Joint Probability Distribution Average Degree <k>=2M/N Degree Distribution (k)=m(k)/N(k)=∑Pk) Joint Probability Distribution: prob. to find a node with degree j and k at the two ends of a randomly selected link PU,k) (j,k) 2M ≠A)(,)=m(元,j P(,k)=m(,k)(jk) 2M
Joint Probability Distribution = k M N 2 / P k n k N ( ) ( ) / = 0 ( ) k k kP k = = ( , ) ( , ) ( ) 2 m j k P j k j k M = Joint Probability Distribution: prob. to find a node with degree j and k at the two ends of a randomly selected link Average Degree Degree Distribution ( , ) ( , ) ( , ) 2 m j k j k P j k M = ( , ) ( , ) m j j P j j M =
Degree Correlation P(k)=P.)∑PA P2(k) P(,k) Excess Degree Dis Pn(k): prob. to have a degree k node at the end of a link P(k) qk= P(k) ek=P(, k) q k k If the network has no degree correlations(neutral) j q14 M.E.J. Newman, Phys. Rev. Lett. 89, 208701(2002)
( ) k p P k ( ) k n q P k ( , ) jk e P j k max min k k jk k j k k k p e q k k = = = If the network has no degree correlations (neutral): M. E. J. Newman, Phys. Rev. Lett. 89, 208701 (2002) jk j k e q q = P j k P k j ( , ) ( , ) = max min , ( , ) 1 k j k k P j k = = max min ( ) ( , ) k n j k P k P j k = = Excess Degree Dis. Pn (k): prob. to have a degree k node at the end of a link
Assortativity Disassortativity Assortative. Neutral Disassortative: hubs tend to link to Hubs tend to avoid each other linking to each other
Assortative : hubs tend to link to each other. Neutral Disassortative: Hubs tend to avoid linking to each other
Assortativity Full Statistical Description Astrophysics co-authorship network Yeast PPl 250300350 200 98765432 ik 05050504 0.005 0.00 0.002 0.001 Assortative Disassortative
jk e Astrophysics co-authorship network Yeast PPI Assortative Disassortative
Problem with Full Statistical Description (1)Difficult to extract (2) Requires a large number of information from a visual elements to inspect Inspection of a matrIX. Nr of max max 1-k independent max e/ements Undirected network max XK max matrix Constraints We need to find a way to reduce the information contained in eik E.J. Newman, Phys. Rev. Lett. 89, 208701(2002)
M. E. J. Newman, Phys. Rev. Lett. 89, 208701 (2002) Undirected network: kmax x kmax matrix Nr. of independent elements Constraints max max ( ) max 1 1 2 k k k − − − , 1 jk j k e = max 1, jk k j k e q = = (2) Requires a large number of elements to inspect: (1) Difficult to extract information from a visual inspection of a matrix. We need to find a way to reduce the information contained in ejk