Click efficiency: a unified optimal ranking for online Ads and documents

Abstract

Ranking of search results and ads has traditionally been studied separately. The probability ranking principle is commonly used to rank the search results while the ranking based on expected profits is commonly used for paid placement of ads. These rankings try to maximize the expected utilities based on the user click models. Recent empirical analysis on search engine logs suggests unified click models for both ranked ads and search results (documents). These new models consider parameters of (i) probability of the user abandoning browsing results (ii) perceived relevance of result snippets. However, current document and ad ranking methods do not consider these parameters. In this paper we propose a generalized ranking function—namely Click Efficiency (CE)—for documents and ads based on empirically proven user click models. The ranking considers parameters (i) and (ii) above, optimal and has the same time complexity as sorting. Furthermore, the CE ranking exploits the commonality of click models, hence is applicable for both documents and ads. We examine the reduced forms of CE ranking based upon different underlying assumptions, enumerating a hierarchy of ranking functions. Interestingly, some of the rankings in the hierarchy are currently used ad and document ranking functions; while others suggest new rankings. Thus, this hierarchy illustrates the relationships between different rankings, and clarifies the underlying assumptions. While optimality of ranking is sufficient for document ranking, applying CE ranking to ad auctions requires an appropriate pricing mechanism. We incorporate a second price based mechanism with the proposed ranking. Our analysis proves several desirable properties including revenue dominance over Vickrey Clarke Groves (VCG) for the same bid vector and existence of a Nash equilibrium in pure strategies. The equilibrium is socially optimal, and revenue equivalent to the truthful VCG equilibrium. As a result of its generality, the auction mechanism and the equilibrium reduces to the current mechanisms including Generalized Second Price Auction (GSP) and corresponding equilibria. Furthermore, we relax the independence assumption in CE ranking and analyze the diversity ranking problem. We show that optimal diversity ranking is NP-Hard in general, and a constant time approximation algorithm is not likely. Finally our simulations to quantify the amount of increase in different utility functions conform to the results, and suggest potentially significant increase in utilities.

Appendix

1.1 A-1 Proof of theorem 1

Theorem 1

The expected utility in ( 3 ) is maximum if the entities are placed in the descending order of the value of the ranking function CE,

$$CE(e_{i})=\frac{U(e_{i})C(e_{i})}{C(e_{i})+\gamma(e_{i})} $$

Proof

Consider results e _i and e _{i + 1} in positions i and i + 1 respectively. Let μ _i = γ(e _i )+C(e _i ) for notational convenience. The total expected utility from e _i and e _{i + 1} when e _i is placed above e _{i + 1} is

$$\prod\limits_{j=1}^{i-1}{(1-\mu_{j})}\left[ U(e_{i}) C(e_{i}) + (1- \mu_{i}) U(e_{i+1}) C(e_{i+1})\right] $$

If the order of e _i and e _{i + 1} are inverted by placing e _i above e _{i + 1}, the expected utility from these entities will be,

$$\prod\limits_{j=1}^{i-1}(1-\mu_{j})\left[\left. U(e_{i+1}) C(e_{i+1}) + (1- \mu_{i+1}) U(e_{i}) C(e_{i})\right)\right] $$

Since utilities from all other results in the list will remain the same, the expected utility of placing e _i above e _{i + 1} is greater than inverse placement iff

$$\begin{array}{@{}rcl@{}} U(e_{i}) C(e_{i}) + (1- \mu_{i}) U(e_{i+1}) C(e_{i+1}) &\ge& U(e_{i+1}) C(e_{i+1}) + (1- \mu_{i+1}) U(e_{i}) C(e_{i}) \\ & \Updownarrow & \\ \frac{U(e_{i}) C(e_{i})}{\mu_{i}} &\ge& \frac{U(e_{i+1}) C(e_{i+1})}{\mu_{i+1}} \end{array} $$

This means if entities are ranked in the descending order of $\frac {U(e) C(e)}{C(e)+\gamma (e)}$ any inversions will reduce the profit. Since any arbitrary order can be effected by a number of inversions on the ranking by CE, this implies that ranking by $\frac {U(e) C(e)}{C(e)+\gamma (e)}$ is optimal. □

1.2 A-2 Proof of theorem 2

Theorem 2

The order proposed in Theorem 1 is optimal for multiple clicks if the user restarts browsing at the position one below the last clicked entity.

