Part 6 - Doob's h-transform and Girsanov's theorem

Introduction

In this part we’ll introduce Doob’s $h$ -transform and Girsanov’s theorem.

Recall our definition of the SDE,

dx(t) = \underbrace{f(x(t), t) \, dt}_{\text{drift}} + \underbrace{L(x(t), t) \, d\beta(t)}_{\text{diffusion}},

By the end of this, we’ll have a tool in our toolbox to transform this SDE into a new SDE with the following form,

\begin{equation} dx(t) = f(x(t), t) \, dt + L^2(x(t), t) \nabla \log h(x(t), t) \, dt + L(x(t), t) \, d\beta(t), \end{equation}

and we’ll already propose that,

dx(t) = \frac{v - x(t)}{T - t} dt + d\beta(t), t \leq T,

is already a special case of Equation (1). For clarity, we read the above SDE as, as we move towards time $T$ , we’ll get closer to the point $v$ .

So, essentially, we’re taking a Brownian motion and forcing it to a point $v$ at time $T$ .

Figure 1: Brownian motion forced to a point

Recap: Density and Bayes’ Rule

Let’s firstly recap what we know about densities and Bayes’ rule.

If we have a random value $x$ , we say that $x$ has a density $p_x$ if,

P(x \in A) = \int_A p_x(x) \, dx,

or in a sloppier notation,

p_x(x) = \frac{P(x \in A)}{dx}.

We say $(x,y)$ has a joint density $p_{x,y}$ if,

P(x \in A, y \in B) = \int_A \int_B p_{x,y}(x,y) \, dy \, dx,

Further, Bayes’ rule states that,

P(A \mid B) = \frac{P(A \cap B)}{P(B)}.

We say that $(x,y)$ has a conditional density $p_{y \mid x}$ if,

P(y \in A \mid x \in B) = \frac{P_{x, y}(x, y)}{p_x(x)} = \frac{p_{x,y}(x,y)}{\int p_{x, y}(x, y) dy},

where we in the second step used marginalization to get the denominator.

One can also write Bayes’ rule in terms of conditionals,

P(A \mid B) = \frac{P(A) P(B \mid A)}{P(B)},

or in terms of densities,

P(x \in A \mid y \in B) = \frac{p_x(x) p_{y \mid x}(y \mid x)}{p_y(y)}.

Transition Density

Now, let’s try to formulate what we’ll call transition density.

If $x(t)$ , for $t \geq 0$ is a Markov process ¹,

p(x(t) \mid x(s)) = \frac{P(x(t) \in dy \mid x(s) = x)}{dy} \biggr\rvert_{\substack{y = x(t) \newline x = x(s)}}

or in words, the conditional density of $x(t)$ given $x(s) = x$ .

Now, think of this in the general case, since it is Markov, we now that,

\begin{align*} & P(x(t_0), x(t_1), \ldots, x(t_n)), \text{ joint of } n \text{ random variables} \newline & = P(x(t_0)) P(x(t_1) \mid x(t_0)) \ldots P(x(t_n) \mid x(t_{n-1})). \end{align*}

Champan-Kolmogorov

We’ll make use of a equation called Chapman-Kolmogorov equation ² which states,

P(x(t) \mid x(s)) = \int P(x(t) \mid x(u)) P(x(u) \mid x(s)) \, dx(u), \quad s < u < t,

We will not prove or further discuss this equation, but it is important to know that it exists.

Example: Transition Density of a Brownian Motion

Let’s now try to apply what we’ve learned so far to a Brownian motion.

Tip (Example: Transition density of a Brownian motion)

So, we have,

P(\beta(t) \mid \beta(s)) = \frac{1}{\sqrt{2 \pi (t - s)}} \exp\left(-\frac{(\beta(t) - \beta(s))^2}{2(t - s)}\right),

and if we visualize this.

Figure 3: Transition density of a Brownian motion

Doob’s $h$ -transform (for transition densities)

Let’s first start with an example and see how we can solve it.

Tip (Example: Doob’s

h

-transform on transition densities)

What is the transition density of $\beta(\cdot)$ (which one reads as, at any arbitrary time), given that $\beta(T) = v$ ? We’ll done this transition density with $p^\star$ , so we have,

p^\star(\beta(t) \mid \beta(s)) = \frac{P(\beta(t) \in dy \mid \beta(s) = x, \beta(T) = v)}{dy} \biggr\rvert_{y = \beta(t)}

Which we can write as,

\begin{equation} = \frac{P(\beta(s)) P(\beta(t) \mid \beta(s)) h(\beta(t), t)}{\int P(\beta(s)) P(\beta(t) \mid \beta(s)) h(\beta(t), t) d \beta(t)}, \end{equation}

where

h(s, x) = \frac{P(\beta(T) \in dy \mid \beta(s) = x)}{dy} \biggr\rvert_{y = v}

Figure 4: Transition density of a Brownian “bridge”

Now, you may already see that, if $h(s, x)$ is Gaussian, then maybe $\nabla \log h(s, x) \implies \frac{v - x}{T - s}$ .

By the Chapman-Kolmogorov equation, we can write,

\frac{P(\beta(t) \mid \beta(s)) h(\beta(t), t)}{h(\beta(s), s)},

which is precisely what we’ll call Doob’s $h$ -transform.

