4
$\begingroup$

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?

$\endgroup$

1 Answer 1

3
$\begingroup$

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.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.