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Deriving Instantaneous Forward Rates from Zero-Coupon Bond Prices

Article Quant Q&A · Author: NC520
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Summary

The document asks how to derive the instantaneous forward rate from zero-coupon bond prices. It begins with the relation between a forward bond price over a maturity interval and the integral of the forward curve across that interval. The question compares differentiating with respect to the interval's starting maturity against differentiating with respect to its ending maturity, then taking the interval length toward zero.

The first approach applies the Leibniz rule to both moving integration limits; in the zero-length limit, the resulting forward-rate terms cancel. The alternative differentiates with respect to the upper endpoint, which yields the integrand at that endpoint, and then takes the limit. This points to the familiar relation between the instantaneous forward rate and the maturity derivative of the log discount bond price. The supplied document contains the question and derivation attempts but no accepted answer, so it does not resolve the subtleties of notation or explicitly confirm the second approach's assumptions.

Key ideas

  • A forward bond price over a maturity interval is determined by integrating the instantaneous forward curve over that interval.
  • Differentiating an integral with respect to its starting maturity moves both endpoints and produces two boundary terms.
  • As the interval shrinks, the two terms from differentiating with respect to the starting maturity cancel.
  • Differentiating with respect to the upper endpoint isolates the forward rate at that endpoint.

Tags

Fixed incomeDerivatives pricingStatistics
Full text
# Instantaneous forward rate and bond prices


# Instantaneous forward rate and bond prices












Let $P(t,T,T+\tau)$ be the time $t$ forward price for a zero coupon bond (ZCB) spanning $[T, T+\tau]$, and $f(t,T)$ be the time $t$ instantaneous forward rate to time $T$. The relationship between the two is given by:

$$P(t,T,T+\tau) = exp \left( - \int_T^{T+\tau} f(t,u) du \right)$$

I want to derive the following result for the price of a ZCB $P(t,T) \equiv P(t,T,T)$ from the previous equation:

$$f(t,T) = - \frac{\partial \ln P(t,T)}{\partial T}$$

My intuition is to first compute the derivative wrt $T$, and then take the limit $\tau \rightarrow 0$:

\begin{align} - \ln P(t,T,T+\tau) &= \int_T^{T+\tau} f(t,u) du \\ - \frac{\partial}{\partial T} \ln P(t,T,T+\tau) &= \frac{\partial}{\partial T} \int_T^{T+\tau} f(t,u) du \quad \text{By Liebniz rule} \\ - \frac{\partial}{\partial T} \ln P(t,T,T+\tau) &= f(t,T+\tau) \frac{\partial}{\partial T} (T+\tau) - f(t,T) \frac{\partial}{\partial T} (T) \\ - \frac{\partial}{\partial T} \ln P(t,T,T+\tau) &= f(t,T+\tau) - f(t,T) \quad \text{Taking the limit} \\ - \frac{\partial}{\partial T} \ln P(t,T) &= f(t,T) - f(t,T) = 0 \end{align}

1) Could you explain why this approach leads to the wrong result?

Alternatively, first compute the derivative wrt $T+\tau$, and then take the limit $\tau \rightarrow 0$:

\begin{align} - \ln P(t,T,T+\tau) &= \int_T^{T+\tau} f(t,u) du \\ - \frac{\partial}{\partial (T+\tau)} \ln P(t,T,T+\tau) &= \frac{\partial}{\partial (T+\tau)} \int_T^{T+\tau} f(t,u) du \quad \text{By Fundamental Theorem of Calculus} \\ - \frac{\partial}{\partial (T+\tau)} \ln P(t,T,T+\tau) &= f(t,T+\tau) \quad \text{Taking the limit} \\ - \frac{\partial}{\partial T} \ln P(t,T) &= f(t,T) \end{align}

2) Is this approach correct?

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.