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