Suponemos que $\mathbb{Q}$ es una medida de avance. $$\mathbb{E^Q}\left[\frac{P(t,S)}{P(t,T)}|\mathcal{F}_s\right]=\mathbb{E^P}\left[\frac{P(t,S)}{P(t,T)}\frac{e^{-\int_{s}^{T}r_u\,du}}{P(s,T)}\,|\,\mathcal{F}_s\right]$$ $$\hspace{5cm}=\frac{1}{P(s,T)}\mathbb{E^P}\left[\frac{P(t,S)}{P(t,T)}{e^{-\int_{s}^{T}r_u\,du}}\,|\,\mathcal{F}_s\right]$$ $$\hspace{6.9cm}=\frac{1}{P(s,T)}\mathbb{E^P}\left[\mathbb{E^P}\left[\frac{P(t,S)}{P(t,T)}{e^{-\int_{s}^{T}r_u\,du}}\,|\,\mathcal{F}_t\right]|\mathcal{F}_s\right]$$ tenemos $$\hspace{0.3cm}\mathbb{E^Q}\left[\frac{P(t,S)}{P(t,T)}|\mathcal{F}_s\right]=\frac{1}{P(s,T)}\mathbb{E^P}\left[{e^{-\int_{s}^{t}r_u\,du}}\frac{P(t,S)}{P(t,T)}\mathbb{E^P}\left[e^{-\int_{t}^{T}r_u\,du\,}\,\,|\,\mathcal{F}_t\right]|\mathcal{F}_s\right]$$ $$\hspace{1.5cm}=\frac{1}{P(s,T)}\mathbb{E^P}\left[{e^{-\int_{s}^{t}r_u\,du}}\frac{P(t,S)}{P(t,T)}P(t,T)|\mathcal{F}_s\right]$$
$$=\frac{1}{P(s,T)}\mathbb{E^P}\left[{e^{-\int_{s}^{t}r_u\,du}}P(t,S)|\mathcal{F}_s\right]$$ entonces $$\mathbb{E^Q}\left[\frac{P(t,S)}{P(t,T)}|\mathcal{F}_s\right]=\frac{1}{P(s,T)}{e^{\int_{0}^{s}r_u\,du}}\,\,\mathbb{E^P}\left[{e^{-\int_{0}^{t}r_u\,du}}P(t,S)|\mathcal{F}_s\right]$$ conocemos el proceso de descuento del precio de los bonos $\{e^{-\int_{0}^{t}r_u\,du}P(t,S)\}$ es una martingala bajo $\mathbb{P}$ por lo que tenemos $$\hspace{1cm}\mathbb{E^Q}\left[\frac{P(t,S)}{P(t,T)}|\mathcal{F}_s\right]=\frac{1}{P(s,T)}{e^{\int_{0}^{s}r_u\,du}}{e^{-\int_{0}^{s}r_u\,du}}P(s,S)$$ $$=\frac{P(s,S)}{P(s,T)}$$
$$$$ EDITAR: Alternativa, \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*}
0 votos
Tal vez ver referencia (p. 271, 275, 336 o algo más) en mi pregunta
0 votos
Sí, creo que el capítulo 11 en la referencia y luego aquí es el solución