1
$\begingroup$

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).

$\endgroup$
  • 1
    $\begingroup$ 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 Feb 21 '15 at 22:23
  • $\begingroup$ @Adam Sorry. I forgot to include a du at the end of the dv expression $\endgroup$ – BCLC Feb 22 '15 at 5:29
  • $\begingroup$ @Adam Oh anyway I just differentiated the expression. Chain rule. $\endgroup$ – BCLC Feb 22 '15 at 5:30
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.