Proof

Induction on number of clicks.

Base Case::

Single click, proved in Theorem 1.

Inductive Hypothesis::

The proposed ordering is optimal for n clicks.

Let there be total of n ranked entities and e _c be the n ^th clicked entity. The user will browse down starting next to e _c . Since there is only one click remaining, optimal ordering of entities is in the descending order of CE by the base case. Since the relevance and abandonment probabilities e _{c + 1} to e _n remain unchanged by the independence assumption above, the optimal sequence will be the sub-sequence of e _{c + 1} to e _n in the ranking. □

1.3 A-3 Proof of theorem 3

Theorem 3

The order by w _i b _i is the same as the order by CE _i for the auction i.e.

$$w_{i} b_{i} \ge w_{j}b_{j} \Longleftrightarrow CE_{i} \ge CE_{j} $$

Proof

Without loss of generality, we assume that a _i refers to ad in the position i in the descending order of w _i b _i .

$$\begin{array}{@{}rcl@{}} CE_{i}& = & \frac{p_{i}c_{i}}{\mu_{i}} \\ & = & \frac{b_{i+1}c_{i+1} \mu_{i} }{\mu_{i+1} c_{i}} \frac{c_{i}}{\mu_{i}} \\ & = & \frac{b_{i+1}c_{i+1}}{\mu_{i+1}} \\ & = & w_{i+1}b_{i+1} \\ & \ge & w_{i+2}b_{i+2} = CE_{i+1} \end{array} $$

□

1.4 A-4 Proof of theorem 4

Theorem 4 (Nash Equilibrium)

: Without the loss of generality assume that the advertisers are ordered in the decreasing order of $\frac {c_{i}v_{i}}{\mu _{i}}$ where v _i is the private value of the i ^th advertiser. The advertisers are in a pure strategy Nash Equilibrium if

$$b_{i} = \frac{\mu_{i}}{c_{i}}\left[ v_{i} c_{i} + (1-\mu_{i}) \frac{b_{i+1}c_{i+1}}{\mu_{i+1}} \right] $$

This equilibrium is socially optimal for advertisers as well as optimal for search engines for the given CPC’s.

Proof

Let there are n advertisers. Without loss of generality, let us assume that advertisers are indexed in the descending order of $\frac {v_{i} c_{i}}{\mu _{i}}$. We prove equilibrium in two steps.

Step 1: :

Prove that

$$w_{i} b_{i} \ge w_{i+1}b_{i+1} $$

(1)

$$w_{i} b_{i} = \frac{b_{i} c_{i}}{\mu_{i}} $$

Expanding b _i by (19),

$$\begin{array}{@{}rcl@{}} w_{i} b_{i} & = & v_{i}c_{i}+(1-\mu_{i})\frac{b_{i+1}c_{i+1}}{\mu_{i+1}} \\ & = & v_{i}c_{i}+(1-\mu_{i})w_{i+1}b_{i+1} \\ & = & \frac{v_{i}c_{i}}{\mu_{i}} \mu_{i}+(1-\mu_{i})w_{i+1} b_{i+1} \end{array} $$

Notice that w _i b _i is a convex linear combination of w _{i + 1} b _{i + 1} and $\frac {v_{i}c_{i}}{\mu _{i}}$. This means that the value of w _i b _i is in between (or equal to) the values of w _{i + 1} b _{i + 1} and $\frac {v_{i}c_{i}}{\mu _{i}}$. Hence to prove that w _i b _i ≥w _{i + 1} b _{i + 1} all we need to prove is that $\frac {v_{i}c_{i}}{\mu _{i}} \ge w_{i+1}b_{i+1}$. This inductive proof is given below.

