# Doob's $$h$$-transform

doob_transform

## Loosely speaking

Let $$A$$ be a non-negative kernel on $$S$$ and $$h$$ that is $$A$$-harmonic positive function, then we call the transition kernel

$Q_h(x, y) = {h(y) \over h(x)} A(x, y),$

the Doob $$h$$-transform of $$A$$.

Examples:

process $$h$$
Schur processes Schur functions
Whittaker processes Whittaker functions
$$q$$-Whittaker processes $$q$$-Whittaker functions
Macdonald processes Macdonald polynomials

In Markov doob integrability and the notes that depend on it we use this "loosely speaking" version of Doob's $$h$$-transform.

In the following we mostly follow exposition in Bloemendal's draft notes "Doob's $$h$$-transform: theory and examples".

## Derivation

Consider a (continuous or discrete time) Markov process $$X_t$$ on state $$S$$ with kernel $$P_t$$. We associate with it a path space $$\Omega$$ with probability measures $$(P_x)_{x \in S}$$, and let $$\mcf_t$$ be the natural filtration of $$(X_t)$$.

Let

$\mci := \{A: \theta_t^{-1} A = A \forall t\ge 0\},$

then any $$\mci$$-measurable function is shift-invariant: $$H \circ \theta_t = H \forall t \ge 0$$, since

$(H \circ \theta_t)^{-1} B = \theta_t^{-1} H^{-1} (B).$

Claim 1. There is a one-one correspondence between bounded $$\mci$$-measurable functions and bounded $$P_t$$-harmonic functions.

Proof. For any $$H \in \mci$$, let

$h(x) = \expe_x H$

Then by the Markov property

$h(X_t) = \expe_{X_t} H = \expe_x (H \circ \theta_t | \mcf_t) = \expe_x (H | \mcf_t)$

is a martingale, hence by Corollary 4 in infinitesimal_generator $$h$$ is $$P_t$$-harmonic.

Conversely, given $$P_t$$-harmonic $$h$$, by martingale convergence theorem

$H = \lim_{t \to \infty} h(X_t)$

exists. And obviously $$H \in \mci$$. $$\square$$

$$L^\infty \mci$$ is called the Poisson boundary.

For $$H = \ind_A$$ for some $$A \in \mci$$, we define the conditional probability $$Q_x = P_x(\cdot | A)$$ (see Claim 1 in radon_nikodym_derivative)

${d Q_x \over d P_x} = {\ind_A \over h(x)}.$

By Claim 2 in radon_nikodym_derivative and Pf of Claim 1

${d Q_x \over d P_x}|_{\mcf_t} = {h(X_t) \over h(x)}$

Let $$\tilde S = \{x \in S: h(x) > 0\}$$ be accessible states.

Claim. $$Q_x$$ is a probability measure on paths in $$\tilde S$$

Proof. $$\forall t$$, $$Q_x (h(X_t) > 0) = \expe_x \ind_{h (X_t) > 0} {h (X_t) \over h(x)} = \expe_x {h(X_t) \over h(x)} = 1$$, thus $$\forall t X_t \in \tilde S, P_x$$-a.s. $$\square$$

Define kernel $$Q_t$$

$Q_t(x, d y) = {h(y) \over h(x)} P_t(x, dy) \qquad (1)$

Claim. $$Q_t$$ is a stochastic semigroup.

Proof.

• $$Q_t$$ is stochastic because $$h$$ is $$P_t$$-harmonic.
• $$Q_t$$ is a semigroup because it is a conjugate of $$P_t$$

$$\square$$

Theorem 1. Under $$Q_x$$ with $$x \in \tilde S$$, $$X_t$$ is Markov on $$\tilde S$$ with kernel $$Q_t$$

Proof. Formally using Bayes theorem

$P_x(X_{t + s} \in dy | A, \mcf_t) = {P_x(A | X_{t + s} = y, \mcf_t) P_x(X_{t + s} \in dy | \mcf_t) \over P_x(A | \mcf_t)} = {\expe_y(\ind_A \circ \theta_{t + s}) P_{X_t}(X_{t + s} \in dy) \over \expe_x (\ind_A | \mcf_t)} = {h(y) P_s(X_t, dy) \over h(X_t)}.$

More precisely we want to show

$\expe^Q_x (f(X_{t + s}) | \mcf_t) = h(X_t)^{-1} \expe_{X_t} f(X_s) h(X_s) \qquad Q_x \text{-a.s.}$

