Proof about discounted zero coupon bond

Question

Hey guys I am having trouble finishing this proof:

Proposition 5.1 Under the above assumptions, the process $r$ satisfies under $\mathbb{Q}$ $$ d r(t)=\left(b(t)+\sigma(t) \gamma(t)^{\top}\right) d t+\sigma(t) d W^{*}(t) $$

where $W^{*}(t)=W(t)-\int_{0}^{t} \gamma(s)^{\top} d s$ denote the Girsanov transformed $\mathbb{Q}$ -Brownian motion.

Now I know that $\frac{P(t, T)}{B(t)}$ is the discounted zero-coupon bond and is a $\mathbb{Q}$ -martingale where $P(t, T)=\mathbb{E}_{\mathbb{Q}}\left[e^{-\int_{t}^{T} r(s) d s} \mid \mathcal{F}_{t}\right]$.

Now I need to show that: For any $T>0$, there exists adapted process $\mathbb{R}^{d}$ -valued process $v(t, T), t \leq T$ such that $$ \frac{d P(t, T)}{P(t, T)}=r(t) d t+v(t, T) d W^{*}(t) . $$ $$ \frac{P(t, T)}{B(t)}=P(0, T) \mathcal{E}_{t}\left(\int_{0} v(s, T) d W^{*}(s)\right) $$

My attempt so far:

Recall $d\left(\frac{P(t, T)}{B(t)}\right)$ is a martingale hence there exists $ k(t, T)$ such that $d\left(\frac{P(t, T)}{B(t)}\right)$$= k(t, T) w_{t}^{*} .$

Let $V(t, T)=\frac{K(t, T)}{\frac{P(t, T)}{B(t)}}$ then:

$\frac{d\left(\frac{P(t, T)}{B(t)}\right)}{\frac{P(t, T)}{B(t)}}=V(t, T) d w^{*}(t)$

I am not sure how to go about solving the differential equation to get to the end of the proof.

Answer 1

Your attempt is correct as far as the second equation is concerned. Nonetheless I will include this in my answer: You know that $$ d\left(\frac{P(t,T)}{B(t)}\right)=k(t,T)\,dW^*_t\quad\quad\quad\quad\quad\text{(1)} $$ (where I fixed your notation). Applying Ito's lemma to the LHS of this relation gives $$ \frac{dP}{B}-P\frac{dB}{B^2}=\frac{P}{B}\left(\frac{dP}{P}-\frac{dB}{B}\right)=\frac{P}{B}\left(\frac{dP}{P}-r\right)\,. $$ Setting $v(t,T)=\frac{B(t)}{P(t,T)}k(t,T)$ gives obviously $$ \frac{dP}{P}-r=v(t,T)\,dW^*_t\,. $$ This is the first equation you wanted to show. To see the second equation observe that from (1) we get directly $$ \frac{d\left(\frac{P(t,T)}{B(t)}\right)}{\frac{P(t,T)}{B(t)}}=v(t,T)\,dW^*_t\,. $$ By the Ito formula this is known to be equivalent to $$ \frac{P(t,T)}{B(t)}=P(0,T)\,{\cal E}\left(\int_0^tv(s,T)\,dW^*_s\right)\,. $$