*Note:* This is a crosspost from <http://math.stackexchange.com>; the original question may be found [here](https://math.stackexchange.com/questions/2036746/comprehension-question-within-script-on-bessel-process).


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.



Comprehension question within script on Bessel process
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:
$$
d X\_t =\frac{d-1}{2 X\_t} \; dt + dZ\_t \;, \quad X\_0 = \sqrt{x} \;.
$$


