A Gradient Estimate for PageRank

Abstract

Personalized PageRank has found many uses in not only the ranking of webpages, but also algorithmic design, due to its ability to capture certain geometric properties of networks. In this paper, we study the diffusion of PageRank: how varying the jumping (or teleportation) constant affects PageRank values. To this end, we prove a gradient estimate for PageRank, akin to the Li-Yau inequality for positive solutions to the heat equation (for manifolds, with later versions adapted to graphs).

Appendix

The proof of Theorem 1 includes some rather lengthy computations, and is deferred for the full paper. For the benefit of readers, however, we have included a sketch here which highlights the initial part of the proof where one relates the quantity to be bounded with its own square using CDE.

Proof

(Proof sketch for Theorem 1). Let $\displaystyle {H=\frac{t\varGamma (\sqrt{f})}{\sqrt{f\cdot M}}}$. Fix $t>0$. Let $(x^{\star },t)$ be a point in $V\times \{t \}$ such that H(x, t) is maximized. We desire to bound $H(x^{\star },t)$. Our goal, then is to apply the CDE inequality to $\varDelta (\sqrt{f}H)$. In order to do this, we must ensure that $\varDelta \sqrt{f} < 0$, but a computation shows that if $\varDelta \sqrt{f} \ge 0$, then $H \le \frac{1}{2}$ so this is allowable.

Following this, one computes by bounding the arising $\varDelta \varGamma (\sqrt{f})$ by CDE. One bounds the ensuing terms; clearly $\frac{2}{t}\varGamma (\sqrt{f})-\frac{2}{t}\varGamma \left( \sqrt{f},\frac{u}{\sqrt{f}} \right) \ge -\frac{2}{t}\varGamma \left( \sqrt{f},\frac{u}{\sqrt{f}}\right)$ and by Lemma 3, $\varDelta \sqrt{f} = \left( \frac{f-u}{2t\sqrt{f}} - \frac{\sqrt{M}H}{t}\right) .$ Then one bounds:

$$\begin{aligned} \left( \varDelta \sqrt{f}\right) H&\ge \frac{2}{\sqrt{M}nt}\left( \frac{(f-u)^2}{2\sqrt{f}}-\frac{(f-u)\sqrt{M}H}{\sqrt{f}}+MH^2 \right) \\&~~~~~~~~~~~~~ ~~~~~~~~~~~~ +\frac{2\varGamma (\sqrt{f})}{\sqrt{M}} -\frac{2}{\sqrt{M}}\varGamma \left( \sqrt{f},\frac{u}{\sqrt{f}} \right) \\&\ge \frac{2\left( MH^2-\sqrt{f}\sqrt{M}H \right) }{\sqrt{M}nt} \\&~~~~ -\frac{1}{\sqrt{M}}\widetilde{\sum _{y\sim x}}\left( u(x)\left( 1-\sqrt{\frac{f(y)}{f(x)}}\right) +u(y)\left( 1-\sqrt{\frac{f(x)}{f(y)}}\right) \right) \\&\ge \frac{2\left( MH^2-\sqrt{f}\sqrt{M}H \right) }{\sqrt{M}nt}-\sqrt{M}, \end{aligned}$$

Now one proceeds carefully, noting that we have related H and its square and thus, in principle at least, have recorded an upper bound for H. Now we continue to compute to recover the result.

Remark: In a typical application of the maximum principle, one maximizes over [0, T] and then uses information from the time derivative. Here, we don’t do this. This is important because one obtains an inequality of the form

$$ H^2 \le C_1 \cdot H + C_2 \cdot t $$

Because of the dependence of this inequality on the time where the maximum occurs, if the $t^{\star }$ maximizing the function over all [0, T] is considered, then the result will depend on $t^{\star }$, giving a bound like $\displaystyle H\le \frac{\sqrt{2C_2t^{\star }}}{t}$. However, since we are able to do the computation at t, this problem does not arise.