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For independently driven nonnegative squared Bessel processes $X,Y$, put $Z=X+Y$. Where $Z>0$, use Brownian weights $\sqrt{X/Z}$ and $\sqrt{Y/Z}$. Where $Z=0$, fill the weights with $(1,0)$. Their squared norm is always one and their cross-variation is zero, so their stochastic integral is Brownian motion by the Lévy characterization of Brownian motion. Multiplying by $2\sqrt{Z}$ recovers the sum's martingale part even at zero, and the drifts add to $\alpha +\beta$.

A Bessel process of dimension $d$ is the radial part of $d$-dimensional Brownian motion. Away from zero it solves the stochastic differential equationThis equation also defines noninteger dimensions through its weak solutions.

For $Z_t=e^{B_t+at}$, use the clock $C_t={\int }_0^t{Z}_s^{2}ds$. After the inverse time change, the Dambis-Dubins-Schwarz theorem givesso the time-changed process is a Bessel process of dimension $d=2a+2$. Its lifetime is $C_{\infty }$, which is finite almost surely exactly when $d<2$.

Because $W$ is continuous and adapted, the Borel function $\mathrm{sgn}(W)$ is predictable. Its value at zero is $-1$, so its square is exactly one everywhere. Consequently $A$ is a zero-starting continuous local martingale withThe Lévy characterization of Brownian motion therefore makes $A$ a Brownian motion in the same filtration. Here that characterization is the deterministic-clock case of the conditional-exponential argument in question 2.

Apply the Itô formula to $V=W^2$:Since $\sqrt{V_t}=|W_t|$ and $|W_t|\mathrm{sgn}(W_t)=W_t$, including at zero, substitution yieldsThus $W^2$ is a squared Bessel process of dimension one.

The native PDF repeats $d{W}_y$ in its list of independent noise terms, whereas its third stochastic differential equation uses $d{W}_z$. We use the intended three independent coordinate Brownian motions $W_x,W_y,W_z$.

Put $\mathbf{X}=(X,Y,Z)$, $r=|\mathbf{X}|$ and $S=r^2$. Before the origin is reached, Itô's formula givesThe $6$ is the sum of three quadratic variation terms, each $2dt$. The drift contribution is $-2\alpha dt$ because the numerator is $|\mathbf{X}{|}^2$. The radial stochastic integral can be written using a scalar Brownian motion $B$:Thus $S/2$ is a squared Bessel process of dimension $\delta =3-\alpha$ until it reaches zero. If this equation is taken globally with the usual nonnegative squared Bessel process continuation, taking expected values givesThe stochastic integral has zero expected value: stopping first at large levels bounds the integrands, yields a finite first moment for $S$, and then makes $\mathbb{E}{\int }_0^{t}S(u)du$ finite, so the unrestricted integral is a square-integrable martingale.

Boundary qualification. The punctured domain in the question does not itself specify that continuation. The boundary classification of a Bessel process says that the origin is inaccessible from $r_0>0$ when $\delta \ge 2$, equivalently $\alpha \le 1$. In this parameter range the original stochastic differential equation is global in the stated domain and the boxed result holds as written.

For $1<\alpha <3$, we have $0<\delta <2$: the origin is reached in finite time with probability one. One can also see accessibility directly from the radial generator of an SDE diffusionFor $r<b$, the harmonic function $1-(r/b{)}^{\alpha -1}$ is the probability of reaching zero before $b$. Finite-time access follows by using the nonnegative functionin the stopped Itô formula; it bounds the expected value of the time to reach zero or $b$. Letting $b\to \infty$ gives almost sure finite hitting of zero. The coefficients of the original vector stochastic differential equation are undefined there, so its stated punctured-domain version has no prescribed long-time evolution in this range.

The natural continuing radial squared Bessel process gives $6-2\alpha$, but stopping and holding the particle at the origin would give a different answer. Indeed, with $T_0$ the first hit of zero, Itô's formula for the stopped process givesSince $T_0<\infty$ almost surely, dominated convergence applied to $(t\wedge T_0)/t\le 1$ givesTherefore $6-2\alpha$ is the usual continued-radial answer; for $1<\alpha <3$ a boundary convention is necessary. It is not a uniquely defined limit for the punctured-domain model alone.