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xorshammer.com | ||
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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... |
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www.jeremykun.com
This proof assumes knowledge of complex analysis, specifically the notions of analytic functions and Liouville's Theorem (which we will state below). The fundamental theorem of algebra has quite a few number of proofs (enough to fill a book!). In fact, it seems a new tool in mathematics can prove its worth by being able to prove the fundamental theorem in a different way. This series of proofs of the fundamental theorem also highlights how in mathematics there are many many ways to prove a single theorem... |
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almostsuremath.com
Given a sequence $latex {X_1,X_2,\ldots}&fg=000000$ of real-valued random variables defined on a probability space $latex {(\Omega,\mathcal F,{\mathbb P})}&fg=000000$, it is a standard result that the supremum $latex \displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle X\colon\Omega\rightarrow{\mathbb R}\cup\{\infty\},\smallskip\\ &\displaystyle X(\omega)=\sup_nX_n(\omega). \end{array} &fg=000000$ is measurable. To ensure that this is well-defined, we need to allow X to have values in $latex... |
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allouttabubblegum.wordpress.com
Keep an eye here, going to reactivate this blog. |
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