# Girsanov theorem and Bayes rule

We will review Girsanov theorem, and related change of measure.

Given two probability measures ${\mathbb{P}}$ and ${\mathbb{Q}}$ on ${(\Omega,\mathcal{F})}$ and a filtration ${\{\mathcal{F}_t\}}$, they are also probability measures on ${(\Omega, \mathcal{F}_t)}$, and we will use notation ${\mathbb{P}_t}$ and ${\mathbb{Q}_t}$ when the measures are restriced by ${\mathcal{F}_t}$. Suppose ${\mathbb{Q} <<\mathbb{P}}$, then ${\mathbb{Q}_t <<\mathbb{P}_t}$, and write

$\displaystyle \Lambda_t = \frac{d \mathbb{Q}}{d \mathbb{P}} \Big|_t = \frac{d \mathbb{Q}_t}{d \mathbb{P}_t}.$

Lemma 1 Suppose ${\mathbb{Q} <<\mathbb{P}}$, then ${\{\Lambda_t, t\ge 0 \}}$ is a ${\mathbb{P}}$-martingale.

Proof: For all ${0\le s \le t}$, ${A\in \mathcal{F}_s}$,

$\displaystyle \begin{array}{ll} \mathbb{E}_P ({\bf 1}_A \mathbb{E}_P (\Lambda(t)|\mathcal{F}_s)) &= \mathbb{E}_P ({\bf 1}_A \Lambda(t)) = \mathbb{E}_{P_t} ({\bf 1}_A \Lambda(t)) \\ &= \mathbb{E}_{Q_t}({\bf 1}_A) = \mathbb{E}_{Q_s} ({\bf 1}_A) = \mathbb{E}_{P_s} ({\bf 1}_A \Lambda(s)) = \mathbb{E}_P ({\bf 1}_A \Lambda(s)). \end{array}$

$\Box$

Proposition 2 (Bayes rule) Let ${\mathbb{ P} \sim \mathbb{Q}}$ be equivalent two probability measures on the probability space ${\{\Omega, \mathcal{F}\}}$. Let ${\mathcal{F}_t}$ be a filtration with ${\mathcal{F}_T = \mathcal{F}}$. Then,

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

Proof: By Lemma 1, ${\Lambda_t = \mathbb{E}_{\mathbb{P}} [ \Lambda_T|\mathcal{F}_t]}$ is a ${\mathbb{P}}$ martingale. 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

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

$\Box$

Lemma 3 Under the same assumptions of Proposition 2, ${M_t}$ is a ${\mathbb{Q}}$-martingale if and only if ${M_t \Lambda_t}$ is ${\mathbb{P}}$-martingale.

Proof: By Proposition 2 and Lemma 1, we have the following equivalent identities: For any ${t\ge s}$,

1. ${\mathbb{E}^\mathbb{Q}[M_t |\mathcal{F}_s] = M_s}$;
2. ${\mathbb{E}^\mathbb{P}[M_t \Lambda_T|\mathcal{F}_s] = M_s \Lambda_s}$;
3. ${\mathbb{E}^\mathbb{P}[ \mathbb{E}^\mathbb{P} [M_t \Lambda_t|\mathcal{F}_t] |\mathcal{F}_s] = M_s \Lambda_s}$;
4. ${\mathbb{E}^\mathbb{P}[M_t \Lambda_t|\mathcal{F}_s] = M_s \Lambda_s}$;

Hence, we conclude the result. $\Box$

Indeed, Lemma 1 is a sepcial case of Lemma 3 with ${M_t \equiv 1}$.

Example 1 Let ${(W^1, W^2)}$ be two independent standard Brownian motion on ${(\Omega, \mathcal{F}, \mathbb{P}, \mathcal{F}_t)}$. Given

$\displaystyle d \bar W_t = \theta_t dt + d W^1_t, \ \bar W_0 = 0.$

Find all ${\mathbb{P}}$ equivalent probability measure ${\mathbb{Q}}$ under which ${\{\bar W_t: 0\le t\le T\}}$ is a Brownian motion.

