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}.

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One comment

  1. Pingback: Bayes rule and forward measure « 01law's Blog


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