# Bayes rule and forward measure

Bayes rule is an important theory in statistics. In this following, we discuss one of the presentation, which can be applied to interest rate model, see [MR05].

Proposition 1 (Bayes rule) Let ${\mathbb{ P}>>\mathbb{Q}}$ be two probability measures on the probability space ${\{\Omega, \mathcal{F}\}}$, satisfying ${ \frac{d \mathbb{Q}}{d \mathbb{P}} := \Lambda_T. }$ Let ${\mathcal{F}_t}$ be a filtration with ${\mathcal{F}_T = \mathcal{F}}$. Then,

$\displaystyle \mathbb{E}_{\mathbb{Q}} [X |\mathcal{F}_t] = \frac{\mathbb{E}_{\mathbb{P}}[\Lambda_T X|\mathcal{F}_t]}{\mathbb{E}_{\mathbb{P}}[\Lambda_T |\mathcal{F}_t]}, \quad \hbox{ for all } \mathcal{F}_T \hbox{ measurable random variable } X$

Proof: Note that ${\mathbb{P}}$ and ${\mathbb{Q}}$ are also probability measures on ${(\Omega, \mathcal{F}_t)}$, we denote the restricted probabilities by ${\mathbb{P}_t}$ and ${\mathbb{Q}_t}$ respectively. Then, ${\Lambda_t = \frac{d \mathbb{Q}_t}{d \mathbb{P}_t}}$ are ${\mathbb{P}}$ martingale, i.e. ${\Lambda_t = \mathbb{E}_{\mathbb{P}} [ \Lambda_T|\mathcal{F}_t]}$, see Lemma 1 of here.

It’s enough to show,

$\displaystyle \mathbb{E}_{\mathbb{P}} [ {\bf 1}_A \Lambda_t \mathbb{E}_{\mathbb{Q}}[X|\mathcal{F}_t]] = \mathbb{E}_{\mathbb{P}}[{\bf 1}_A \mathbb{E}_{\mathbb{P}}[\Lambda_T X|\mathcal{F}_t]] \hbox{ for all } A\in \mathcal{F}_t.$

It follows by

$\displaystyle \begin{array}{ll} LHS & = \mathbb{E}_{\mathbb{P}_t} [ {\bf 1}_A \Lambda_t \mathbb{E}_{\mathbb{Q}}[X|\mathcal{F}_t]] = \mathbb{E}_{\mathbb{Q}_t} [ {\bf 1}_A \mathbb{E}_{\mathbb{Q}}[X|\mathcal{F}_t]] \\ & = \mathbb{E}_{\mathbb{Q}} [ {\bf 1}_A \mathbb{E}_{\mathbb{Q}}[X|\mathcal{F}_t]] = \mathbb{E}_{\mathbb{Q}} [ {\bf 1}_A X] = \mathbb{E}_{\mathbb{P}} [ {\bf 1}_A X \Lambda_T] = RHS. \end{array}$

$\Box$

In particular, with ${\mathcal{F}_t = \mathcal{F}_0}$ in the above proposition, we have

$\displaystyle \mathbb{E}_{\mathbb{Q}} [X] = \mathbb{E}_{\mathbb{P}}[\Lambda_T X], \quad \hbox{ for all } \mathcal{F}_T \hbox{ measurable random variable } X$

In this below, we consider an application of Bayes rule to the interest rate theory. Suppose ${r_t}$ is the short rate, and ${\mathbb{P}}$ is the unique equivalent martingale measure. Then, the price at time ${t}$ of the derivative with maturity ${T}$ and payoff ${\mathcal{F}_T}$ measurable ${F}$ is written by

$\displaystyle \mathbb{E}_{\mathbb{P}} [ e^{-\int_t^T r_s ds} F | \mathcal{F}_t]. \ \ \ \ \ (1)$

$\displaystyle$

In particular, the price of zero bond is, with ${F= 1}$,

$\displaystyle P(t,T) = \mathbb{E}_{\mathbb{P}} [ e^{-\int_t^T r_s ds} | \mathcal{F}_t].$

The forward measure is the probability ${\mathbb{\tilde P}}$ defined as

$\displaystyle \frac{d \mathbb{\tilde P}}{d \mathbb{P}} = \frac{1}{P(0,T)} e^{-\int_0^T r_s ds} := \Lambda_T. \ \ \ \ \ (2)$

Then, by the Bayes rule, the derivative price of (1) can be written w.r.t. forward measure:

$\displaystyle \mathbb{E}_{\mathbb{P}} [ e^{-\int_t^T r_s ds} F | \mathcal{F}_t] = P(t,T) \mathbb{E}_{\mathbb{\tilde P}} [ F | \mathcal{F}_t], \ 0\le t\le T. \ \ \ \ \ (3)$

Proposition 2 For all ${0\le T \le S}$, the process

$\displaystyle t\rightarrow \frac{P(t,S)}{P(t,T)}, \ 0\le t \le T$

is an ${\mathcal{F}_t}$-martingale under ${\mathbb{\tilde P}}$.

Proof: It’s enough to show

$\displaystyle \mathbb{E}_{\mathbb{\tilde P}} \Big[ \frac{P(t,S)}{P(t,T)} \Big| \mathcal{F}_u \Big] = \frac{P(u,S)}{P(u,T)}, \quad \forall u

which is indeed a consequence of (3) with ${F = \frac{P(t,S)}{P(t,T)}}$. $\Box$