/explore
|
You are here 
|
mkatkov.wordpress.com | ||
| | | | |
thehighergeometer.wordpress.com
|
|
| | | | | Here's a fun thing: if you want to generate a random finite $latex T_0$ space, instead select a random subset from $latex \mathbb{S}^n$, the $latex n$-fold power of the Sierpinski space $latex \mathbb{S}$, since every $latex T_0$ space embeds into some (arbitrary) product of copies of the Sierpinski space. (Recall that $latex \mathbb{S}$ has underlying... | |
| | | | |
djalil.chafai.net
|
|
| | | | | This post provides the solution to a tiny exercise of probability theory, answering the question asked by a student during the MAP-432 class yesterday. Let \( {(\Omega,\mathcal{F},\mathbb{P})} \) be a probability space equipped with a filtration \( {{(\mathcal{F}_n)}_{n\geq0}} \). Recall that a random variable \( {\tau} \) taking values in \( {\mathbb{N}=\{0,1,\ldots\}} \) is a stopping time when \( {\{\tau=n\}\in\mathcal{F}_n}... | |
| | | | |
jmanton.wordpress.com
|
|
| | | | | If $latex Y$ is a $latex \sigma(X)$-measurable random variable then there exists a Borel-measurable function $latex f \colon \mathbb{R} \rightarrow \mathbb{R}$ such that $latex Y = f(X)$. The standard proof of this fact leaves several questions unanswered. This note explains what goes wrong when attempting a "direct" proof. It also explains how the standard proof... | |
| | | | |
arkadiusz-jadczyk.eu
|
|
| | | We continue Becoming anti de Sitter. Every matrix $\Xi$ in the Lie algebra o(2,2) generates one-parameter group $e^{\Xi t}$ of linear transformations of $\mathbf{R}^4.$ Vectors tangent to orbits of this group form a vector field. Let us find the formula for the vector field generated by $\Xi. | ||