Induction hypothesis: :

Assume that

$$\forall_{i \ge j } \frac{v_{i}c_{i}}{\mu_{i}} \ge w_{i+1} b_{i+1}$$

Base case: :

Prove for i = N i.e. for the bottommost ad.

$$\frac{v_{N-1} c_{N-1}}{\mu_{N-1}} \ge w_{N} b_{N}$$

Assuming ∀ _i>N b _i = 0

$$w_{N}b_{N} = v_{N} c_{N} \le \frac{v_{N} c_{N}}{\mu_{N}} (\text{as}~\mu_{N} \le 1) \!\le\! \frac{v_{N-1}c_{N-1}}{\mu_{N-1}}~(\text{by the assumed order i.e. by}~\frac{v_{i} c_{i}}{\mu_{i}})$$

Induction: :

Expanding w _j b _j by (19),

$$w_{j} b_{j} = \frac{v_{j} c_{j}}{\mu_{j}} \mu_{j}+(1-\mu_{j})w_{j+1} b_{j+1} $$

w _j b _j is the convex linear combination, i.e $\frac {v_{j} c_{j}}{\mu _{j}} \ge w_{j}b_{j} \ge w_{j+1} b_{j+1}$, as we know that $\frac {v_{j}c_{j}}{\mu _{j}} \ge w_{j+1}b_{j+1}$ by induction hypothesis. Consequently,

$$w_{j} b_{j} \le \frac{v_{j} c_{j}}{\mu_{j}} \le \frac{v_{j-1}c_{j-1}}{\mu_{j-1}}~\text{ (by the assumed order)}$$

This completes the induction.

□

Since advertisers are ordered by w _i b _i for pricing, the above proof says that the pricing order is the same as the assumed order in this proof (i.e. ordering by $\frac {v_{i} c_{i}}{\mu _{i}}$). Consequently,

$$p_{i} = \frac{b_{i+1}c_{i+1} \mu_{i}}{\mu_{i+1}c_{i}} $$

As corollary of Theorem 3 we know that C E _i ≥C E _{i + 1}.

In the second step we prove the equilibrium using results in Step 1.

Step 2: :

No advertiser can increase his profit by changing his bids unilaterally

Proof (of lack of incentive to undercut to advertisers below)

In the first step let us prove that ad a _i can not increase his profit by decreasing his bid to move to a position j≥i below.

Inductive hypothesis: :

Assume true for i ≤ j ≤ m.

Base Case: :

Trivially true for j = i.

Induction: :

Prove that the expected profit of a _i at m + 1 is less or equal to the expected profit of a _i at i.

Let ρ _k denotes the amount paid by a _i when he is at the position k. By inductive hypothesis, the expected profit at m is less or equal to the expected profit at i. So we just need to prove that the expected profit at m + 1 is less or equal to the expected profit at m. i.e.

$$\frac{(v_{i} - \rho_{m})}{(1-\mu_{i})}\prod\limits_{l=1}^{m}(1-\mu_{l}) \ge \frac{(v_{i} - \rho_{m+1})}{(1-\mu_{i})}\prod\limits_{l=1}^{m+1}(1-\mu_{l}) $$

Canceling the common terms,

$$ v_{i} - \rho_{m} \ge (v_{i} - \rho_{m+1})(1 - \mu_{m+1}) $$

(2)

ρ _m —the price charged to a _i at position m—is based on the Equations 16 and 19. Since the a _i is moving downward, a _i will occupy position m by shifting ad a _m upwards. Hence the ad just below a _i is a _{m + 1}. Consequently, the price charged to a _i when it is at the m ^th position is,

$$\rho_{m} = \frac{b_{m+1}c_{m+1}\mu_{i}}{\mu_{m+1}c_{i}} = \frac{\mu_{i}}{c_{i}}\left[ v_{m+1}c_{m+1}+ (1-\mu_{m+1})\frac{b_{m+2}c_{m+2}}{\mu_{m+2}} \right] $$

Substituting for ρ _m and ρ _{m + 1} in (2),

$$\begin{array}{@{}rcl@{}} &&v_{i} \! - \! \frac{\mu_{i}}{c_{i}}\left[v_{m+1}c_{m+1}+ (1-\mu_{m+1})\frac{b_{m+2}c_{m+2}}{\mu_{m+2}}\right]\\ &&{\kern1pc}\ge\!\! \left( v_{i}-\frac{\mu_{i}}{c_{i}}\left[v_{m+2}c_{m+2}+ \frac{}{} (1-\mu_{m+2})\frac{b_{m+3}c_{m+3}}{\mu_{m+3}}\right]\right)\!(1 \! -\!\mu_{m+1}) \end{array} $$

