Comprehension question within script on Bessel process

Question

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.

Answer 1

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: $$ d X_t =\frac{d-1}{2 X_t} \; dt + dZ_t \;, \quad X_0 = \sqrt{x} \;. $$