = 0$). What is the difference between $W_{t}$ and $\sqrt{W_{t}^{2}}$?

Using the Ito formula (in differential notation), $dW_{t}^{2} = dt + 2W_{t}dW_{t}$, so $d\sqrt{W_{t}^{2}} = [0 + (\frac{1}{2})(-\frac{1}{4}(W_{t}^{2})^{-\frac{3}{2}})(2W_{t})^{2})]dt + [\frac{1}{2}(W_{t}^{2})^{-\frac{1}{2}}]dW_{t}^{2} = (0)dt + (\frac{W_{t}}{\sqrt{W_{t}^{2}}})dW_{t}$, which doesn't really help (we need to assume they are different to show they are different, and visa versa)...

Edit: Of course they cannot be the same process, since $\sqrt{W_{t}^{2}}$ can never go below 0. But then why does it look like $E[\sqrt{W_{t}^{2}}] = E[\int_{0}^{t} \frac{W_{t}}{\sqrt{W_{t}^{2}}}dW_{t}] = 0$?

share|cite|improve this question
Obviously the formula for $d \sqrt{W_t^2}$ doesn't work at a point where $W_t = 0$. Ito's lemma assumes a twice-differentiable function, and $\sqrt{x^2}$ is not even once differentiable at $x=0$. – Robert Israel Jan 22 '12 at 7:44
Robert has pointed out why the Ito Lemma can't be applied to your case. In all the rest, the difference between $W_t$ and $\sqrt{W^2_t}$ is the same as the difference between $x$ and $|x|$. – Ilya Jan 22 '12 at 13:12
up vote 1 down vote accepted

As noted in comments to your question, you can't apply blindly Itô here as $\sqrt{x}$ isn't twice differentiable everywhere. Though there is an extension of Itô's formula called Tanaka's formula that allows you represent $\sqrt{W^2_t}=|W_t|$ in the following way :

$|W_t|=\int_0^t sign(W_s)dW_s + L_t$

Where $L_t$ is the local time at 0 of the Brownian motion $W_t$.

So to answer your remark, taking expectation $E[|W_t|]=E[\int_0^t sign(W_s)dW_s + L_t]=E[L_t]$ and $E[L_t]$ has non zero value.

The law of local time of Brownian motion at 0 is known to be the same as the law of the maximum of a Brownian motion which density is known to be for $x>0$ :

$f(x)=\frac{2}{\sqrt{2\pi t}}.e^{-\frac{x^2}{2t}}$

So $E[L_t]=\frac{2}{\sqrt{2\pi t}}.\int_{\mathbb{R}^+}xe^{-\frac{x^2}{2t}}dx=\sqrt{\frac{t}{2.\pi}}$

You might say that this a quite involved way to derive it as the law of $|W|_t$ is also known explicitly and give rise to the same calculations and result, but this was only to make appear the compensator part of $|W_t|$ and its martingale part.

Best regards

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.