<t>The claim is wrong, I am sorry. I got a counterexample where the sum grows as sqrt( log(k) ). This counterexample can be constructed as follows: Simply fill only sqrt( log( k ) ) of the columns, and let the other ones empty. Then we only obtain cost for these sqrt( log(k) ) many columns. We can a...

<t>QuoteOriginally posted by: Justin2000....My guess is that the matrix giving the extreme value for the sum on the left hand side is of the form m_j=(0,...,0,1/j,1/(j+1),...,1/k) (i think if it's true the induction wrt k might be used to prove it)...For this choice of the matrix, the sum is 7/6 for...

Given is a real positive lower triangular matrix with sum 1 in each row. Proof that I do not have a solution, and I think that this question is quite hard.

<r>Concerning the response of pmtpr:Sure it is possible to express this process in terms of such an equation and then let N go to infinity. The following page gives a nice introduction to this topic: <URL url="http://www.mathpages.com/home/kmath084/kmath084.htm">http://www.mathpages.com/home/kmath08...

<t>First, yes, we can assume that x and y are integers.Concerning the reply of betso:I think that the setting described in the paper you posted is quite different: They assume that a constant fraction of the current fortune is invested in each bet. Hence, there is a "relative" change in each round a...

<t>Consider the following game: you start with one unit, and in each round, you win one unit with probability p and you loose one unit with probability 1-p.This example of a random walk is listed in basically every book about probability theory, and it is well-known that the ruin probability is < 1 ...

<t>I am not very familiar with covariance matrixes. But it is probably not possible to write every symmetric matrix as a covariance matrix of some random values?However, I think that it is definitely possible to write every symmetric matrix M in the form A B A^t, where B is diagonal and contains the...

<t>For those who care about computational complexity, this problem is NP-hard.This can be shown by a simple reduction of maximum clique: Assume that the correlation matrix contains only 0s ans 1s. Then this matrix is basically the adjacency matrix of an undirected graph of the size of the matrix. In...