<t>Quote1. The article is implicitly suggesting that not all what-if cases have been taken care of; is there a nasty parameter value lurking in the wings?No nasty parameters that I am aware of.Quote2. The algorithm is iterative (== recursive?) so convergence is not deterministic nor always assured? ...

<r>QuoteOriginally posted by: AVtYes, Genz. Or Vasicek. I have my own. Or <URL url="http://arxiv.org/abs/1004.3616">http://arxiv.org/abs/1004.3616</URL> - which I would prefer.Many thanks... :)An extended version has been published in JSS: <URL url="http://www.jstatsoft.org/v52/i10">http://www.jstat...

SPOILER: Hadwiger-Nelson problem. Both the Moser Spindle and Solomon W. Golomb's ten-vertex graph (cf. Wikipedia) fit into the unit radius circular disk. Both require four colors.

Which models? Pricing models, or risk models, or models in general?

<t>First of all, consider the EBA Guidelines on IRC. At the end of the day, IRC is a regulatory exercise, and some academically appealing concepts will not be allowed anyway. For instance, you will not be allowed to generate hypothetical spreads (what will be the spread for a CCC-rated Pfandbrief?) ...

mod p:[$](p-3)^{p-1} - (p-1)^{p-3} = (-3)^{p-1} - (-1)^{p-3} = 3^{p-1} - 1 = 3^p / 3 - 1 = 3 / 3 - 1 = 0[$]in step 2: p is odd and p > 3in step 3: p > 3in step 4: Fermat

Yes, 106 is correct. You can check this by hand, by observing that for 10 <= n <= 99, a[n] = n(n+1)/2 + 4 (mod 11), which is never zero (mod 11).

A little easier: find, not using a computer, the smallest n such that a[n] is divisible by 11.

How dou you measure "size"? n*m, or n+m, or min(m,n), or max(m,n), or...?What's a submatrix? A block, or any (subset of rows)x(subset of columns)?

0.5m; or didn't I get the joke?

Basically, this is Expected Shortfall for a standard normal variable. This is dealt with, for example, in Embrechts/Frey/McNeil - Quantitative Risk Measurement, Example 2.18.

<t>The number of 10-digit numbers modulo reordering of digits (i.e., the number of orbits of the action of the symmetric group S_10 by permutation of digits) is equal to the number of monomials of degree 10 in 10 variables, i.e. C(19,10) = 92378.Therefore, by the pigeon hole principle, there must be...

<t>QuoteProbability of permutation cycle...is all u need to know...which gives answer > 30%Let me see... the probability of NOT everybody finding his ring is then equal to the fraction of permutations with a cycle of length > 50, which isNice!Just for completeness: in my variant of the problem, wher...

<t>A variant to the problem: what if people are allowed to peek into n-1 boxes? The probability of all people finding their rings can then be made at least ( (n-1)/n) )^n, which converges to 1/e.Again, one can do slightly better by considering permutations. If everybody peeks into all boxes but the ...

<t>QuoteOriginally posted by: animeshsaxenaThe worst is (1/2)^50.Why that? Do you mean (1/2)^100?In the general case with n people, n boxes, and n/2 peeks per person, one can do slightly better than (1/2)^n:- people with number <= n/2 peek into boxes 1, ..., n/2- people with number > n/2 peek into b...