Andrew Salmon
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 Mar 6 comment If $m^*(E)=\infty$, then $E=\bigcup_{k=1}^{\infty}E_k$, $E_k$ measurable and $m^*(E_k)<+\infty$ @Kurome yes it's a general result, but it's trivial that if $E$ has finite measure, then $E$ can be decomposed into finitely many sets of finite measure. The only interesting case is the case where $m(E) = +\infty$. Perhaps I misunderstand what you're getting at. Mar 5 comment If $m^*(E)=\infty$, then $E=\bigcup_{k=1}^{\infty}E_k$, $E_k$ measurable and $m^*(E_k)<+\infty$ @Kurome The ceiling function $\lceil x \rceil$ has the property that $\lceil x \rceil - 1 < x \le \lceil x \rceil$. Feb 12 awarded Yearling Jan 4 answered $\frac{2^{2n}}{2n} \le {2n \choose n}$? Dec 27 reviewed Reject The elements in a finite field. Dec 27 answered Why m*(E) <= m*(En) if E = union of(En)? Dec 27 accepted Estimating the number of books in the world from randomly chosen overlapping lists Dec 25 comment Estimating the number of books in the world from randomly chosen overlapping lists I'm not sure I quite follow. I don't see the correspondence to the problem. Will this work if all the lists have drastically different sizes? Suppose one list has 600 entries and another list has only 50. Dec 25 asked Estimating the number of books in the world from randomly chosen overlapping lists Dec 23 awarded Nice Answer Dec 19 comment Conditional probability and a normal distribution @A.S. How does one interpret $P(\mu)$ if $\mu$ is not a proposition but a random variable? Dec 19 revised Conditional probability and a normal distribution edited tags Dec 19 comment Conditional probability and a normal distribution @A.S. I am mostly looking for a good resource that I can find that approaches questions very similar to this; that would certainly be the most helpful thing in the long run. Dec 19 comment Conditional probability and a normal distribution @A.S. Observations are orderings of $X_{i,k}$ but you are not necessarily given a complete ranking, only a partial ordering over a subset of all $X_{i,k}$ for fixed $i$. Dec 18 comment Conditional probability and a normal distribution @A.S. No, I mean that $X_{i,k}$ is drawn from $N(\mu_k,\sigma)$ and similarly $X_{i,k'} \sim N(\mu_{k'},\sigma)$ and if we are given an "observation," it is in the form $X_{i,k} > X_{i,k'}$ or $X_{i,k'} > X_{i,k}$, from which we can make an inference about the values of $\mu_k$ and $\mu_{k'}$. Dec 18 asked Conditional probability and a normal distribution Sep 21 awarded Enlightened Sep 21 awarded Nice Answer Aug 14 awarded Notable Question Jun 11 awarded Revival