Use of Girsanov's theorem in bond pricing

Question

Assume that we want to calculate the time $t=0$ price of a bond: $B(0,T) = E_P[\exp(-\int_0^T r_s ds)]$, where $r$ is the interest rate following the SDE $dr_t=k(\theta-r_t)dt+\sigma dB_t=b(r_t)dt+\sigma dB_t$.

I was shown that one could write the price as

$B(0,T) = E_{\hat{P}}[\exp(-\int_0^TB^{*}_sds)\exp(\int_0^Tb(B^{*}_s)dB^{*}s-\frac{1}{2}\int_0^Tb^2(B^{*}_s)ds)]$

where $B_t^{*}$ is the "new" Brownian motion from the Girsanov theorem.

However, when I try to implement it, it results in prices that are too low. Here is what I did:

Since $dr_t=b(r_t)dt+\sigma dB_t$, Girsanov's theorem gives a new Brownian motion with dynamics $dB_t^{*}=\frac{1}{\sigma}b(r_t)dt+dB_t$, and the dynamics of $r_t$ becomes $dr_t=\sigma dB_t^{*}$. So $r_t=r_0+\sigma B_t^{*}$ and $dB_t^{*}= \frac{1}{\sigma}b(r_0+\sigma B_t^{*})dt+dB_t$. With this last expression I tried to use Euler discretization to find $B_t^{*}$, then finally I approximated the three integrals as sums.

What am I doing wrong? Secondly, I also wonder what the starting point of $B_t^{*}$ should be, i.e. $B_0^{*}$.

Answer 1

Bond Price Dynamics

I do not know the source of the bond dynamics you show above but seeing how we are dealing with an affine model there is a very elegant way to derive those.

Due to the model being affine the bond price is given by $$P(t,T)=A(t,T)e^{-r(t)B(t,T)}$$ you can find the exact formulas for $A(t,T)$ and $B(t,T)$ in this document (or just read the relvant chapter in Brigo Mercurio. If you are curious to know how one arrives at above pricing formula I suggest this paper.

Now that we know above closed form formua, getting the dynamics of $P(t,T)$ is just a matter of applying Itô and using some algebra

$$ dP(t,T)=r_tP(t,T)dt-\sigma B(t,T)P(t,T)dW_t $$

Getting a solution for $r_t$

The generalized Vasicek model is given by $$ dr_t=\kappa(t)(\theta(t)-r_t)d t+\sigma(t)dW_t $$

The unique solution is

$$ r_t=A^{-1}(t)\left[r_0+\int_0^tA(s)\kappa(s)\theta(s)ds+\int_0^t A(s)\sigma(s)dW_s\right] $$ with $A(t)=\exp\left(\int_0^t\kappa(s)ds\right)$

Seeing how you have constant and not time dependant parameters the above simplifies to

$$ r_t=e^{-\kappa t}\left[r_0+\int_0^t\kappa\theta e^{\kappa s} ds+\int_0^t \sigma e^{\kappa s} dW_s\right] $$

P.S.: To arrive at the above solution one simply applies Ito's Formula to $d(r_t \cdot e^{\int_0^t\kappa(s)ds})$