Deriving Bond Dynamics from a Discounted Martingale
Summary
This note shows how to derive the diffusion of a zero-coupon bond price from the martingale property of its discounted value under a risk-neutral measure. Starting with the discounted bond as a stochastic integral against the transformed Brownian motion, it applies Itô’s formula to relate that martingale’s diffusion coefficient to the bond’s return after subtracting the short rate. This identifies the bond volatility process in the desired price dynamics.
The second result follows by dividing the discounted bond’s differential by its value and using the stochastic exponential solution for a process with proportional Brownian increments. The initial value is the bond price at time zero. The argument is concise and assumes the relevant processes are well-defined, the bond price is nonzero, and the martingale representation applies. It establishes the stated identities; it does not derive the underlying short-rate model or discuss conditions ensuring the stochastic exponential is a true martingale.
Key ideas
- The discounted zero-coupon bond is represented as a martingale driven by transformed Brownian motion.
- Applying Itô’s formula links its diffusion coefficient to the bond’s return net of the short rate.
- The bond volatility is obtained by dividing the discounted martingale’s coefficient by its value.
- The discounted bond price can be written as its initial price times a stochastic exponential.
- The proof relies on regularity and martingale representation assumptions.
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Full text
# Proof about discounted zero coupon bond
# Proof about discounted zero coupon bond
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)$$= k(t, T) w_{t}^{*} .$
Let $V(t, T)=\frac{K(t, T)}{\frac{P(t, T)}{B(t)}}$ then:
$\frac{d\left(\frac{P(t, T)}{B(t)}\right)}{\frac{P(t, T)}{B(t)}}=V(t, T) d w^{*}(t)$
I am not sure how to go about solving the differential equation to get to the end of the proof.
## Answer by Kurt G. (score 1, accepted)
https://quant.stackexchange.com/a/66474
Your attempt is correct as far as the second equation is concerned. Nonetheless I will include this in my answer: You know that $$ 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)\,. $$Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
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