# Change of numeraire to the forward measure in the Vasicek model

I am working through the Brigo/Mercurio book on Interest Rate Models (Second Edition) and I am having some trouble with the change of numeraire in chapter 3.2.1, page 59 to be exact, formula 3.9. It is about the Vasicek model with dynamic $$dr(t) = k[\theta -r(t)]dt + \sigma dW(t)$$ and $$P(t,T) = A(t,T) exp(-B(t,T)r(t))$$.

They write that using the change-of-numeraire toolkit and formula 2.12 in particular with $$S_t = B(t)$$ the bank-account numeraire, $$U_t = P(t,T)$$ the T-forward numeraire and $$X_t=r_t$$, you can obtain $$dr(t) = [k\theta-B(t,T)\sigma^2 -kr(t)]dt + \sigma dW^T(t)$$ with $$dW^T(t) = dW(t)+\sigma B(t,T)dt$$. I can see how plugging the $$Q^T$$-Brownian motion into the new dynamic yields back the original dynamic. However i cant figure out how to obtain the new dynamic through the formula 2.12, specifically how $$B(t,T)\sigma=\rho \left(\frac{\sigma_t^S}{S_t} - \frac{\sigma_t^U}{U_t}\right)$$ Can someone help me or provide a link to a more detailed explanation?

Let's assume you are working with 1-dimensional Brownian motion, the instantaneous correlation matrix $$\rho$$ drops to 1. $$C$$ and $$C'$$ both are 1.

Now, referring to Proposition 2.3.1, in particular with $$S_t = B(t)$$ and $$U_t = P(t,T)$$, you can write out the two processes as:

$$dB(t) = (...)dt$$

$$dP(t,T) = (...)dt - \sigma B(t,T)P(t,T)dW_t^{T}$$,

Note that there's no volatility coefficient in the first equation, so $$\sigma_t^B=0$$. Also note that the $$B(t,T)$$ in the second equation is not the bank-account numeraire, but is the $$B$$ function in Brigo's book.

Now, the drift in $$\mathbb{Q}^T$$, denoted by $$\mu_t^{P}$$, can be derived using equation (2.12):

$$\mu_t^{P}(r(t))=\mu_t^{B}(r(t))-\sigma\left(\frac{0}{B(t)}-\frac{\sigma B(t,T)P(t,T)}{P(t,T)}\right) = \mu_t^{B}(r(t))+\sigma^2B(t,T)$$

Hence the additional drift due to measure change is $$\sigma^2B(t,T)$$ and then you have equation (3.9).

Apply the same logic to equation (2.13) you'll get:

$$dW^T(t)=dW(t)+\sigma B(t,T)dt$$

Hope this helps.