Since, ${\langle \bar W\rangle_t = t}$, it’s enough to find all density function ${\Lambda_t = \frac{d \mathbb{Q}_t}{d \mathbb{P}_t}}$ satisfying

1. ${\Lambda_t}$ is a ${\mathbb{P}}$ martingale with ${\Lambda_0 = 1}$;
2. ${\Lambda_t \bar W_t}$ is a ${\mathbb{P}}$ martingale with ${\Lambda_0 = 1}$;

Since ${\Lambda_t}$ is a ${\mathbb{P}}$ martingale, it must takes form of

$\displaystyle d \Lambda_t = \lambda_t^1 dW_t^1 + \lambda_t^2 dW_t^2, \Lambda_0 = 1.$

By Ito’s product formula, we also have

$\displaystyle d \Lambda_t \bar W_t = d M_t + \Lambda_t \theta_t dt + \lambda_t^1 dt$

for some ${\mathbb{P}}$ martingale ${M_t}$. Thus, ${\lambda_t^1 = -\Lambda_t \theta_t}$, and ${\lambda_t^2}$ could be any. As a conclusion, ${\mathbb{Q}}$ is a ${\mathbb{P}}$ equivalent probability measure under which ${\bar W}$ is a Brownian motion, if and only if, ${\Lambda_t = \frac{d \mathbb{Q}_t}{d \mathbb{P}_t}}$ is a solution of

$\displaystyle d \Lambda_t = \Lambda_t (- \theta_t dW_t^1 - \lambda_t^2 dW_t^2), \ \Lambda_0 = 1,$

for some ${\lambda^2}$ satisfying integrability condition ${\mathbb{E} [ \int_0^T(\lambda^2_s)^2 ds ] <\infty.}$

Theorem 4 (Girsanov) Let ${W}$ be a standard Brownian motion w.r.t ${(\Omega, \mathcal{F}, \mathbb{P}, \mathcal{F}_t)}$. ${e^Y}$ is exponential martingale with ${Y}$ being of the form

$\displaystyle Y(t) = \int_0^t F(s) d W(s) - \frac 1 2 \int_0^t F^2(s) ds.$

Then, a new process ${\tilde W}$ defined by

$\displaystyle \tilde W(t) = W(t) - \int_0^t F(s) ds$

is a ${Q}$-Brownian motion.

Proof: First, we use Lemma 3 to show that ${\tilde W}$ is a centered ${\mathbb{Q}}$-martingale. Then, by the fact ${\langle \tilde W\rangle (t) = t}$, we can complete the proof using Levy’s characterization. $\Box$

The following partial extension of Girsanov’s theorem will be of use in applications to finance. First recall that, if Levy-type stochastic integral ${Y}$ be given by

$\displaystyle d Y(t) = G(t) dt + F(t) dW(t) + H(t,x) \tilde N(dt, dx) + K(t,x) N(dt, dx). \ \ \ \ \ (1)$

then ${e^Y}$ is a local martingale if and only if for ${t\ge 0}$,

$\displaystyle \begin{array}{ll} e^{Y(t)} & = 1 + \int_0^t e^{Y(s-)} F(s) d W(s) + \int_0^t \int_{|x|<1} e^{Y(s-)} (e^{H(s,x)} -1) \tilde N(ds, dx) \\ & \hspace{2in} + \int_0^t \int_{|x|\ge 1} e^{Y(s-)} (e^{K(s,x)} -1) \tilde N(ds, dx). \end{array} \ \ \ \ \ (2)$

Proposition 5 Let ${M}$ be a ${\mathbb{P}}$ local martingale of the form

$\displaystyle M(t) = \int_0^t \int_{|x|<1} L(x,s) \tilde N(ds,dx),$

where ${L\in \mathcal{P}_2(t,B_1(0)\setminus\{0\})}$. Let ${e^Y}$ be an exponential ${\mathbb{P}}$ martingale of (2). Then, the new process ${\tilde M}$ given by

$\displaystyle \tilde M(t) = M(t) - \int_0^t\int_{|x|<1} L(x,s) (e^{H(s,x)} - 1) \nu(dx) ds$

is a ${\mathbb{Q}}$ local martingale, provided that the integral exists.

Proof: Use Lemma 3. $\Box$

A sufficient condition for the existence of the integral in ${\tilde M}$ is that ${\int_0^t \int_{|x|<1} |e^{H(s,x)} - 1|^2 \nu(dx) ds <\infty}$.