2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $$ \frac{P(t,T)}{B(t)}=P(0,T)\,{\cal E}\left(\int_0^tv(s,T)\,dW^*_s\right)\,. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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