Infinitesimal generator

Imagine we have an arbitrary function $g$ that maps,

g \mapsto f \cdot g^\prime + \frac{1}{2} L^2 \cdot g^{\prime \prime},

or in full notation,

g(x) \mapsto f(x) \cdot g^{\prime}(x) + \frac{1}{2} L^2(x) \cdot g^{\prime \prime}(x),

We’ll call this mapping/operator $Ag$ .

The operator $A$ is called “infinitesimal generator” of the process $x(t), t \geq 0$ with drift $f$ and diffusion $L$ .

Let’s write down Itô’s formula with $A$ .a

Informally,

\frac{\left[ \mathbb{E}\left[ g(x) \mid x(s) = x \right] - g(x) \right] }{t - s} \xrightarrow{t \to s} (Ag)(x),

note that $\mathbb{E}[g(x) \mid x(s) = x] = g(x)$

Danger (Remark)

Note that $h(\beta(s), s)$ is a Martingale ³, because,

h(\beta(s), s) = \int P(\beta(t) \mid \beta(s)) h(\beta(t), t) d \beta(t),

or in other words, the drift term in Itô’s formula disappears. Hence, $(Ag)(x) + \frac{\partial}{\partial s} h(s, x) = 0$ .

So, $h$ solves the backward Kolmogorov equation!

Recall that Equation (2) holds for $x(t)$ in,

dx(t) = \underbrace{f(x(t), t) \, dt}_{\text{drift}} + \underbrace{L(x(t), t) \, d\beta(t)}_{\text{diffusion}},

where $\beta(s)$ is replaced by $x(s)$ .

Time derivative of $A$

Idea: Compute the time derivative of the conditional expectation,

\mathbb{E}[g(x(t)) \mid x(s) = x, x(T) = v],

We’ll call the new $A$ operator $A^\star$ .

\begin{align*} (\frac{\partial}{\partial s} g + A^\star g)(x) & = \lim_{t \to s} \frac{\mathbb{E}[g(x(t)) \mid x(s) = x, x(T) = v] - g(x)}{t - s} \newline & = \lim_{t \to s} \frac{\mathbb{E}[g(x(t)) - g(x(s)) \mid x(s) = x, x(T) = v]}{t - s} \newline & = \lim_{t \to s} \int g(x(t)) - g(x(s)) p^\star(x(t) \mid x(s)) \, dx(t) \newline & = \lim_{t \to s} \int \left[ g(x(t)) - g(x(s)) \right] \frac{h(x(t), t)}{h(x(s), s)} p(x(t) \mid x(s)) \, dx(t) \newline & = \lim_{t \to s} \frac{\mathbb{E}[[g(x(t)) - g(x(s))] h(x(t), t) \mid x(s) = x]}{h(x(s), s)} \newline & = \frac{1}{h(s, x)} \left( \left( \frac{\partial}{\partial s} + A \right) (h g) \right)(s, x). \end{align*}

This holds in general for Markov processes.

Thus, we have,

\begin{align*} A^\star g & = \frac{1}{h} A(h g) = \ldots = Ag + L^2 \frac{\nabla h}{h} \cdot g^\prime \newline & = (f + L^2 \nabla \log h) g^\prime + \frac{1}{2} L^2 g^{\prime \prime} \end{align*}

The $h$ -transformed process has drift $f + L^2 \nabla \log h$ and diffusion $L^2$ .

Girsanov’s theorem

Girsanov’s theorem states that, if we have a process $x(t)$ with drift $f$ and diffusion $L$ , then the $h$ -transformed process $x^\star(t)$ is defined as,

dx^\star(t) = f(x^\star(t), t) \, dt + L^2(x^\star(t), t) \nabla \log h(x^\star(t), t) \, dt + L(x^\star(t), t) \, d\beta(t),

and the original process $x(t)$ is defined as,

dx(t) = f(x(t), t) \, dt + L(x(t), t) \, d\beta(t),

Further, $\mathrm{Law}(x^\star(t))$ has density,

\int \frac{P(x(0)) P(x(t) \mid x(0)) h(x(t), t)}{h(x(0), 0)} \, dx(0),

Let’s say now that, if we choose a smart $h$ , we can determine the density of $x^\star(t)$ ,

\int \frac{P(x(0)) P(x(t) \mid x(0)) h(x(t), t)}{h(x(0), 0)} \, dx(0) = \pi(x(t)).

That is what we’ll do in the next part ;).

Also, note that $\mathrm{Law}(x(t))$ has density,

\int P(x(0)) P(x(t) \mid x(0)) \, dx(0).

Footnotes

Part 6 - Doob's $h$ -transform and Girsanov's theorem

Introduction

Recap: Density and Bayes’ Rule

Transition Density

Champan-Kolmogorov

Example: Transition Density of a Brownian Motion

Doob’s $h$ -transform (for transition densities)

Infinitesimal generator

Time derivative of $A$

Girsanov’s theorem

Search

Part 6 - Doob's hhh-transform and Girsanov's theorem

Introduction

Recap: Density and Bayes’ Rule

Transition Density

Champan-Kolmogorov

Example: Transition Density of a Brownian Motion

Doob’s hhh-transform (for transition densities)

Infinitesimal generator

Time derivative of AAA

Girsanov’s theorem

Part 6 - Doob's $h$ -transform and Girsanov's theorem

Doob’s $h$ -transform (for transition densities)

Time derivative of $A$