Comprehension question within script on Bessel process Note: This is a crosspost from http://math.stackexchange.com; the original question may be found here.
I have a question regarding a script by Greg Lawler on Bessel processes: 
http://www.math.uchicago.edu/~lawler/bessel.pdf
There I encounter difficulties in understanding the last sentence on page 2.
Let $ W_t = (W_t^1,\dots,W_t^d) $ be a (standard) d-dimensional Brownian motion and 
$$ X_t = |W_t| = |W_t|_2 = \left(\sum\limits_{j=1}^d \left(W_t^j\right)^2\right)^{1/2} $$ 
its (Euclidean) norm. 
Now it is noted: 
$$ dX_t^2=\sum_{j=1}^d d[(W_t^j)^2] = 2\sum_{j=1}^d W_t^j dW_t^j + d \;dt $$ 
and we are supposed to be allowed to write the later as
$$ dX_t^2 = d \; dt + 2X_t dZ_t $$ 
with
$$ Z_t = \sum_{j=1}^d\int_0^t\frac{|W_s^j|}{X_s} dW_s^j .$$
It is not clear to me why we may rewrite it like that. I think
$$ X_t dZ_t = \sum_{j=1}^d |W_t^j| dW_t^j $$ holds true. 
Wouldn't this imply, e.g.,
$$ |W_t^1| dW_t^1 = W_t^1 dW_t^1? $$ 
But this is not true, is it? Do I have a lapse of thought here?
Thank you for any hints :) 
Edit: So on SE someone suggested the $|\cdot|$ in the definition of $Z_t$ to be a typo. That is also what I was thinking lately.
 A: Here is another way to arrive at the result.  Set $Y_t = X_t^2$ (for clarity). As the OP shows,
$$
dY_t = d \; dt + 2 \sum_{j=1}^d W_t^j dW_t^j \;, \quad Y_0 = x > 0 \;.
$$ By definition of the infinitesimal generator of $Y_t$, we have
\begin{align*}
L f(x) &= \lim_{t \to 0^+} \frac{1}{t} \mathbb{E}_x \left\{ f(Y_t) - f(x) \right\}  \\
&= d \; f'(x) + f''(x) \lim_{t \to 0^+}  \frac{2}{t} \sum_{j} \mathbb{E}_x \left\{ \left( \int_0^{t} W_s^j dW_s^j \right)^2 \right\} \\
&= d \; f'(x) + f''(x) \lim_{t \to 0^+}  \frac{2}{t} \sum_{j} \mathbb{E}_x \left\{ \int_0^{t} (W_s^j)^2 ds \right\} \\
&= d \; f'(x) + f''(x)  \lim_{t \to 0^+} \frac{2}{t} \mathbb{E}_x \left\{ \int_0^{t} Y_s ds \right\} \\
&= d \; f'(x) + 2 x \; f''(x) 
\end{align*}
where we used a Taylor series in $f$ about $t=0$ and the Itô isometry. 
With a slight abuse of notation, this $L$ implies that $Y_t$ is a weak solution of the SDE:
$$
d Y_t = d \; dt + 2 \sqrt{Y_t} \; dZ_t
$$ where $Z$ is a one-dimensional Brownian motion, and by Itô's formula, $X_t = \sqrt{Y_t}$ satisfies:
$$
d X_t =\frac{d-1}{2 X_t} \; dt +  dZ_t \;, \quad X_0 = \sqrt{x} \;.
$$
