Deriving the Zero-Coupon Bond Dynamics from Forward Rates
Summary
The document derives the price dynamics of a zero-coupon bond from a stochastic model for forward rates. It defines the logarithm of the bond price as the negative integral of forward rates from the current time to maturity, then uses the given forward-rate evolution to express that integral in terms of its initial value and volatility terms. Differentiating this representation yields the dynamics of the log price, including the short rate as its drift contribution and maturity-specific forward-rate volatility as its diffusion.
Applying Itô’s lemma to the exponential of the log price gives the bond-price process: its drift is the short rate and its diffusion is the bond’s volatility exposure. The derivation makes the intermediate steps explicit, including the quadratic-variation correction that offsets the volatility-related drift in the log price. It is a symbolic derivation under the stated forward-rate dynamics; the document does not specify a numerical volatility model or discuss calibration, market data, or broader model assumptions.
Key ideas
- Represent the log of a zero-coupon bond price as the negative integral of forward rates to maturity.
- Integrating the forward-rate dynamics separates initial-curve terms from stochastic volatility terms.
- The log bond price has a volatility-related drift correction and a maturity-specific diffusion term.
- Itô’s lemma converts the log-price dynamics into bond-price dynamics with short-rate drift.
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# Zero coupon bond calculations
# Zero coupon bond calculations
I am given the following forward rate dynamics $df(t,u)=\frac{\partial}{\partial u}(\frac{\sigma^2}{2})dt-\frac{\partial}{\partial u}\sigma dW$ and want to calculate the dynamics of the ZCB $p$ via the computations below. The second equality is clearly Ito lemma but equality three and four throws me off completely, can someone see whats going on?
## Answer by Gordon (score 6, accepted)
https://quant.stackexchange.com/a/49455
The notations in the snapshot are pretty messy. I prefer to proceed as follows.
Let $X_t = -\int_t^T f(t, u)du$. Note that \begin{align*} f(t, u) - f(0, u) = \frac{\partial }{\partial u}\left(\int_0^t \frac{\sigma^2(s, u)}{2} ds - \int_0^t \sigma(s, u) d W_s \right). \end{align*} Then \begin{align*} r_t = f(t, t) = f(0, t) + \frac{\partial }{\partial u}\left(\int_0^t \frac{\sigma^2(s, u)}{2} ds - \int_0^t \sigma(s, u) d W_s \right)\Big|_{u=t}. \end{align*} Moreover, \begin{align*} \int_t^T f(t, u)du - \int_t^T f(0, u)du &= \left(\int_0^t \frac{\sigma^2(s, T)}{2} ds - \int_0^t \sigma(s, T) d W_s \right) \\ &\qquad -\left(\int_0^t \frac{\sigma^2(s, t)}{2} ds - \int_0^t \sigma(s, t) d W_s \right). \end{align*} That is, \begin{align*} X_t &= - \int_t^T f(0, u)du + \left(\int_0^t \frac{\sigma^2(s, t)}{2} ds - \int_0^t \sigma(s, t) d W_s \right) \\ &\qquad\qquad\qquad\qquad - \left(\int_0^t \frac{\sigma^2(s, T)}{2} ds - \int_0^t \sigma(s, T) d W_s \right). \end{align*} Consequently, \begin{align*} dX_t &= f(0, t) dt + \frac{\sigma^2(t, t)}{2} dt - \sigma(t, t) d W_t \\ &\qquad + \frac{\partial }{\partial u}\left(\int_0^t \frac{\sigma^2(s, u)}{2} ds - \int_0^t \sigma(s, u) d W_s \right)\Big|_{u=t} dt -\frac{\sigma^2(t, T)}{2} dt + \sigma(t, T) d W_t\\ &= r_t dt -\frac{\sigma^2(t, T)}{2} dt + \sigma(t, T) d W_t. \end{align*} Therefore, \begin{align*} dP(t, T) &= d\big(e^{X_t} \big)\\ &=P(t, T)\Big(dX_t + \frac{1}{2} d\langle X, X\rangle_t \Big)\\ &=P(t, T)\big(r_t dt + \sigma(t, T) d W_t \big). \end{align*}Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
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