How can I convert the following process to a standard Brownian Motion?
$$\mathrm{d}r_t=(a-br_t)\mathrm{d}t+\sigma\sqrt{r_t}\mathrm{d}W_t$$
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I am not quite sure I get your question. You cannot solve the model in closed-form. What you get is that \begin{align*} r_t=r_0e^{-a t}+\frac{b}{a}(1- e^{-a t})+\sigma e^{-at}\int_0^te^{a u} \sqrt{r_u}\mathrm{d}W_u. \end{align*} Furthermore, you can get that $r_t$ follows a (non-central) chi-squared distribution and can compute the (conditional) moments of $r_t$ and obtain closed-form solutions for the prices of zero-coupon bonds and European-style zero-coupon bond options.
You may want to have a look at "Interest rate models" from Brigo and Mercurio which is an excellent reference.
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$\begingroup$ Thank you for your response. Yes, I am trying to obtain closed-form for zero-coupon bonds but to do that I need to work under P*, probability that converts that given $W_u$ to standard Brownian motion, say, $W*$, I am not sure how to get $r_t$ in the form of $W*$. Thank you for the reference btw. $\endgroup$ Feb 12, 2020 at 12:14
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$\begingroup$ @Stochasticlearner You cannot get $r_t$ in explicit form regardless what measure you use. If you want to price bonds, a common approach is to solve the corresponding bond price PDE. Guessing that bond price is affine-linear in $r_t$, the PDE reduces to a system of ODEs. The solution is given in the book I mentioned above. $\endgroup$– KevinFeb 12, 2020 at 12:22