Simplifying, and multiplying both sides by −1

$$\begin{array}{@{}rcl@{}} &&\frac{\mu_{i}}{c_{i}}\left[v_{m+1}c_{m+1}+(1-\mu_{m+1})\frac{b_{m+2}c_{m+2}}{\mu_{m+2}}\right] \le v_{i}\mu_{m+1}+ \frac{\mu_{i}}{c_{i}} (1 - \mu_{m+1})\\&&{\kern1pc}\times \left[\frac{}{}v_{m+2}c_{m+2}+ (1-\mu_{m+2})\frac{b_{m+3}c_{m+3}}{\mu_{m+3}}\right] \end{array} $$

Substituting by b _{m + 2} from (19) on RHS.

$$\frac{\mu_{i}}{c_{i}}\left[v_{m+1}c_{m+1}+(1-\mu_{m+1})\frac{b_{m+2}c_{m+2}}{\mu_{m+2}}\right] \le v_{i}\mu_{m+1}+ \frac{\mu_{i}}{c_{i}} (1 - \mu_{m+1}) \frac{b_{m+2} c_{m+2}}{\mu_{m+2}} \\ $$

Canceling out the common terms on both sides,

$$\begin{array}{@{}rcl@{}} \frac{\mu_{i}}{c_{i}} v_{m+1}c_{m+1}& \le &v_{i}\mu_{m+1} \\ &\Updownarrow& \\ \frac{v_{m+1}c_{m+1}}{\mu_{m+1}} & \le & \frac{v_{i}c_{i}}{\mu_{i}} \end{array} $$

Which is true by the assumed order as m≥i □

Inductive proof for m ≤ i is somewhat similar and enumerated below.

Inductive hypothesis: :

Assume true for j ≤ m.

Base Case: :

Trivially true for j = i.

Proof (of lack of incentive to overbid ad one above)

The case in which a _i increases his bid to move one position up i.e. to i − 1 is a special case and need to be proved separately. In this case, by moving a single slot up, the index of the ad below a _i will change from i + 1 to i − 1 (a difference of two). For all other movements of a _i to a position one above or one below, the index of the advertisers below will change only by one. Since the amount paid by a _i depends on the ad below a _i , this case warrants a slightly different proof,

$$\begin{array}{@{}rcl@{}} (v_{i} - \rho_{i})\prod\limits_{l=1}^{i-1}(1-\mu_{l}) & \ge & (v_{i} - \rho_{m-1})\prod\limits_{l=1}^{i-2}(1-\mu_{l}) \\ &\Updownarrow& \\ (v_{i} - \rho_{i})(1-\mu_{i-1}) & \ge & v_{i} - \rho_{i-1} \end{array} $$

Expanding ρ _i is straight forward.To expand ρ _{i − 1}, note that when a _i has moved upwards to i − 1, the ad just below a _i is a _{i − 1}. Since a _{i − 1} has not changed its bids, the ρ _{i − 1} can be expanded as $\frac {\mu _{i}}{c_{i}}\left [ v_{i-1}c_{i-1}+ (1-\mu _{i-1})\frac {b_{i}c_{i}}{\mu _{i}}\right ]$. Substituting for ρ _i and ρ _{i − 1},

$$\begin{array}{@{}rcl@{}} \left( v_{i} \,-\, \frac{\mu_{i}}{c_{i}}\left[v_{i+1}c_{i+1}+ \ge v_{i}\,-\,\frac{\mu_{i}}{c_{i}}\left[ v_{i-1}c_{i-1}+ (1\,-\,\mu_{i+1})\frac{b_{i+2}c_{i+2}}{\mu_{i+2}}\right] \right) (1-\mu_{i-1}) (1-\mu_{i-1})\frac{b_{i}c_{i}}{\mu_{i}}\right] \end{array} $$

