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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).

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    $\begingroup$ How do you get from a natrual log to a ratio in the last two lines of your solution before the substitution of v,dv? In one line, you have ln P(..) and the next you have P(t,u)/p(t,u). What identity is this? $\endgroup$
    – Adam Hughes
    Commented Feb 21, 2015 at 22:23
  • $\begingroup$ @Adam Sorry. I forgot to include a du at the end of the dv expression $\endgroup$
    – BCLC
    Commented Feb 22, 2015 at 5:29
  • $\begingroup$ @Adam Oh anyway I just differentiated the expression. Chain rule. $\endgroup$
    – BCLC
    Commented Feb 22, 2015 at 5:30

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

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