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July 16

Statistics: Sampling a copula

Hello,

I am a graduate student at Old Dominion University in Norfolk, VA. I am trying to run a Monte Carlo simulation that includes the generation of pseudo-random correlated bivariate numbers (or points). The Wikipedia entry for copulas (http://en.wikipedia.org/wiki/Copula_%28probability_theory%29#Empirical_copulas) lists a procedure for generating random samples from a multivariate distribution. The first step is:

1. Draw a sample (U_1^k,...,U_d^k) ~ C (k=1,...,n) of size n from the copula C

I don't how to do this and have unable to find any information anywhere else about it. Could you provide additional information about how to this, or point me to a site that describe how to do this?

Thank you for your help.

Wilfred

173.50.125.48 (talk) 02:42, 16 July 2013 (UTC)[reply]

Successive differences on the sequence of number of functions on a finite set, that is, the sequence {n^n}

1,1,4,27,256,3125,46656,823543,8^8,9^9,...if you take enough successive differences, you will get the sequence 0,3,17,169,2079,31261,554483,... I learned today on OEIS that the second sequence is called the binomial transform of the first sequence. Question 1) Analogy with results on symmetric group: The {n!} sequence gets the binomial transform sequence 0,1,2,9,44,.. which are # of permutations in Symm(n) with no fixed points (derangements). So my first question is, as an analogy with permutations, are the numbers, 0,3,17,169,.. likewise an enumeration per n of a distinguished subset of functions on the set {1,2,3,..n}? I've fiddled with various sets of functions, but, for example, I haven't yet found a special 17 element subset of the 27 functions on {1,2,3}. Of course I've looked at functions with no, or at least few, fixed points. Question 2) It is wellknown that the ratio |Symm(n)|/#derangements goes to e ~ 2.718281828 as n goes to infinity. That is, the sequence of ratios 2/1, 6/2, 24/9, 120/44... has e as the limit. So, my second question is, as an anology with the permutation ratio going to e, what the sequence of ratios approach in the functions case? It's about 1.47 and decreasing with n, perhaps it's agm(1,2)? (To be clearer, I'm talking about the ratios 4/3,27/17,256/169,3125/2079,46656/31261,...). So, thanks in advance.Rich (talk) 11:40, 16 July 2013 (UTC)[reply]

There are binomial_coeff(n, k)(n - 1)^{n - k} functions from n -> n with k fixed points. You can twist around the sequences used and take absolute values in a way to get a sequence whose difference sequence will connect up the above sequence of n and k; but you could, probably, do this with all sorts of things. Getting to the meat of the issue, is their any reason to expect a connection? Is there some specific context to what you're looking at; or just curiosity? As for the ratios, I can't see any reason to expect that the limit would be to a "known" number; is there one? What reason do you expect agm(1, 2)? Your questions seem interesting, but what led you to them? This would greatly help in searching out answers.Phoenixia1177 (talk) 09:54, 17 July 2013 (UTC)[reply]

Well it was just curiosity and the analogy with the e ratio with bijections. Thanks for your input.76.218.104.120 (talk) 12:04, 22 July 2013 (UTC)[reply]

If you look up OEIS it says it is the number of functions whose digraph has no isolated vertex. I take that to mean there is no k such that f(k)=k and no other value equals k. Thus for the 2x2 case it excludes 01->01 and for the 3x3 case it excludes 012,011,021,022,010,210,212,002,102,112. Dmcq (talk) 10:49, 17 July 2013 (UTC)[reply]

Thanks, that's helpful.76.218.104.120 (talk) 13:04, 17 July 2013 (UTC)[reply]

In case anyone's interested, I broke down and took a closer look at it symbolically, and it wasnt too hard to see the limit ratio is e^(1/e), which is slightly less than agm(1,2).-Thanks to both of you,-Rich170.170.59.138 (talk) 08:52, 10 August 2013 (UTC)[reply]

IMDb Weighted Average

Good Morning. I hardly ever visit this section of the reference desk because it so quickly goes over my head. I don't really get advanced math, but I overwhelmingly respect it. Today I meekly approach the oracles for some help.

I enjoy movies and try to rate everything I watch on IMDb. As a personal challenge this morning, I downloaded my vote history and tried to figure out how IMDb weighs user ratings to come up with the Top 250. Below the list, at the bottom of the page, they give the formula as

weighted rating (WR) = (v ÷ (v+m)) × R + (m ÷ (v+m)) × C

where
R = average for the movie (mean) = (Rating)
v = number of votes for the movie = (votes)
m = minimum votes required to be listed in the Top 250 (currently 25000)
C = the mean vote across the whole report (currently 7.0)

So, using The Shawshank Redemption as an example, which has received the following 1,001,818 user ratings:

10: 580929
9: 229299
8: 105677
7: 36817
6: 11765
5: 6270
4: 2999
3: 2527
2: 2523
1: 23012

you get a raw average rating of 9.109131.

Plugging that into the given formula,

WR = (1001818 ÷ (1001818 + 25000)) × 9.109131 + (25000 ÷ (1001818 + 25000)) × 7.0

however, does not give me 9.3. It gives me 9.062. Am I missing something (like more clarifying parentheses)? Or is the formula incorrect? IMDb itself does say:

IMDb publishes weighted vote averages rather than raw data averages. Various filters are applied to the raw data in order to eliminate and reduce attempts at 'vote stuffing' by individuals more interested in changing the current rating of a movie than giving their true opinion of it.

The exact methods we use will not be disclosed. This should ensure that the policy remains effective. The result is a more accurate vote average.

But if that's true, why publish a formula at the bottom of the Top 250 page at all? For disinformation? Just curious. Thanks in advance. Kingsfold (Quack quack!) 13:52, 16 July 2013 (UTC)[reply]

They probably discard multiple votes from one location, maybe with a timeout. They might also do things like rate limit votes so they aren't so affected by some star telling his followers that something is good or bad. Nothing wrong with that and it makes for a better figure. You don't count multiple votes in elections. There is also the possibility that they have changed their formula and not updated the page about it. For instance it looks like their formula is trying to regress the value to the average unless there are lots of votes, I do something similar with something but use the square root of the number instead of a base figure like 25000, but their way may be better for what they're rating instead of what I'm doing. Dmcq (talk) 15:07, 16 July 2013 (UTC)[reply]

Partial derivative notation

Hi, I visited your website at http://en.wikipedia.org/wiki/Partial_derivative. And I found the notations for partial derivative of a function which can be expressed into 4 different terms. I'm just wondering, if you could provide me some reference books or additional information that describe the 2nd term i.e., "f,x". This expression can be found in the 1st set of equation from your website. I look forward to hearing from you. GER 119.236.121.156 (talk) 18:00, 16 July 2013 (UTC)[reply]

This seems to be a fairly unusual notation for the partial derivative. Much more typical than

f_{,x}

is

f_{x}

. I've changed the article to reflect this. In some differential geometry and general relativity textbooks, you can find notation for the partial derivative that looks like

f_{,i}

, meaning the partial derivative of f with respect to the ith coordinate. But I think you'll rarely see (if ever)

f_{,x}

used. Sławomir Biały (talk) 18:49, 16 July 2013 (UTC)[reply]

The comma, by the way, is used in differential geometry and general relativity to distinguish the partial derivative from the covariant derivative for which one uses a semicolon. YohanN7 (talk) 21:20, 16 July 2013 (UTC)[reply]

Yes, you may wish to look at Ricci calculus#Differentiation. — Quondum 15:36, 17 July 2013 (UTC)[reply]