Simplifying and multiplying by −1

$$\begin{array}{@{}rcl@{}} v_{i}\mu_{i-1}+\frac{\mu_{i}}{c_{i}}\left[v_{i+1}c_{i+1} + \le \frac{\mu_{i}}{c_{i}} \left[v_{i-1}c_{i-1}+ (1-\mu_{i-1})\frac{b_{i}c_{i}}{\mu_{i}}\right] (1-\mu_{i+1})\frac{b_{i+2}c_{i+2}}{\mu_{i+2}}\right] (1-\mu_{i-1}) \end{array} $$

Substituting b _{i + 1} from (19)

$$\begin{array}{@{}rcl@{}} v_{i}\mu_{i-1}+\frac{\mu_{i}}{c_{i}} \frac{b_{i+1}c_{i+1}}{\mu_{i+1}} (1-\mu_{i-1})& \le & \frac{\mu_{i}}{c_{i}} \left[v_{i-1}c_{i-1}+ (1-\mu_{i-1})\frac{b_{i}c_{i}}{\mu_{i}}\right] \\ &\Updownarrow& \\ v_{i}\mu_{i-1}+\frac{\mu_{i}}{c_{i}} (1-\mu_{i-1}) \frac{b_{i+1}c_{i+1}}{\mu_{i+1}} & \le & \frac{\mu_{i} v_{i-1}c_{i-1} }{c_{i}} + \frac{\mu_{i}}{c_{i}}(1-\mu_{i-1})\frac{b_{i}c_{i}}{\mu_{i}} \end{array} $$

We now prove that both the terms in RHS are greater or equal to the corresponding terms in LHS separately.

$$\begin{array}{@{}rcl@{}} v_{i}\mu_{i-1} & \le & \frac{\mu_{i} v_{i-1}c_{i-1} }{c_{i}} \\ & \Updownarrow & \\ \frac{v_{i}c_{i}}{\mu_{i}} & \le & \frac{ v_{i-1}c_{i-1} }{\mu_{i-1}} \end{array} $$

Which is true by our assumed order.

Similarly,

$$\begin{array}{@{}rcl@{}} \frac{\mu_{i}}{c_{i}} (1-\mu_{i-1}) \frac{b_{i+1}c_{i+1}}{\mu_{i+1}} & \le & \frac{\mu_{i}}{c_{i}}(1-\mu_{i-1})\frac{b_{i}c_{i}}{\mu_{i}} \\ & \Updownarrow & \\ \frac{b_{i+1}c_{i+1}}{\mu_{i+1}} & \le & \frac{b_{i}c_{i}}{\mu_{i}} \end{array} $$

Which is true by (1) above. This completes the proof for this case. □

Induction: :

Prove that the expected profit at m − 1 is less or equal to the expected profit at m. The proof is similar to the induction for the case m>i.

Proof

Base case is trivially true.

$$(v_{i} - \rho_{m})\prod\limits_{l=1}^{m-1}(1-\mu_{l}) \ge (v_{i} - \rho_{m-1})\prod\limits_{l=1}^{m-2}(1-\mu_{l}) $$

Canceling common terms,

$$(v_{i} - \rho_{m})(1-\mu_{m-1}) \ge v_{i} - \rho_{m-1} $$

In this case, note that a _i is moving upwards. This means that a _i will occupy position m by pushing the ad originally at m one position downwards. Hence the original ad at m is the one just below a _i now. i.e.

$$\rho_{m} = \frac{b_{m}c_{m}\mu_{i}}{\mu_{m}c_{i}} = \frac{\mu_{i}}{c_{i}}\left[ v_{m}c_{m}+ (1-\mu_{m})\frac{b_{m+1}c_{m+1}}{\mu_{m+1}} \right] $$

Substituting for ρ _m and ρ _{m − 1}

$$\begin{array}{@{}rcl@{}} \left( v_{i} \,-\, \frac{\mu_{i}}{c_{i}}\left[v_{m}c_{m}+ \ge v_{i}\,-\,\frac{\mu_{i}}{c_{i}}\left[ v_{m-1}c_{m-1}+ (1\,-\,\mu_{m})\frac{b_{m+1}c_{m+1}}{\mu_{m+1}}\right] \right) (1\,-\,\mu_{m-1}) (1-\mu_{m-1})\frac{b_{m}c_{m}}{\mu_{m}}\right] \end{array} $$

