martingale - 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

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:

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