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| | jasoncrane.org
3.6 parsecs away

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| | [AI summary] A personal reflection by Jason Crane on sound, vision, and nostalgia inspired by David Bowie's Low album
| | almostsuremath.com
2.8 parsecs away

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| | I will give a proof of the measurable section theorem, also known as measurable selection. Given a complete probability space $latex {(\Omega,\mathcal F,{\mathbb P})}&fg=000000$, we denote the projection from $latex {\Omega\times{\mathbb R}}&fg=000000$ by $latex \displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle\pi_\Omega\colon \Omega\times{\mathbb R}\rightarrow\Omega,\smallskip\\ &\displaystyle\pi_\Omega(\omega,t)=\omega. \end{array} &fg=000000$ By definition, if $latex {S\subseteq\Omega\times{\mathbb R}}&fg=000000$ then, for every $latex {\omega\in\pi_\Omega(S)}&fg=000000$, there...
| | mkatkov.wordpress.com
1.9 parsecs away

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| | For probability space $latex (\Omega, \mathcal{F}, \mathbb{P})$ with $latex A \in \mathcal{F}$ the indicator random variable $latex {\bf 1}_A : \Omega \rightarrow \mathbb{R} = \left\{ \begin{array}{cc} 1, & \omega \in A \\ 0, & \omega \notin A \end{array} \right.$ Than expected value of the indicator variable is the probability of the event $latex \omega \in...
| | www.jeremykun.com
29.1 parsecs away

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| Last time we defined and gave some examples of rings. Recapping, a ring is a special kind of group with an additional multiplication operation that "plays nicely" with addition. The important thing to remember is that a ring is intended to remind us arithmetic with integers (though not too much: multiplication in a ring need not be commutative). We proved some basic properties, like zero being unique and negation being well-behaved.