Simplifying and multiplying by −1

$$\begin{array}{@{}rcl@{}} v_{i}\mu_{m-1}\,+\,\frac{\mu_{i}}{c_{i}}\left[v_{m}c_{m} + \le \frac{\mu_{i}}{c_{i}} \left[v_{m-1}c_{m-1}+ (1\,-\,\mu_{m-1})\frac{b_{m}c_{m}}{\mu_{m}}\right] (1-\mu_{m})\frac{b_{m+1}c_{m+1}}{\mu_{m+1}}\right] (1-\mu_{m-1}) && \end{array} $$

Substituting by b _m from (19)

$$v_{i}\mu_{m-1}+\frac{\mu_{i}}{c_{i}}\frac{b_{m}c_{m}}{\mu_{m}} (1-\mu_{m-1}) \le \frac{\mu_{i}}{c_{i}} \left[v_{m-1}c_{m-1}+ (1-\mu_{m-1})\frac{b_{m}c_{m}}{\mu_{m}}\right] $$

Canceling common terms,

$$\begin{array}{@{}rcl@{}} v_{i}\mu_{m-1} & \le &\frac{\mu_{i}}{c_{i}} v_{m-1}c_{m-1} \\ &\Updownarrow& \\ \frac{v_{i}c_{i}}{\mu_{i}} & \le & \frac{v_{m-1}c_{m-1}}{\mu_{m-1}} \end{array} $$

Which is true by the assumed order as m<i. □

1.5 A-5 Proof of theorem 5

Theorem 5 (Search Engine Revenue Dominance)

: For the same bid values for all the advertisers, the revenue of search engine by CE mechanism is greater or equal to the revenue by VCG.

Proof

VCG payment of the ad at position i (i.e. a _i ) is equal to the reduction in utility of the ads below due to the presence of a _i . For each user viewing the list of ads (i.e. for unit view probability), the total expected loss of ads below a _i due to a _i is,

$$\begin{array}{@{}rcl@{}} p_{i}^{V_{u}} & = & \frac{1}{1-\mu_{i}}\sum\limits_{j=i+1}^{n} b_{j} c_{j}\prod\limits_{k=1}^{j-1}(1-\mu_{k}) - \sum\limits_{j=i+1}^{n} b_{j} c_{j}\prod\limits_{k=1}^{j-1}(1-\mu_{k}) \\ & = & \frac{\mu_{i}}{1-\mu_{i}}\sum\limits_{j=i+1}^{n} b_{j} c_{j}\prod\limits_{k=1}^{j-1}(1-\mu_{k}) \\ & = & \frac{\mu_{i}}{1-\mu_{i}} \prod\limits_{k=1}^{i}(1-\mu_{k})\sum\limits_{j=i+1}^{n} b_{j}c_{j}{\prod}_{k=i+1}^{j-1}(1-\mu_{k}) \\ & = & \mu_{i} \prod\limits_{k=1}^{i-1}(1-\mu_{k})\sum\limits_{j=i+1}^{n} b_{j}c_{j}{\prod}_{k=i+1}^{j-1}(1-\mu_{k}) \end{array} $$

This is the expected lose per user browsing the ad list. Pay per click should be equal to the lose per click. To calculate the pay per click, we divide by the click probability of a _i . i.e.

$$\begin{array}{@{}rcl@{}} {p_{i}^{V}} & = & \frac{ \mu_{i} {\prod}_{k=1}^{i-1}(1-\mu_{k}){\sum}_{j=i+1}^{n} b_{j}c_{j}{\prod}_{k=i+1}^{j-1}(1-\mu_{k})}{ c_{i} {\prod}{k=1}^{i-1}(1-\mu_{k})} \\ & = & \frac{\mu_{i}}{c_{i}} \sum\limits_{j=i+1}^{n} b_{j}c_{j}\prod\limits_{k=i+1}^{j-1}(1-\mu_{k}) \end{array} $$

