### Trang Le

#math graduate. Postdoc fellow with Jason Moore.

Trang Le, PhD

2020-04-15

Guest lecture, Fundamentals of AI

- Random walk
- Stochastic processes
- Markov chain
- Martingale

\(X_0, X_1, X_2, X_3, \cdots\) discrete time

\(\{X_t\}_{t\geq 0}\) continuous time

a **stochastic process** is a *probability distribution* over a space of paths; this path often describes the evolution of some random value/system over time.

- Dependencies in a sequence of values?
- Long term behavior?
- Boundary events (if any)?

\(Y_i\) i.i.d random variables \( \begin{cases} 1 & (\textrm{prob} \frac{1}{2}) \\ -1 & (\textrm{prob} \frac{1}{2}) \end{cases} \)

X_t = \sum_{i = 1}^tY_i

\forall t > 0

X_0 = 0

\(X_0, X_1, X_2, ...\) is called simple random walk.

*hard to visualize the time component*

If \(0 = t_0\leq t_1 \leq \cdots \leq t_k\)

then \(X_{t_{i+1}}- X_{t_i}\) are mutally independent.

- \(E(X_k) = 0\)
- Independent increment

- Stationary

For all \(h \geq 1, t \geq 0\), the distribution of \(X_{t+h} - X_t\) is the same as the distribution of \(X_h\).

At each turn, my balance goes up by $1 or down by $1.

My balance = Simple random walk

*If I play until I win $100 or lose $100,*

*what is the probability of me winning?*

*If I play until I win $100 or lose $50,*

*what is the probability of me winning?*

Let \(Y_0, Y_1, ..., Y_n\) be a simple random walk.

\[Z\left(\frac{t}{n}\right) = Y_t\]

Interpolate linearly and take \(n \to \infty\).

Then, the resulting distribution is the Brownian motion.

A discrete-time stochastic process \(X_0, X_1, X_2, ...\) with state space \(S\) has Markov property if

P(X_{t+1}=s|X_t, \cdots, X_1, X_0)= P(X_{t+1}=s|X_t)

\forall t \geq 0, \forall s \in S

If the state space \(S\) (\(X_t \in S\)) is finite,

all the elements of a Markov chain model

can be encoded in a transition probability matrix.

p_{ij} = P(X_{t+1}=j|X_t = i) \quad \forall i, j \in S

\sum_{j\in S}p_{ij} = 1

*Why?*

\begin{pmatrix}
p_{11} & p_{12} & \cdots & p_{1m}\\
p_{21} & p_{22} & \cdots & p_{2m}\\
\vdots & \vdots & \ddots & \\
p_{m1} & p_{m2} & & p_{mm}
\end{pmatrix}

*transition probability
(jumping from state i to state j)*

P = \begin{pmatrix}
p_{11} & p_{12} & \cdots & p_{1m}\\
p_{21} & p_{22} & \cdots & p_{2m}\\
\vdots & \vdots & \ddots & \\
p_{m1} & p_{m2} & & p_{mm}
\end{pmatrix}

This matrix contains all the information about the stochastic process.

*Why?*

What is the probability of going from state 7 to state 9 in one steps?

What is the probability of going from state 7 to state 9 in two steps?

*\[P_{79}\]*

*\[P^2_{79}\]*

What is the probability of going from state 7 to state 9 in \(n\) steps?

*What is our transition probability matrix?*

*state*

*state*

P = \begin{pmatrix}
0.9& 0.1 \\
0.3 & 0.7\\
\end{pmatrix}

\pi P = \pi \begin{pmatrix}
0.9& 0.1 \\
0.3 & 0.7\\
\end{pmatrix} = \pi

\pi = (? \quad ?)

*Speech recognition*

- Observation: sounds
- hidden states: words

*Face detection*

- Observation: overlapping rectangles of pixel intensities
- hidden states: facial features

__DNA sequence region classification__

- Observation: nucleotides
- Hidden states: introns, exons

Markov Chain Monte Carlo method (MCMC)

A discrete-time stochastic process \(X_0, X_1, X_2, ...\) is a martingale if

X_t = E[X_{t+1}|\{X_0, X_1,\cdots, X_t\}]

\forall t \geq 0.

*i.e.*, expected gain in the process is zero at all times.

A random variable \(X\) is distributed according to either \(f\) or \(g\). Consider a random sample \(X_1, \cdots, X_n\). Let \(Y_n\) be the *likelihood ratio*

\[Y_n = \prod_{i = 1}^n \frac {g(X_i)}{f(X_i)}\]

If X is actually distributed according to the density \(f\) rather than according to \(g\), then \[\{Y_n: n = 1, 2, 3, \cdots\}\] is a **martingale** with respect to\(\{X_n: n = 1, 2, 3, \cdots\}\).

- Random walk
- Stochastic processes
- Markov chain
- Hidden Markov model
- MCMC
- Martingale

By Trang Le

Guest lecture for Penn BMIN 520-401 course, Spring 2020

- 562

Loading comments...