Why Forward Rates Are Martingales Under the T-Forward Measure

# How to prove martingality of forward rate under T-forward measure


# How to prove martingality of forward rate under T-forward measure












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.

## Answer by Magic is in the chain (score 6, accepted)

https://quant.stackexchange.com/a/46104

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] $

## Answer by Prabhnoor Duggal (score 5)

https://quant.stackexchange.com/a/55529

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.

## Answer by Gordon (score 4)

https://quant.stackexchange.com/a/55553

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.