Doob's \(h\)-transform

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doob_transform

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".

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}

\(\square\).

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.

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.

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.

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

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 \(r = {1 - p \over p}\).

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

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

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.

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, N. O'Connell, Journal of Physics A: Mathematical and General, Vol. 36, No. 12, p.3049–3066, March 2003.
• [sawyer97] Martin boundaries and random walks, Stanley A Sawyer, Contemporary Mathematics, Vol. 206, p.17–44 1997.