Comprehension question within script on Bessel process

Question

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: I crossposted this question on http://mathoverflow.net:

https://mathoverflow.net/questions/256039/comprehension-question-within-script-on-bessel-process

Answer 1

I think it is a typo. That is, it should be $$ Z_t = \sum_{j=1}^d\int_0^t\frac{W_s^j}{X_s} dW_s^j ,$$ rather than $$ Z_t = \sum_{j=1}^d\int_0^t\frac{|W_s^j|}{X_s} dW_s^j .$$