About two years ago, I attended a seminar given by Dr. Sid Redner of the Santa Fe Institute titled, “Is Basketball Scoring a Random Walk?” I was certainly skeptical that such an exciting game shared similarities with coin flipping, but, nevertheless, Dr. Redner went on to convince me–and surely many other audience members–that basketball does indeed exhibit behavior akin to a random walk. At the very end of his lecture, Dr....
The Puzzle Inspired by true events Alice is assigned to be the 56th passenger to board a full plane with 60 seats. However, a panic causes all the passengers–including Alice–to arrange themselves radomly in line to board. As Alice was originally 56th, she decides that she would be happy as long as passengers with the assigned spots 57, 58, 59, and 60 are not in front of her. What is the probability that Alice will be happy?...
Definition 5.9: A stochasitc process $\{X_t\}_{t \geq 0}$ is a Gaussian Process if its finite dimensional distributions are consistent Gaussian measures for any $0 \leq t_1 < t_2 < \ldots < t_k$. Recall that a Gaussian random vector $\mathbf{X} = (X_1, X_2,\ldots,X_n)^T$ is completely characterized by its first and second moments $$\mathbf{m} = \mathbb{E}[\mathbf{X}], \quad \mathbf{K} = \mathbb{E}[(\mathbf{X} - \mathbf{m}) (\mathbf{X} - \mathbf{m})^T]$$ Meaning that the characteristic function is expressed only in terms of $\mathbf{m}$ and $\mathbf{K}$ $$\mathbb{E}\left[e^{i \mathbf{\xi} \cdot \mathbf{X}}\right] = e^{i \mathbf{\xi} \cdot \mathbf{m} - \frac{1}{2}\mathbf{\xi}^T \mathbf{K} \mathbf{\xi}} $$ This means that for any $0 \leq t_1 < t_2 < \ldots < t_k$, the measure $\mu_{t_1, t_2, \ldots, t_k}$ is uniquely determined by an $\mathbf{m} = (m(t_1), \ldots, m(t_k))$ and a covariance matrix $\mathbf{K}_{ij} = K(t_i, t_j)$....
Markov processes in continuous time and space Given a probability space $(\Omega, \mathcal{F}, \mathbb{P})$ and the filtration $\mathbb{F} = (\mathcal{F}_t)_{t \geq 0}$, a stochastic process $X_t$ is called a Markov process wrt $\mathcal{F}_t$ if $X_t$ is $\mathcal{F}_t$-adapted For any $t \geq s$ and $B \in \mathcal{R}$, we have $$\mathbb{P}(X_t \in B | \mathcal{F}_s) = \mathbb{P}(X_t \in B | X_s)$$ Essentially, this is saying that history doesn’t matter, only the current state matters....
Filtration Definition 5.3: (Filtration). Given a probability space, the filtration is a nondecreaseing family of $\sigma$-algebras $\{\mathcal{F}_t\}_{t \leq 0}$ such that $\mathcal{F}_s \subset \mathcal{F}_t \subset \mathcal{F}$ for all $0 \leq s < t$. Intuitively, the filtration is a sigma algebra of events that can be determined before time $t$ (we can’t lose information by foing forward in time). A stochastic process is called $\mathcal{F}_t$-adapted if it is measurable with respect to $\mathcal{F}_t$; that is, for all $B \in \mathcal{R}$, $X_t^{-1}(B) \in \mathcal{F}_t$....
Definition of a stochastic process A stochastic process is a parameterized random variable $\{X_t\}_{t\in\mathbf{T}}$ defined on a probability space $(\Omega, \mathcal{F}, \mathbb{P})$ taking on values in $\mathbb{R}$. $\mathbf{T}$ can seemingly be any subset of $\mathbb{R}$. For any fixed $t \in \mathbf{T}$, we can define the random variable $$X_t: \Omega \rightarrow \mathbb{R}, \quad \omega \rightarrowtail X_t(\omega)$$ Thinking of a simple random walk, this means that $X_t$ is a random variable that takes in some subset of $\Omega = \{H,T\}^\mathbb{N}$ and outputs a real valued number (the sum of the first $t$ values in $\omega$): $\{\omega_1, \omega_2, \ldots \} \rightarrow \sum_{n \leq t} X(\omega_n)$...
Here are my notes for E, Li, and Vanden-Eijnden’s Applied Stochastic Analysis Chapter 5 - Stochastic Processes 5.1 - Axiomatic Construction of Stochastic Process 5.2 - Filtration and Stopping Time 5.3 - Markov Processes 5.4 - Gaussian Processes Chapter 6 - Wiener Process 6.1 - The Diffusion Limit of Random Walks 6.2 - The Invariance Principle
I was recently doing a probability puzzle that I can’t quite remember the context of, but I came across the answer that the probability would be $$\mathbb{P}(X) = n p^n \; \quad \forall \: n\in\mathbb{N}, p \in [0,1].$$ But this is obviously wrong! Plug in $p=.9, n=2$, and you get that $\mathbb{P}(X) = 1.62$. Thaat’s not how probability works! However, for $p=0.5$, $\mathbb{P}(X)$ will remain $\leq 1$ for all $n \in \mathbb{N}$....