Converting to recursive form,

$$\begin{array}{@{}rcl@{}} {p_{i}^{V}} & = & \frac{ b_{i+1} \mu_{i}}{c_{i}} c_{i+1} + (1-\mu_{i+1}) \frac{\mu_{i} c_{i+1}}{c_{i}\mu_{i+1}}p_{i+1}^{V} \\ & = & \frac{b_{i+1} \mu_{i} c_{i+1}}{c_{i} \mu_{i+1}} \mu_{i+1} + (1-\mu_{i+1}) \frac{\mu_{i} c_{i+1}}{c_{i}\mu_{i+1}}p_{i+1}^{V} \end{array} $$

For the CE mechanism payment from (16) is,

$$p_{i}^{CE} = \frac{b_{i+1}c_{i+1}\mu_{i}}{\mu_{i+1}c_{i}} $$

Note that ${p_{i}^{V}}$ is convex combination of $P_{i}^{CE}$ and $\frac {\mu _{i} c_{i+1}}{c_{i}\mu _{i+1}}p_{i+1}^{V}$, and hence is between these two values. To prove that $p_{i}^{CE} \ge {p_{i}^{V}}$ all we need to prove is that $P_{i}^{CE} \ge \frac {\mu _{i} c_{i+1}}{c_{i}\mu _{i+1}}p_{i+1}^{V} \Leftrightarrow b_{i} \ge {p_{i}^{V}}$. This directly follows from individual rationality property of VCG. Alternatively, a simple recursion with base case as ${p_{N}^{V}}=0$ (bottommost ad) will prove the same. Note that we consider only the ranking (not selection), and hence the VCG pricing of the bottommost ad in the ranking is zero. □

1.6 A-6 Proof of theorem 6

Theorem 6 (Equilibrium Revenue Equivalence)

: At the equilibrium in Theorem 4, the revenue of search engine is equal to the revenue of the truthful dominant strategy equilibrium of VCG.

Proof

Rearranging (3) and substituting true values for bid amounts,

$$\begin{array}{@{}rcl@{}} {p_{i}^{V}}& = & \frac{\mu_{i}}{c_{i}} \left[ v_{i+1} c_{i+1} + \frac{(1-\mu_{i+1})c_{i+1}}{\mu_{i+1}}p_{i+1}^{V} \right] \end{array} $$

For the CE mechanism, substituting equilibrium bids from (19) in payment (16),

$$p_{i}^{CE} = \frac{b_{i+1}c_{i+1}\mu_{i}}{\mu_{i+1}c_{i}} = \frac{\mu_{i}}{c_{i}} \left[ v_{i+1} c_{i+1} +(1-\mu_{i+1}) \frac{b_{i+2}c_{i+2}}{\mu_{i+2}} \right] $$

Rewriting b _{i + 2} in terms of p _{i + 1},

$$\begin{array}{@{}rcl@{}} p_{i}^{CE} & = & \frac{\mu_{i}}{c_{i}} \left[ v_{i+1} c_{i+1} +\frac{(1-\mu_{i+1})c_{i+1}}{\mu_{i+1}} p_{i+1}^{CE} \right] \\ & = & {p_{i}^{V}} \;\;\;\;(iff~p_{i+1}^{V}=p_{i+1}^{CE}) \end{array} $$

Ad at the bottommost position pays same amount zero, a simple recursion will prove that the payment for all positions for both VCG and the proposed equilibrium is the same. □

1.7 A-7 Proof of theorem 7

Theorem 7

Diversity ranking optimizing expected utility in ( 22 ) is NP-Hard.

Proof

Independent set problem can be formulated as a ranking problem considering similarities. Consider an unweighed graph G of n vertices {e ₁,e ₂,..e _n } represented as an adjacency matrix. This conversion is clearly polynomial time. Now, consider the values in the adjacency matrix as the similarity values between the entities to be ranked. Let the entities have the same utilities, perceive relevances and abandonment probabilities. In this set of n entities from {e ₁,e ₂,..,e _n }, clearly the optimal ranking will have k pairwise independent entities as the top k entities for a maximum possible value of k. But the set of k independent entities corresponds to the maximum independent set in graph G. □