Is $\frac{P(t,S)}{P(t,T)}$ martingale?

Assume $r_t$ follow the CIR process and $P(t,T)=E[exp(-\int_{t}^{T}r_s ds)|F_t]$.I am going to show $\frac{P(t,S)}{P(t,T)}$ ($S<T$) is an $F_t$-martingale under Forward Measure but So confused! Do I need to solve C.I.R process? Should I use the definition of martingale? please guide me! so thanks.

• Maybe see reference (p. 271, 275, 336 or something else) in my question – BCLC Aug 6 '15 at 9:05
• Yeah I think ch 11 in the reference and then here is the solution – BCLC Aug 6 '15 at 9:14

By definition of the $T$-forward measure $P_T$, the process $\Big\{\frac{P(t,S)}{P(t,T)} \mid t\geq 0\Big\}$ is a martingale under the measure $P_T$, without assuming any specific models of the short rate $r_t$. That is, this martingale property is model independent.

However, as a good exercise, you can also do the following:

1. Given the CIR interest rate model under the risk-neutral measure $P$, compute the bond prices.

2. Find the Radon-Nykodim derivative of the $T$-forward measure with respect to the risk-neutral measure, that is, $\frac{dP_T}{dP}\big|_t$, for $0 \leq t \leq T$.

3. Find the bond price formula or SDE under the $T$-forward measure.

4. Show that the process $\Big\{\frac{P(t,S)}{P(t,T)} \mid t\geq 0\Big\}$ is a martingale under the $T$-forward measure.

• precise.But this needs to be clarified – user16891 Aug 5 '15 at 9:42

We assume $\mathbb{Q}$ is forward measure. $$\mathbb{E^Q}\left[\frac{P(t,S)}{P(t,T)}|\mathcal{F}_s\right]=\mathbb{E^P}\left[\frac{P(t,S)}{P(t,T)}\frac{e^{-\int_{s}^{T}r_u\,du}}{P(s,T)}\,|\,\mathcal{F}_s\right]$$ $$\hspace{5cm}=\frac{1}{P(s,T)}\mathbb{E^P}\left[\frac{P(t,S)}{P(t,T)}{e^{-\int_{s}^{T}r_u\,du}}\,|\,\mathcal{F}_s\right]$$ $$\hspace{6.9cm}=\frac{1}{P(s,T)}\mathbb{E^P}\left[\mathbb{E^P}\left[\frac{P(t,S)}{P(t,T)}{e^{-\int_{s}^{T}r_u\,du}}\,|\,\mathcal{F}_t\right]|\mathcal{F}_s\right]$$ we have $$\hspace{0.3cm}\mathbb{E^Q}\left[\frac{P(t,S)}{P(t,T)}|\mathcal{F}_s\right]=\frac{1}{P(s,T)}\mathbb{E^P}\left[{e^{-\int_{s}^{t}r_u\,du}}\frac{P(t,S)}{P(t,T)}\mathbb{E^P}\left[e^{-\int_{t}^{T}r_u\,du\,}\,\,|\,\mathcal{F}_t\right]|\mathcal{F}_s\right]$$ $$\hspace{1.5cm}=\frac{1}{P(s,T)}\mathbb{E^P}\left[{e^{-\int_{s}^{t}r_u\,du}}\frac{P(t,S)}{P(t,T)}P(t,T)|\mathcal{F}_s\right]$$

$$=\frac{1}{P(s,T)}\mathbb{E^P}\left[{e^{-\int_{s}^{t}r_u\,du}}P(t,S)|\mathcal{F}_s\right]$$ then $$\mathbb{E^Q}\left[\frac{P(t,S)}{P(t,T)}|\mathcal{F}_s\right]=\frac{1}{P(s,T)}{e^{\int_{0}^{s}r_u\,du}}\,\,\mathbb{E^P}\left[{e^{-\int_{0}^{t}r_u\,du}}P(t,S)|\mathcal{F}_s\right]$$ we know the discounted bond price process $\{e^{-\int_{0}^{t}r_u\,du}P(t,S)\}$ is a martingale under $\mathbb{P}$, thus we have $$\hspace{1cm}\mathbb{E^Q}\left[\frac{P(t,S)}{P(t,T)}|\mathcal{F}_s\right]=\frac{1}{P(s,T)}{e^{\int_{0}^{s}r_u\,du}}{e^{-\int_{0}^{s}r_u\,du}}P(s,S)$$ $$=\frac{P(s,S)}{P(s,T)}$$

 EDIT: Alternatively, \begin{align*} \mathbb{E^Q}\left[\frac{P(t,S)}{P(t,T)}\mid \mathcal{F}_s\right] &= \frac{1}{P(s,T)}\mathbb{E^P}\left[{e^{-\int_{s}^{t}r_u\,du}}P(t,S)\mid\mathcal{F}_s\right]\\ &=\frac{1}{P(s,T)}\mathbb{E^P}\left[{e^{-\int_{s}^{t}r_u\,du}}\mathbb{E^P}\Big(e^{-\int_t^Sr_u\,du} \mid \mathcal{F}_t \Big)\mid\mathcal{F}_s\right]\\ &=\frac{1}{P(s,T)}\mathbb{E^P}\left[\mathbb{E^P}\Big(e^{-\int_s^Sr_u\,du} \mid \mathcal{F}_t \Big)\mid\mathcal{F}_s\right]\\ &= \frac{1}{P(s,T)}\mathbb{E^P}\left[e^{-\int_s^Sr_u\,du}\mid\mathcal{F}_s\right]\\ &= \frac{P(s,S)}{P(s,T)}. \end{align*}

• Please check the typos. – Gordon Aug 5 '15 at 13:38
• Hi Gordon , I check my solution, I can not see typos unfortunately. please edit it. – user16891 Aug 5 '15 at 15:26
• In most of places, $\mathcal{F}_t$ should be $\mathcal{F}_s$, for $0\leq s \leq t \leq S<T$. – Gordon Aug 5 '15 at 15:28
• You do not really need the fact that $\big\{e^{-\int_{0}^{t}r_u\,du}P(t,S) \mid 0 \leq t \leq S\big\}$ is a martingale under $\mathbb{P}$. – Gordon Aug 5 '15 at 16:07
• I think, this issue is discussed in the Bjork's book – user16891 Aug 5 '15 at 16:22