We start from the RHS. For $$B \in \mcf_t$$,

\begin{align} \expe_x^Q &h(X_t)^{-1} \expe_{X_t} f(X_s) h(X_s) \ind_B \\ &= \expe^Q_x h(X_t)^{-1} \expe_x(f(X_{t + s}) h(X_{t + s}) | \mcf_t) \ind_B\\ &= \expe_x {h(X_t) \over h(x)} h(X_t)^{-1} \expe_x(f(X_{t + s}) h(X_{t + s}) | \mcf_t) \ind_B\\ &= \expe_x h(x)^{-1} f(X_{t + s}) h(X_{t + s}) \ind_B\\ &= \expe^Q_x f(X_{t + s}) \ind_B \end{align}

$$\square$$.

## Continuous-time

Taking generator on both sides of (1) we have

$L^Q = m_{1 / h} L^P m_h \qquad (1.5)$

Claim. Specifically if $$X$$ is a diffusion under $$P$$ with SDE and generator

\begin{align} d X_t &= \sigma(X_t) d B_t + b(X_t) dt\\ L^P &= {1 \over 2} \sigma \sigma^T : \nabla \nabla^T + b \cdot \nabla = {1 \over 2} \sum a_{i j} \partial_{x_i x_j} + \sum b_i \partial_{x_i} (1.7) \end{align}

then

$L^Q = L^P + a \nabla \log h \cdot \qquad (2)$

Proof. plug $$L^P$$ in (1.7) into (1.5), and noting $$L h = 0$$. $$\square$$

Remark. (2) can also be written as

$L^Q f = L^P f + {1 \over h} \sigma^T \nabla h \cdot \sigma^T \nabla f$

We also have that under $$Q$$,

$d X_t = \sigma(X_t) d B_t + (b(X_t) + a \nabla h(X_t)) dt$

namely a drift in the direction of increasing $$h$$ is added.

## Conditioning on a null event

When $$P_x(A) = 0 \forall x$$, one has to resort to the theory of Martin boundary (see perhaps e.g. [{sawyer97}]).

Heuristically, one may use a sequence $$A_n \in \mci$$ (with harmonic function $$h_n$$) such that $$\bigcup A_n = A$$, and a sequence $$c_n \in \real$$ such that $$\lim_n c_n h_n = h$$ exists, is finite and positive.

## Absorbing boundary

A special case is when we consider an absorbing boundary $$\partial S \subset S$$ such that

$P_t(x, dy) = \delta_x(dy), \forall x \in \partial S \forall t \ge 0$

is the delta measure.

Let $$T = \tau_{\partial S}$$, the previous formula means $$X_t = X_{t \wedge T}$$.

We also consider $$Z \subset \partial S$$, and $$H = \{X_T \in Z\}$$.

Then the condition is "the Markov process hits $$Z$$ when exitting $$S \setminus \partial S$$", i.e. $$\{X_T \in Z\}$$.

The harmonic function $$h$$ is then a solution to the Dirichlet problem:

$\begin{cases} L h(x) = 0, \qquad x \in S \setminus \partial S\\ h(x) = \ind_Z, \qquad x \in \partial S \end{cases}$

The solution

$h(x) = P_x(X_T \in Z) = \expe_x h(X_T)$

is a special case of potential theory in probability:

$\begin{cases} L h(x) = 0, \qquad x \in S \setminus \partial S\\ h(x) = f(x), \qquad x \in \partial S. \end{cases}$

has solution

$u(x) = \expe_x f(X_T)$

since $$u(X_t)$$ is a martingale since

$u(X_t) = \expe_{X_t} f(X_T) = \expe(f(X_{T + t}) | \mcf_t) = \expe(f(X_T) | \mcf_t)$

due to the absorbing boundary.

## Examples

### Simple random walk on $$0 : N$$ hitting $$N$$ before $$0$$.

$$S = 0 : N$$, $$\partial S = \{0, N\}$$, $$Z = \{N\}$$.

If $$X_t$$ is symmetric then the equation $$P h = h$$ is

$h(i) = {1 \over 2} (h(i + 1) + h(i - 1)), \qquad i = 1 : N - 1 \qquad (3)$

combined with boundary condition it is a Dirichlet problem for Laplace equation:

$\begin{cases} \Delta h(i) = 0, \qquad i = 1 : N - 1\\ h(0) = 0\\ h(N) = 1 \end{cases}$

where $$\Delta$$ is the discrete Laplacian.

