# Forward Rates are martingal under Forwar Measure detailled proof

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

But I can't see how to get from there to there :

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

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

If anyone could explain in depth this passage I would be really glad.

Thanks

The measure $$\mathbb{Q}$$ is associated to the money market account $$t \mapsto \beta_t = \exp \int_{0}^{t} r_s d s$$. The measure $$\mathbb{Q}^T$$ is associated to the zero coupon bond $$t \mapsto P_{tT}$$ where $$P_{tT}:=\mathbb{E}^{\mathbb{Q}}_t \left[ \frac{\beta_t}{\beta_T} \right]$$. We know that the Radon-Nikodym derivative between $$\mathbb{Q}^{T_1}$$ and $$\mathbb{Q}^{T_2}$$ is given by $$t \mapsto \frac{d \mathbb{Q}^{T_2}}{d\mathbb{Q}^{T_1}}(t)=\frac{P_{tT_2}}{P_{0T_2}} \cdot \frac{ P_{0T_1} }{ P_{tT_1} }$$ We know by the Bayes theorem for a payoff function $$X$$ observed at time $$T$$ that $$\mathbb{E}^{\mathbb{Q}^{T_2}}_{t} \left[ X_T \right] = \frac{\mathbb{E}^{\mathbb{Q}^{T_1}}_{t} \left[ X_T \cdot \frac{d \mathbb{Q}^{T_2}}{d\mathbb{Q}^{T_1}}(T) \right]}{\mathbb{E}^{\mathbb{Q}^{T_1}}_{t} \left[ \frac{d \mathbb{Q}^{T_2}}{d\mathbb{Q}^{T_1}}(T) \right]}$$ Equivalently, $$\mathbb{E}^{\mathbb{Q}^{T_2}}_{t} \left[ X_T \right] = \frac{\mathbb{E}^{\mathbb{Q}^{T_1}}_{t} \left[ X_T \cdot \frac{d \mathbb{Q}^{T_2}}{d\mathbb{Q}^{T_1}}(T) \right]}{\frac{d \mathbb{Q}^{T_2}}{d\mathbb{Q}^{T_1}}(t)}$$ Now that our tools are ready, let us attack. To prove that the process $$F(t;T_1,T_2)=\frac{1}{\tau} \left[ \frac{ P_{tT_1} }{ P_{tT_2} } -1 \right]$$ is a $$\mathbb{Q}^{T_2}$$ martingale, it is equivalent to prove the same statement for the ratio of bonds $$X_T :=P_{TT_1} / P_{T T_2}$$. Proceed as follows: \begin{align} \mathbb{E}^{\mathbb{Q}^{T_2}}_t \left[ \frac{ P_{TT_1} }{ P_{TT_2} } \right] & = \frac{\mathbb{E}^{\mathbb{Q}^{T_1}}_t \left[ \frac{d \mathbb{Q}^{T_2}}{d\mathbb{Q}^{T_1}}(T) \cdot \frac{ P_{TT_1} }{ P_{TT_2} } \right]}{ \frac{d \mathbb{Q}^{T_2}}{d\mathbb{Q}^{T_1}}(t) } \\ & = \frac{\mathbb{E}^{\mathbb{Q}^{T_1}}_t \left[ \frac{ P_{TT_2} P_{0T_1} }{ P_{TT_1} P_{0T_2} } \cdot \frac{ P_{TT_1} }{ P_{TT_2} } \right]}{ \frac{ P_{tT_2} P_{0T_1} }{ P_{tT_1} P_{0T_2} } } \\ & = \frac{ P_{tT_1}}{P_{tT_2} } \end{align}
The first equation is the result of the effort to show that the product of the forward and the relevant zero coupon, $$F \left(t,T_n \right)P \left(t,T_{n+1}\right)$$, can be treated as a traded asset.
$$\frac{V_t}{B_t}=E^Q\left[\left. \frac{V_S}{B_S} \right|\mathcal{F}_t\right]$$
So in the second equation you are just plugging in the F times P, i.e. $$V_t=F \left(t,T_n \right)P \left(t,T_{n+1}\right)$$, for the traded asset V, and choosing $$P \left(t,T_{n+1}\right)$$ as the numeraire (in place of $$B_t$$).