Bond Prices from Forward Rates

Article Quant Q&A · Author: BCLC

Summary

The document explains how to recover a bond price from its instantaneous forward rate when the forward rate is defined as the negative maturity derivative of the log bond price. Integrating that relationship from the valuation date to maturity gives the change in log price; the boundary condition that a bond at maturity is worth one then yields the exponential of the negative integrated forward rate.

The alternative derivation reaches an absolute value because integrating the derivative of a logarithm gives the logarithm of the absolute value. The stated unit-price boundary condition selects the positive solution, so the negative branch cannot satisfy it. The argument presumes the relevant derivatives and integrals exist and that the logarithm is defined along the maturity interval. It is a mathematical clarification rather than an empirical result or a full model of bond-price dynamics.

Key ideas

Integrating the negative maturity derivative of log bond price gives the log-price change across maturities.
The unit bond-price condition at maturity fixes the integration constant.
The absolute value in the alternative derivation does not imply two admissible price solutions.
The derivation assumes sufficient regularity for the derivative and integral to be related.

Tags

Fixed income Derivatives pricing Statistics

Full text

# Differential equation involving bond price and forward rate


# Differential equation involving bond price and forward rate












Given forward rate f(t,T) and bond price P(t,T) where

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

$P(T,T) = 1 = P(t,t)$,

T>0 and

$t \in [0,T]$

Does it follow that $P(t,T) = exp(-\int_{t}^{T} f(t,u) du)$?

My professor gives an argument that suggests it is so, but a different way I tried suggested the instead we have $P(t,T) = \pm exp(-\int_{t}^{T} f(t,u) du)$. Who is right? What is the flaw in the wrong one's reasoning?

My professor's:

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

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

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

$- \int_{t}^{T} f(t,u) du = \ln P(t,T) - \ln P(t,t)$

$- \int_{t}^{T} f(t,u) du = \ln P(t,T)$

$e^{- \int_{t}^{T} f(t,u) du} = P(t,T)$

QED

Mine:

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

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

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

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

$- \int_{t}^{T} f(t,u) du = \int_{t}^{T} \frac{\partial}{\partial u} P(t,u) / P(t,u) du$

Let

$v = P(t,u)$

$dv = \frac{\partial}{\partial u} P(t,u)$

$- \int_{t}^{T} f(t,u) du = \ln |P(t,u)/ P(t,t)|$

$- \int_{t}^{T} f(t,u) du = \ln |P(t,u)|$

$e^{- \int_{t}^{T} f(t,u) du} = |P(t,u)|$

$\pm e^{- \int_{t}^{T} f(t,u) du} = P(t,u)$

QED

P.S. it is assumed we can swap integral and derivative (if even relevant).

## Answer by BCLC (score 0, accepted)

https://quant.stackexchange.com/a/16684

The negative solution does not satisfy $P(T,T)=P(t,t)=1$

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Deriving Bond Prices from Instantaneous Forward Rates

Summary

Key ideas

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