Differentiating a Bond Price in a Simple HJM Model
Summary
The document works through the differentiation of a zero-coupon bond price in a simple Heath–Jarrow–Morton interest-rate model. Starting from an exponential expression involving the initial forward curve, the drift integral, and a Brownian-motion term, the question applies Ito’s lemma and asks how to obtain the bond-price dynamics. The accepted answer focuses on verifying an integral identity implied by the specified drift function.
Substituting that drift into the integral from the current time to maturity yields a quadratic term in time to maturity and a term involving the drift’s time-dependent component. This matches the identity needed to reconcile the questioner’s Ito calculation with the stated bond-price dynamics. The exchange provides an algebraic step rather than a broader derivation of HJM theory. Its conclusion depends on the given model specification and assumptions, and should not be read as a general bond-pricing result beyond that setup.
Key ideas
- The bond price is expressed exponentially using the initial forward curve, drift, and Brownian component.
- Applying Ito’s lemma introduces both a stochastic term and a quadratic variation adjustment.
- The specified HJM drift can be integrated directly over the remaining time to maturity.
- That integral verifies the identity needed in the question’s bond-price calculation.
- The derivation applies to the particular model and drift specification shown.
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Full text
# Simple HJM model, differentiating the bond price
# Simple HJM model, differentiating the bond price
We have the following simple HJM model $$f(t,T)=f(0,T)+\int_0^t\alpha(s,T)ds+\sigma W_t$$ $$r_t=f(0,t)+\int_0^t\alpha(s,t)ds+\sigma W_t$$ $$P(t,T)=\exp-\bigg(\int_t^Tf(0,u)du+\int_0^t\int_t^T\alpha(s,u)duds+\sigma(T-t)W_t\bigg)\tag{1}$$ $$\alpha(t,T)=\sigma^2(T-t)+\sigma\gamma_t$$ Then by Ito's Lemma $$d_tP(t,T)=P(t,T)\Big(-\sigma(T-t)dW_t+(r_t-\sigma(T-t)\gamma_t)dt\Big)\tag{2}$$ My question is, how do we get from (1) to (2)?
I have attempted the following $$P(t,T)=e^{-X_t}$$ $$X_t=\int_t^Tf(0,u)du+\int_0^t\int_t^T\alpha(s,u)duds+\sigma(T-t)W_t$$ \begin{align} \frac{d}{dt}\int_0^t\int_t^T\alpha(s,u)duds&=\int_0^t\bigg(\frac{\partial}{\partial t}\int_t^T\alpha(s,u)ds\bigg)ds+\int_t^T\alpha(t,u)du\\ &=-\int_0^t\alpha(s,t)ds+\int_t^T\alpha(t,u)du\\ \end{align} \begin{align} dX_t&=-f(0,t)dt-\int_0^t\alpha(s,t)ds+\int_t^T\alpha(t,u)du-\sigma W_tdt+\sigma(T-t)dW_t\\ &=\bigg(\int_t^T\alpha(t,u)du-r_t\bigg)dt+\sigma(T-t)dW_t \end{align} \begin{align} d_tP(t,T)&=-P(t,T)dX_t+\frac{1}{2}P(t,T)\sigma^2 (T-t)^2dt\\ &=-P(t,T)\Bigg(\bigg(\int_t^T\alpha(t,u)du-r_t\bigg)dt+\sigma(T-t)dW_t\Bigg)+\frac{1}{2}P(t,T)\sigma^2 (T-t)^2dt\\ &=P(t,T)\Bigg(\bigg(r_t-\int_t^T\alpha(t,u)du+\frac{1}{2}\sigma^2 (T-t)^2\bigg)dt-\sigma(T-t)dW_t\Bigg) \end{align} Thus if I have been correct so far, it would imply that $$\int_t^T\alpha(t,u)du-\frac{1}{2}\sigma^2 (T-t)^2=\sigma(T-t)\gamma_t\tag{3}$$ with the change of measure satisfying the condition $$\alpha(t,T)=\sigma^2(T-t)+\sigma\gamma_t$$ I am unable to show the equality in (3), that is if I haven't made any mistakes in the previous working.
Any help is appreciated.
## Answer by Gordon (score 2, accepted)
https://quant.stackexchange.com/a/30689
Identity $(3)$ obviously holds if $$\alpha(t, T) = \sigma^2(T-t) +\sigma \gamma_t,$$ since \begin{align*} \int_t^T \alpha(t, u)du &= \int_t^T\left(\sigma^2(u-t) +\sigma \gamma_t \right)du\\ &=\frac{1}{2}\sigma^2(T-t)^2 + \sigma \gamma_t (T-t). \end{align*}Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.