So I read an other post about this :

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.



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: $$ \begin{align} \mathbb{E}^{\mathbb{Q}^{T_2}}_t \left[ \frac{ P_{TT_1} }{ P_{TT_2} } \right] & = \frac{\mathbb{E}^{\mathbb{Q}^{T_1}}_t \left[ \frac{d \mathbb{Q}^{T_2}}{d\mathbb{Q}^{T_1}}(T) \cdot \frac{ P_{TT_1} }{ P_{TT_2} } \right]}{ \frac{d \mathbb{Q}^{T_2}}{d\mathbb{Q}^{T_1}}(t) } \\ & = \frac{\mathbb{E}^{\mathbb{Q}^{T_1}}_t \left[ \frac{ P_{TT_2} P_{0T_1} }{ P_{TT_1} P_{0T_2} } \cdot \frac{ P_{TT_1} }{ P_{TT_2} } \right]}{ \frac{ P_{tT_2} P_{0T_1} }{ P_{tT_1} P_{0T_2} } } \\ & = \frac{ P_{tT_1}}{P_{tT_2} } \end{align} $$

| improve this answer | |

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.

And once you have established that it is the price of a traded asset, then you can write its price using the valuation formula, which in way holds for all trades asset. It essentially says that the price of an asset, say V, divided by a numeraire asset, say B, will be a martingale under some probability measure.

$\frac{V_t}{B_t}=E^Q\left[\left. \frac{V_S}{B_S} \right|\mathcal{F}_t\right]$

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 $B_t$).

| improve this answer | |
  • $\begingroup$ the valuation formula is in fact change of numéraire theorem right ? $\endgroup$ – Ramzy Nov 24 '19 at 20:55

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.