Preparations for the upcoming SJTU internship Part 2

In this chapter I am writing about my preparations for the upcoming Summer Research Internship at Shanghai Jiao Tong University.

This Chapter looks at the Brownian Bridge, and Context-Aware Transport.

Brownian Bridge

I do not know what Brownian Motion is, so lets do that first

1: Brownian Motion

A standard Brownian motion is a random process \(\boldsymbol{X} = \{X_t : t \in [0, \infty)\}\) with state space \(\mathbb{R}\) that satisfies the following properties:

  1. \(X_0 = 0\) (with probability 1).
  2. \(\boldsymbol{X}\) has stationary increments. So, for \(s, t \in [0, \infty)\) with \(s < t\), the distribution of \(X_t - X_s\) is the same as the distribution of \(X_{t-s}\).
  3. \(\boldsymbol{X}\) has independent increments. That is, for \(t_1, t_2, \dots, t_n \in [0, \infty)\) with \(t_1 < t_2 < \dots < t_n\), the random variables \(X_{t_1}, X_{t_2} - X_{t_1}, \dots, X_{t_n} - X_{t_{n-1}}\) are independent.
  4. \(X_t\) is normally distributed with mean 0 and variance \(t\) for each \(t \in (0, \infty)\).
  5. With probability 1, \(t \mapsto X_t\) is continuous on \([0, \infty)\).

To understand better these rules, we can look at them more closely:

  1. Suppose that we measure the position of a Brownian particle in one dimension, starting at an arbitrary time which we designate as \(t = 0\), with the initial position designated as \(x = 0\). Then this assumption is satisfied by convention.
  2. This is a statement of time homogeneity: the underlying dynamics do not change over time, so the distribution of the displacement of the particle in a time interval \([s, t]\) depends only on the length of the time interval.
  3. This is an idealized assumption that would hold approximately if the time intervals are large compared to the tiny times between collisions of the particle with the molecules.
  4. This is another idealized assumption based on the central limit theorem: the position of the particle at time \(t\) is the result of a very large number of collisions, each making a very small contribution. The fact that the mean is 0 is a statement of spatial homogeneity.
  5. Finally, the continuity of the sample paths is an essential assumption, since we are modeling the position of a physical particle as a function of time.

Brownian motion is characterized by the Wiener process

When thinking about spatial homogeneity/time homogeneity, it reminds one of the Markov Process. In a time-homogeneous Markov chain, the probability of transitioning from state \(i\) to state \(j\) depends only on the number of steps (time elapsed), not on the absolute time.

Consider a Brownian motion starting from \(W(0) = 0\) and ending at \(W(u) = x\). Conditioned on fixed \(W(0)\) and \(W(u)\), what is the distribution of \(W(t)\)? By definition, for \(t < u\), conditioned on \(W(u) = x\), \(W(t)\) is a Gaussian. So it is sufficient to compute its mean and variance. As Figure 2 shows, a natural conjecture is that for \(0 \leq t \leq u\), \(\mathbf{E}[W(t) \mid W(u)] = \frac{t}{u}W(u)\).

To verify this, we first prove the following proposition.

Proposition 4 For any \(0 \leq t \leq u\), \(W(t) - \frac{t}{u}W(u)\) is independent of \(W(u)\).

Proof.

$$ \begin{aligned} \text{Cov}\left(W(t) - \frac{t}{u}W(u), W(u)\right) &= \text{Cov}(W(t), W(u)) - \frac{t}{u}\text{Var}[W(u)] \\ &= t - \frac{t}{u} \cdot u = 0. \end{aligned} $$

The proposition follows from the fact that the two Gaussians are independent iff their covariance is zero. \(\square\)

Context Aware Transport

I am not familiar (mathematically) with Optimal Transport, so let's study that first.

1: Optimal Transport

Optimal transport compares distributions by asking how one distribution must physically move to become the other. It is fundamentally Lagrangian because it tracks displacement of mass. It tracks trajectories. Not still points in space. The optimal transport problem is how to transport \( \mu \) to \( \nu \) whilst minimizing the cost \(c\)

Definition 1.1. One says that \( T : X \to Y \) transports \( \mu \in \mathcal{P}(X) \) to \( \nu \in \mathcal{P}(Y) \), and one calls \( T \) a transport map, if

$$ \nu(B) = \mu\left(T^{-1}(B)\right) \qquad \text{for all } \nu\text{-measurable sets } B. $$

at the Cambridge Lecture on Optimal Transport, there seem to be 2 ways to formulate this:

  1. The Monge Formulation:

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