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. We can associate a family of probability measures $\{\mathbb{P}^x\}_{x\in\mathbb{R}}$ for the processes starting at $x$ by defining $\mu_0$ to be the point mass at $x$. Then, we still have $$\mathbb{P}^x(X_t \in B | \mathcal{F}_s) = \mathbb{P}^x(X_t \in B | X_s), \quad t \geq s$$ and $\mathbb{E}[f(X_0)] = f(x)$ for any function $f \in C(\mathbb{R})$. ⚠️ I am not fully confident on what the above section is saying. Specifically, I am having trouble with understanding how we are defining $\mathbb{P}^x$. However, I can understand the strong markov property, so I think I should be okay moving forward. ...

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 $B \in \mathcal{R}$, $X_t^{-1}(B) \in \mathcal{F}_t$. We can always assume that the $\mathcal{F}_t$ contains $F_t^{X}$ and all sets of measure zero, where $F_t^{X} = \sigma(X_s, s \leq t)$ is the sigma algebra generated by the process $X$ up to time $t$. ...

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

2.2.1 - Definition and first examples Definition 2.2: A symmetric monoidal structure on a preoirder $(X, \leq)$ consists of (i) a monoidal unit, $I \in X$ (ii) a monoidal product $\otimes: X \times X \rightarrow X$ And the monoidal product $\otimes(x_1,x_2) = x_1 \otimes x_2$ must also satisfy the following properties (assume all elements are in $X$) (a) $x_1 \leq y_1$ and $x_2 \leq y_2 \implies x_1 \otimes x_2 \leq y_1 \otimes y_2$ (b) $I \otimes x = x \otimes I = x$ (c) associativity (d) commutivity/symmetry (a) is called monotnoicity and (b) is unitality ...

This is a collection of my notes for Brendan Fong and David Spivak’s An Invitation to Appied Category Theory. The first chapter was done through LaTeX, but the rest should be markdown with mathjax. Chapter 1 - Generative effects: Orders and adjunctions Chapter 2 - Resource theories: Monoidal preorders and enrichment Section 2.2 - Symmetric monoidal preorders

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 $n \in \mathbb{N}$. So, somewhere in the interval $(0.5,0.9)$, we reach a critical value where any $p$ greater than that will result in a probability greater than one, and any value less than it will be a bit more reasonable. ...

I have no idea what I am doing. Anyways, here’s a cool equation: $$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\right) - \frac{\partial L}{\partial q} = 0$$