# Difference between a Brownian Motion and the root of its square

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 $E[\sqrt{W_{t}^{2}}] = E[\int_{0}^{t} \frac{W_{t}}{\sqrt{W_{t}^{2}}}dW_{t}] = 0$?

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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

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

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