Solving this equation gives

$h(i) = i / N$

and the kernel $$Q$$ of the conditioned process by (1) is

$Q(i, i + 1) = {i + 1 \over 2 i},\qquad Q(i, i - 1) = {i - 1 \over 2 i}, \qquad Q(i, \text{other j}) = 0 \qquad (4)$

If $$X_t$$ is asymmetric, with probability $$p$$ of jumping from $$i$$ to $$i + 1$$, then the equation (3) is now

$h(x) = p h(x + 1) + (1 - p) h(x - 1)$

with the same boundary condition. Solving it gives

$h(i) = {1 - r^i \over 1 - r^N}$

where $$r = {1 - p \over p}$$.

### Brownian motion on $$[0, M]$$ hitting $$M$$ before $$0$$

Same as in the RW example, but continuous time.

$$S = [0, M]$$, $$\partial S = \{0, M\}$$, $$Z = \{M\}$$.

The equation $$L h = 0$$ with boundary condition is a Dirichlet problem

$\begin{cases} {1 \over 2} h_{xx} (x) = 0 \qquad x \in (0, M)\\ h(0) = 0\\ h(M) = 1 \end{cases}$

solving which gives

$h(x) = x / M.$

Hence by (2) we have

$L^Q = {1 \over 2} \partial_{xx} + {1 \over x} \partial_x.$

### Brownian motion on $$[0, \pi]$$ conditioned to remain in $$(0, \pi)$$ up to time $$t_1$$

We use the space-time trick. $$S = [0, t_1] \times [0, \pi]$$, $$\partial S = [0, t) \times \{0, \pi\} \cup \{t_1\} \times [0, \pi]$$, $$Z = \{t_1\} \times [0, \pi]$$.

Since the generator of $$(t, X_t)$$ is $$\hat L = \partial_t + L$$, combining this with boundary condition we obtain the backward heat equation:

$\begin{cases} \partial_t \hat h + {1 \over 2} \partial_{xx} \hat h = 0, \qquad (x, t) \in [0, t_1) \times (0, \pi) \\ h(t, 0) = h(t, \pi) = 0\qquad t \in [0, t_1)\\ h(t_1, x) = 1 \qquad x \in [0, \pi] \end{cases}$

Solving this equation we have

$\hat h(t, x) = \sum_{k \ge 1} c_k e^{- k^2 (t_1 - t)} \sin(k x)$

where $$c_k = {4 \over k \pi} \ind_{2 \nmid k}$$.

So the new generator is

$\hat L^Q = \partial_t + L + {\partial_{t, x} \hat h(x, t) \over \hat h(x, t)} \cdot \nabla_{t, x} = \partial_t + L + {\sum c_k k e^{-k^2(t_1 - t)} \cos(k x) \over \sum c_k e^{- k^2 (t_1 - t)} \sin(k x)} \partial_x + {\sum c_k e^{- k^2 (t_1 - t)} (- k^2) \sin k x \over \sum c_k e^{- k^2 (t_1 - t)} \sin k x} \partial_t.$

Other examples include Bessel processes, dyson_brownian_motion, brownian_bridge, brownian_excursion, and various RS(K)-related models.

## Discrete case (possibly Martin boundary)

This section follows exposition in [{oconnell03a}], let $$P$$ be a substochastic transition matrix on a countably infinite state sapce $$S$$, and $$G$$ to be the associated Green's function:

$G(x, y) = \sum_{n \ge 0} P^n (x, y).$

Assuming $$\exists x^* \in S$$ s.t. $$0 < G(x^*, y) < \infty \forall y \in S$$. Suppose $$h$$ is a positive harmonic function w.r.t. $$P$$, then the Doob $$h$$-transform of $$P$$ is the stochastic kernel $$P_h$$ defined by

$P_h(x, y) = {h(y) \over h(x)} P(x, y).$

Theorem (Doob). If $$\exists f$$ s.t.

$\lim_{n \to \infty} {G(x, X(n)) \over G(x^*, X(n))} = f(x) \qquad a.s.$

then $$h = C f$$ for some $$C$$.

When conditioning on an event $$B$$ with positive probability, we take

$h(x) = P_x(B).$

## References

• [oconnell03a] Conditioned random walks and the RSK correspondence, , Journal of Physics A: Mathematical and General, Vol. 36, No. 12, p.3049–3066, March 2003.
• [sawyer97] Martin boundaries and random walks, , Contemporary Mathematics, Vol. 206, p.17–44 1997.