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In a simple graph, G=(V, E), for each pair of distinct vertices u,v, edge(u,v) appears with probability 1/2. A triangle is a cycle of length three. X is a random variable denoting the number of triangles in a random graph G. Given the expected value of X,
E(X) = (n choose 3)*(1/8)
for n = # of nodes, and variance,
Var(X) = (n choose 3) * (n +5) / 64
give an upper bound using Chernoff bounds on the probability that X exceeds its expected value by 10%. Give general expression in terms of n. (Keep in mind that X is the sum of DEPENDENT random variables, as each triangle's probability of being present is affect by those around it. i.e. if triangle (u, v, w) is present, triangle (u, v, x) has a high probability of also being present as one of its edges is already known to exist. If this assumption is wrong, please explain why.)

2007-01-21 12:48:05 · 1 answers · asked by Anonymous in Science & Mathematics Mathematics

1 answers

I do not see how you can use Chernoff here since Chernoff only works for a random variable that is the sum of independent Bernouli trials -- and this is not the case here. However, since you have the variance, you should be able to use a Chebyshev bound.

2007-01-25 07:40:06 · answer #1 · answered by Phineas Bogg 6 · 0 0

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