Conditioned Brownian motion, 1st Laplacian eigenfunction, etc…

A few days ago, a student asked the following question:

Consider a bounded domain ${D \subset \mathbb{R}^d}$ and a Brownian motion conditioned to stay inside ${D}$. How does it look like ? What is the invariant distribution ?

This is very simple, and surprisingly interesting. This involves the first eigenfunction of the Laplacian on ${D}$. To make everything simple, and because this does not change anything, it suffices to study the situation where ${D=[0,1] \subset \mathbb{R}}$. In other words, what does a real Brownian motion conditioned to stay inside the segment ${[0,1]}$ look like.

Discretisation

One could directly do the computations in a continuous setting but, as it is often the case, it is simpler to consider the usual random walk discretisation of a Brownian motion. For this purpose, consider a time discretisation ${\delta>0}$ and a standard random walk ${X_k}$ with increment ${\pm \sqrt{\delta}}$: with probability ${\frac12}$ the random walk goes up ${+\sqrt{\delta}}$ and with probability ${\frac12}$ it goes down ${-\sqrt{\delta}}$. For clarity, assume that there exists an integer ${m_{\delta} \in \mathbb{N}}$ such that ${m_{\delta} \sqrt{\delta} = 1}$. Suppose as well that it is conditioned on the event ${X_k \in [0,1]}$ for ${k=1, \ldots, N}$. Later, we will consider the limiting case ${N \rightarrow \infty}$ and ${\delta \rightarrow 0}$, in this order, to recover the Brownian motion case.

Conditioning

First, let us compute the probability transitions of the random walk ${\{X_k\}_{k=0}^N}$ conditioned on the event ${X_k \in [0,1]}$ for ${k=1, \ldots, N}$. For clarity, let us denote the conditioned random walk by ${\hat{X} = \big\{ \hat{X}_k \big\}_{k \geq 0}}$. This is a Doob h-transform, and the resulting process is a non-homogenous Markov chain. In this simple case the computations are straightforward. The conditioned Markov chain has transition probabilities given by ${\hat{P}(k,x,y) = \mathbb{P}[\hat{X}_{k+1}=y \,|\hat{X}_k=x]}$ with ${x,y \in D_{\delta}}$ where ${D_{\delta} = \sqrt{\delta} \mathbb{N} \; \cap \; [0,1]}$. One can compute the probability that the conditioned Markov chain follows a given trajectory ${(x_0, \ldots, x_N)}$ of ${D_{\delta}}$, $\displaystyle \begin{array}{rcl} \mathbb{P}[\hat{X}_0 = x_0, \ldots, \hat{X}_N=x_N] = \frac{1}{Z(N,\delta)} P(x_0,x_1) \times \ldots \times P(x_{N-1}, x_N) \end{array}$

where ${P(x,y)}$ is the transition kernel of the unconditioned Markov chain and ${Z(N,\delta) = \mathbb{P}[X_k \in D_{\delta} \; \text{for} \; k=0, \ldots, N]}$ is a normalisation constant. The Doob h-transform simply consists in noticing that this also reads $\displaystyle \begin{array}{rcl} \mathbb{P}[\hat{X}_0 = x_0, \ldots, \hat{X}_N=x_N] = \hat{P}(0,x_0,x_1) \times \ldots \times \hat{P}(N-1,x_{N-1}, x_N) \end{array}$

where the conditioned Markov kernel is ${\hat{P}(k,x_k,x_{k+1}) = \frac{P(x,y) \, h(k+1,y)}{h(k,x)}}$ and the function ${h(\cdot, \cdot)}$ is defined by $\displaystyle \begin{array}{rcl} h(k,x) = \mathbb{P}[X_{j} \in D_{\delta} \; \text{for} \; k+1 \leq j \leq N \, |X_k=x]. \end{array}$

Of course we have ${h(N,x)=1}$ for all ${x \in D_{\delta}}$. The quantity ${h(x,k)}$ is the probability that a random walk starting at ${x \in D_{\delta}}$ at time ${k}$ remains inside ${D_{\delta}}$ for time ${k,k+1, \ldots, N}$. Consequently, in order to find the transition probabilities of the conditioned kernel, it suffices to compute the quantities ${h(k, x)}$ for all ${x \in D_{\delta}}$. Since we are interested in the limiting case ${N \rightarrow \infty}$, it actually suffices to consider the case ${k=0}$. It can be computed recursively since ${h(k,x) = \frac12 \, h(k+1, x+\sqrt{\delta}) + \frac12 \, h(k+1, x-\sqrt{\delta})}$ with the appropriate boundary conditions. In other words, adopting the obvious matrix notations, the vector ${h(k,\cdot)}$ satisfies $\displaystyle \begin{array}{rcl} h(k,\cdot) = A h(k+1, \cdot) \end{array}$

