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For x(E [A]) = x(1/2), we find 11 T∞11 T x(E[A])= v+ [0 1/6 5/6] +∑( − )[0 0 1] n=2 2n 2n+1 (4.15) Thus, for this example, E [x(A)] =/ x(E [A]). For this case, the RAPr solution satisfies eT [ 1/6 7/36 23/36 ] = 1. This prop- erty is general and we next show that the vector E [x(A)] is always a proba- bility distribution. Corollary 11. If A ∼ Beta(a, b, [l , r]) with 0 ≤ l < r ≤ 1 and probability density function ρ, then E [xi (A)] > 0 and ∥E [x(A)]∥ = 1. Proof. First,E[xi(A)]≥0isbecause0≤A≤1andvi ≥0.Then,wehave ∥E[x(A)]∥ = eT 1 x(α)ρ(α)dα = 1 eTx(α)ρ(α)dα = 1, (4.16) 00 because eT x = 1 for each α and ∫ 1 ρ(α) dα = 1. 0 Finally, we show that for a certain class of pages, the expectation of RAPr is equal to PageRank with α = E [A]. Theorem12. LetA∼Beta(a,b,[l,r])with0≤lPDF Image | MODELS AND ALGORITHMS FOR PAGERANK SENSITIVITY
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