Let $P(t,T)=\mathbb{E}_{Q_{R}}[e^{\int^{T}_{t}r(u)du}|\mathcal{F}_{t}]$ be the price of a 1-euro zero-coupon bond with maturity $T$ and $r(u)$ the interest rate process. Consider the the forward rate $\frac{-\partial \log P(t,T)}{\partial T}$. How to prove that the forward is a martingale under $Q_{T}$? $Q_{T}$ is the T-forward measure with $P(t,T)$ as the numeraire.

It feels like a very basic question, however I truly cannot find any proofs on the internet.

  • 2
    $\begingroup$ Heuristically $d(logP) = dP/P$ so the forward is the limit of $(P(t,T)-P(t,T+dT))/P(t,T)$. This latter is the ratio of asset prices with P(t,T) in the denominator so is a martingale under the Q measure. $\endgroup$ – dm63 Jun 14 '19 at 19:58
  • $\begingroup$ Can you clarify if you are interested in a) the forward, b) the forward rate, or c) the instantaneous forward rate? The rate you have defined is the instantaneous forward rate, whereas in the answer below the forward rate is used. $\endgroup$ – Daneel Olivaw Jun 14 '19 at 20:47
  • $\begingroup$ My lecture notes tated forward rate, but I guess it is about the instantaneous forward rate! $\endgroup$ – user40884 Jun 15 '19 at 8:48

For the instantaneous forward, please see the last page of this note: T-Forward Measure by Fabrice Douglas Rouah (http://www.frouah.com/finance%20notes/The%20T-Forward%20Measure.pdf).

For the simple forward, you know the relationship between the price of the zero coupon and the simple forward:

$ \frac{P \left(t,T_{n}\right)}{P \left(t,T_{n+1}\right) }=1+\tau F \left(t,T_n \right)$

Which you can rearrange to get:

$F \left(t,T_n \right)P \left(t,T_{n+1}\right) = \frac{1}{\tau} \left(P \left(t,T_{n}\right)-P \left(t,T_{n+1}\right)\right)$

So the left hand side is the price of an asset as it is a difference of the price of two bonds divided by the time fraction (accrual factor). And if you use $P \left(t,T_{n+1} \right)$ as a numeraire, then you get from the general valuation formula:

$ \frac{F \left(t,T_n \right)P \left(t,T_{n+1}\right)}{P \left(t,T_{n+1}\right)}=E^{T} \left[ \left. \frac{F \left(S,T_n \right)P \left(S,T_{n+1}\right)}{P \left(S,T_{n+1}\right)} \right| \mathcal{F}_t\right] $

And simple algebra gives:

$F \left(t,T_n \right)=E^{T} \left[ \left. F \left(S,T_n \right)\right| \mathcal{F}_t\right] $

| improve this answer | |

Your Answer

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