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
    Commented Jun 14, 2019 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. $\endgroup$ Commented Jun 14, 2019 at 20:47
  • $\begingroup$ My lecture notes tated forward rate, but I guess it is about the instantaneous forward rate! $\endgroup$
    – user40884
    Commented Jun 15, 2019 at 8:48

3 Answers 3


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] $


By definition, $$Fo(t,T)=E^T[S_T|F_t]$$ Note that expectation is taken under $T$-forward measure. Now, evaluating at $s<T$: $$E^T[Fo(t,T)|F_s] = E^T[E^T[S_T|F_t]|F_s] = E^T[S_T|F_s] = Fo(s,T)$$ (using the tower property of expectations). Hence Forwards rate is a martingale under the T-forward measure.

  • $\begingroup$ What is $S_T$ here? $\endgroup$ Commented Jun 8, 2021 at 18:22
  • $\begingroup$ @ReneRecktenwald it refers to the Equity price at time 'T' $\endgroup$ Commented Jun 21, 2021 at 21:58

The answer by @Prabhnoor Duggal is correct. Here, I would like to further expand to make it more streamlined (see also Section 2.5 of the book Interest Rate Models - Theory and Practice). Let $Q$ and $Q^T$ be the risk-neutral and the $T$-forward respective probability measures. Then, for $0\le t \le T$, \begin{align*} \frac{dQ}{dQ^T}\big|_{[t, T]} = \frac{B_TP(t, T)}{B_t}. \end{align*} Moreover, \begin{align*} f(t, T) &= \frac{-\frac{\partial }{\partial T}P(t, T)}{P(t, T)}\\ &=\frac{E_Q\left(e^{-\int_t^Tr_s ds}\, r_T\,|\, \mathscr{F}_t \right)}{P(t, T)}\\ &=\frac{E_{Q^T}\left(\frac{dQ}{dQ^T}\big|_{[t, T]}\,e^{-\int_t^Tr_s ds}\, r_T\,|\, \mathscr{F}_t \right)}{P(t, T)}\\ &=E_{Q^T}(r_T \,|\,\mathscr{F}_t). \end{align*} Therefore, $\{f(t, T), \, 0\le t \le T\}$ is a martingale under the $T$-forward probability measure.


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