where ${\big( A_{i,j} \big)_{0 \leq i,j \leq m_{\delta}}}$ is the usual tridiagonal matrix given by ${A_{i,j} = 1}$ if, and only if, ${|i-j|=1}$ and ${A_{i,j} = 0}$ otherwise. It is related to the discrete Laplacian operator. Indeed, because all the eigenvalues of ${A}$ are real with modulus strictly inferior to ${1}$, it follows that ${h(0,\cdot) = A^{N} \textbf{1} \approx \lambda^N \, \varphi_{\delta}(\cdot)}$ where ${\lambda_{\delta}}$ is the highest eigenvalue of ${A}$ and ${\varphi_{\delta}(\cdot)}$ the associated eigenfunction. The eigenvalues of ${A}$ are well-known, and as ${\delta \rightarrow 0}$, the highest eigenvalue converges to ${1}$ and the associated eigenfunction converges to the first eigenfunction of the Laplacian on the domain ${D=[0,1]}$ with Dirichlet boundary. In our case it is ${\varphi(u) = \sin(\pi \, u)}$ and $\displaystyle \begin{array}{rcl} \lim_{N \rightarrow \infty} \frac{ h(1,x \pm \sqrt{\delta})}{h(0,x)} = \lambda_{\delta}^{-1} \frac{ \varphi_{\delta}(x \pm \sqrt{\delta})}{\varphi_{\delta}(x)}. \end{array}$

In other words, the random walk with increments ${\pm \sqrt{\delta}}$ conditioned to stay inside ${D_{\delta}}$ has probability transitions given by $\displaystyle \begin{array}{rcl} \hat{P}(x, x \pm \sqrt{\delta}) = \frac{{\lambda_{\delta}}^{-1}}{2} \frac{ \varphi_{\delta}(x \pm \sqrt{\delta})}{\varphi_{\delta}(x)}. \end{array}$

Next section investigates the limiting case ${\delta \rightarrow 0}$.

Conclusion

We have computed the dynamics of the conditioned random walk with space-increments ${\sqrt{\delta}}$. To obtain the dynamics of the conditioned Brownian motion it suffices to consider the limiting case ${\delta \rightarrow 0}$. The drift of the resulting diffusion is given by $\displaystyle \begin{array}{rcl} \text{(drift)} &=& \lim_{\delta \rightarrow 0} \frac{1}{\delta} \Big\{ \hat{P}(0,x,x+\sqrt{\delta})\sqrt{\delta} - \hat{P}(0,x,x-\sqrt{\delta})\sqrt{\delta} \Big\} \\ &=& \lim_{\delta \rightarrow 0} \frac{{\lambda_{\delta}}^{-1}}{2 \sqrt{\delta}} \Big\{ \frac{ \varphi_{\delta}(x + \sqrt{\delta})}{\varphi_{\delta}(x)} - \frac{ \varphi_{\delta}(x - \sqrt{\delta})}{\varphi_{\delta}(x)} \Big\} = \frac{1}{2} \frac{\varphi'(x)}{\varphi(x)}. \end{array}$

The same computation gives the volatility of the resulting diffusion. It is given by $\displaystyle \begin{array}{rcl} \text{(volatility)} &=& \lim_{\delta \rightarrow 0} \frac{1}{\delta} \Big\{ \hat{P}(0,x,x+\sqrt{\delta})\delta + \hat{P}(0,x,x-\sqrt{\delta})\delta \Big\} \\ &=& \lim_{\delta \rightarrow 0} \frac{{\lambda_{\delta}}^{-1}}{2} \Big\{ \frac{ \varphi_{\delta}(x + \sqrt{\delta})}{\varphi_{\delta}(x)} + \frac{ \varphi_{\delta}(x - \sqrt{\delta})}{\varphi_{\delta}(x)} \Big\} = 1. \end{array}$

As the consequence, this shows that a Brownian motion conditioned to stay inside ${D=[0,1]}$ follows the stochastic differential equation ${dX = \frac{1}{2} \frac{\varphi'(X)}{\varphi(X)} \, dt + dW_t}$ where ${\varphi(\cdot)}$ is the first eigenfunction of the Laplacian on ${D}$ with Dirichlet boundary conditions. More generaly, the same argument would show that a Brownian motion in ${\mathbb{R}^d}$ conditioned to stay inside a nice bounded domain ${D \subset \mathbb{R}^d}$ evolves according to the stochastic differential equation $\displaystyle \begin{array}{rcl} dX = \frac{1}{2} \frac{\nabla \varphi(X)}{\varphi(X)} \, dt + dW_t \end{array}$

where ${\varphi:D \rightarrow \mathbb{R}}$ is the first eigenfunction of the Laplacian on ${D}$. This is a Langevin diffusion and one can immediately see that the invariant distribution of this diffusion is given by $\displaystyle \begin{array}{rcl} \pi_{\infty}(x) \, \propto \varphi(x). \end{array}$

For example, the following plot depicts the first eigenfunction of the Laplacian on the domain ${D=[0,1]^2 \subset \mathbb{R}^2}$.