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4.4 ⋅ the random alpha pagerank model 75 To apply theorem 9, we need E [An ]. If A ∼ Beta(0, 0, [0, 1]), then A is uniform over[0,1]andE[An]= 1 .Finally, n+1 11T∞11T E[x(A)]= v+ [0 1/6 5/6] +∑( − )[0 0 1] 2 2 = [1/6 7/36 23/36] For x(E [A]) = x(1/2), we find T n=2 n+1 n+2 . 11 T∞11 T x(E[A])= v+ [0 1/6 5/6] +∑( n − n+1)[0 0 1] 2 4 = [1/6 5/24 5/8] T n=2 2 2 . 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)] ≥ 0 is because 0 ≤ A ≤ 1 and vi ≥ 0. Then, we have ∥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≤l0,wheree isthevector x(α)=(1−α)v aswell. iiii with a 1 in the ith component. Taking the Neumann series for x(A) gives ∞ x (A)=eT ∑(Aj −Aj+1)Pjv=eT(A0 −A1)v=(1−A)v . (4.17) iiii j=0 Equality of the statistics follows from the linearity of the expectation opera- tor. (4.14) (4.15) When a state i in P has no

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