I am trying to work out on the way to express the following process $Z_t$ as a form of squared Bessel process, if it is at all possible. I have a Brownian motion, $W_t$, with a drift $\alpha_t$ and $Z_t$ is defined by $$Z_t=\int_0^td(W_s+\alpha_s)^2$$
Addendum after Kurt G.'s comment
I can write $$dZ_{t}=dt+2|W_{t}+\alpha_{t}|\mathrm{sgn}(W_{t}+\alpha_{t})d(W_{t}+\alpha_{t})$$ and introduce the BM $$B_t=\int_0^t\mathrm{sgn}(W_{s}+\alpha_{s})d(W_{s}+\alpha_{s})$$ and then I am done with $$dZ_{t}=dt+2\sqrt{Z_{t}}dB_{t},$$ which is the expected result. Is it correct?
Now, I am a bit lost. What can you say about the starting point of this process and the non-centrality parameter of the non-central chi-squared between 0 and the horizon $t$? Is it $\int_0^t \alpha^2_{s}ds$, $\alpha^2_{t}$ or $\alpha^2_{0}$?
Idea after Kurt G.'s comment
Can I say, starting from $W_t$ under the measure $\mathbb P$, that there is another process $W'_t$ defined as: $$dW'_t=dW_t+\alpha'_tdt$$ such that $W'_t$ is a BM under a different measure $\mathbb P'$? Then, $Z_t$ is defined under $\mathbb P'$ as $Z_{t}=W'_t{}^2$. Obviously, $$dZ_{t}=dt+2\sqrt{Z_{t}}dB'_t$$ where $B'_t$ is a new BM (martingale with mean 0 and variance $t$) defined by: $$B'_t=\int_0^t\mathrm{sgn}W'_s dW'_s.$$ Now, going back to the measure $\mathbb P$, if I define $$\frac{d\mathbb P'}{d\mathbb P} = \exp\left(-\frac{1}{2}\int_0^t \alpha'_s{}^2ds+\int_0^t \alpha'_s dB_s\right)$$ then I can claim that $$dZ_{t}=dt+2\sqrt{Z_{t}}dB'_t=dt+2\sqrt{Z_{t}}(dB_t+\alpha'_t dt)=(1+2\alpha'_t\sqrt{Z_{t}})dt+2\sqrt{Z_{t}}dB_t.$$ which is actually not a squared Bessel process but a squared reflected Brownian motion...
The Bessel process is defined as the Euclidean norm of an $n$-dimensional ($n\ge 2$) Brownian motion, $X_t=\|\mathbf{W}_t\|$ and is known to solve the SDE $$ dX_t=dB_t+\frac{n-1}{2}\frac{1}{X_t}\,dt\tag{1} $$ where $B$ is a one-dimensional BM. Ito's formula shows that the square of this, $Y_t=X^2_t$ solves \begin{align} dY_t&=2X_t\,dX_t+d\langle X\rangle_t=2\sqrt{Y_t}\,dB_t+(n-1)\,dt+dt\\[2mm] &=2\sqrt{Y_t}\,dB_t+n\,dt\,.\tag{2} \end{align} Similarly, the $Z_t=(W_t+\alpha_t)^2$ solves \begin{align} dZ_t&=2(W_t+\alpha_t)\,d(W_t+\alpha_t)+d\langle W\rangle_t\\[2mm] &=2(W_t+\alpha_t)\,dW_t+2(W_t+\alpha_t)\,d\alpha_t+dt\,.\tag{3} \end{align} When we introduce the BM $$ \widetilde{W}_t=\int_0^t{\rm sgn}(W_s+\alpha_s)\,dW_s $$ we can write (3) as \begin{align} dZ_t&=2\sqrt{Z_t}\,d\widetilde{W}_t+2(W_t+\alpha_t)\,d\alpha_t+dt\,.\tag{4} \end{align}
For no choice of $\alpha$ (with bounded variation) the terms in (4) match those in (2).
Even if they did the $n$ in (2) must be at least $2$ which obstructs the match with the $dt$ term in (3).