| bio | website | quantdec.com |
|---|---|---|
| location | ||
| age | ||
| visits | member for | 2 years, 8 months |
| seen | yesterday | |
| stats | profile views | 767 |
Consultant (environmental stats a specialty) and teacher.
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Oct 5 |
reviewed | Close Find the number of specific permutations |
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Oct 5 |
reviewed | Leave Open Formula for Poisson process |
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Aug 25 |
awarded | Yearling |
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Aug 17 |
comment |
Does there exist a nicer form for $\beta(x + a, y + b) / \beta(a, b)$? I don't see any distributions in this question. Are you just asking for a simplification of this ratio of values of the Beta function? Are there any restrictions on $x$ and $y$ (possibly they are integers? Non-negative numbers? Real numbers)? |
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Aug 7 |
comment |
How do I integrate this distribution? In light of the inherent contradictions in the question, could you provide more context? I'm referring specifically to the fact that the first equation does not describe a distribution (presumably for $\beta$) per se because it is not normalized and, if it were normalized, then a fortiori the integral over $\beta$ would be $1$. |
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Jul 16 |
answered | Differentiating the posterior distribution function |
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Jul 13 |
revised |
Differentiating the posterior distribution function edited tags |
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Jun 25 |
comment |
Proof that $\mathbb N $ is finite @Thomas I thought as much--there's no difference of opinion here, just different interpretations of English. I would like to clarify that in my previous comment I should have referred to the cardinality of all descriptions that can be made with $15$ English words, which will be far less than (say) $(26^{100})^{15}$. |
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Jun 25 |
comment |
Proof that $\mathbb N $ is finite @Thomas, Anixx is correct: the argument "works" for any finite set guaranteed to have greater cardinality than the set of English words. Even if one cannot even agree on what the set of English words is, I think there would be general agreement that the set of existing English words is no greater than, say, $26^{100}$. |
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Jun 16 |
revised |
Finding an invariant under Group operations deleted 5 characters in body |
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Jun 16 |
revised |
Finding an invariant under Group operations added 690 characters in body |
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Jun 16 |
answered | Finding an invariant under Group operations |
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Jun 15 |
comment |
Queries regarding saddle point Images at google.com/… |
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Jun 11 |
comment |
Ratio Distribution: Poisson Random Variables Cross-posted on stats. |
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Jun 8 |
awarded | Constituent |
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Jun 8 |
awarded | Caucus |
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May 30 |
comment |
Can the borders of a map be deformed to give arbitrary area to any region? Thanks, @Rahul. That brings up the next question. :-) What precisely does "innermost" mean and how do you know you can eventually work outwards? (Bear in mind that the regions need not be simply connected and that their union can be multiply connected, too.) I'm really not trying to give you a hard time here; it's merely that experience shows that an intuitively obvious algorithm can sometimes fail right at these extreme or exceptional cases, so care is needed to demonstrate correctness. (But maybe, on the other hand, I'm just being more obtuse than usual...) |
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May 30 |
comment |
Can the borders of a map be deformed to give arbitrary area to any region? Lovely diagram! Concerning the algorithm: how do we know it terminates? Couldn't it cycle endlessly? E.g., if in your example region A were not part of the figure--creating an annulus--why wouldn't the algorithm "pour" C into B, D into C, E into D, and then B into E, creating a convoluted figure where the relative areas of B,C,D,E were no closer to the intended ratios than before? |
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May 30 |
comment |
Can the borders of a map be deformed to give arbitrary area to any region? No, mainly because I don't have a clear picture of exactly what you're doing. (Please note I'm not disputing the correctness of your solution; I just am unsure exactly what it is!) |
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May 30 |
comment |
Can the borders of a map be deformed to give arbitrary area to any region? @Rahul Too sketchy--I wrote it because it's the kind of thing I thought you were suggesting in your first comment :-). |