Incentivizing advertiser networks to submit multiple bids

Abstract

This paper illustrates a method for making side payments to advertiser networks that creates an incentive for the advertiser networks to submit the second-highest bids they received to an ad exchange and simultaneously ensures that the publishers will make more money on average in the short run as a result of adopting this scheme. We also illustrate how this payment scheme affects publisher payoffs in the long run after advertisers have a chance to modify their strategies in response to the changed incentives of the mechanism.

Appendix

Lemma 1

For any fixed strategies of the other bidders, an individual advertiser can always achieve at least as high an expected payoff by making a bid $b = v$ rather than a bid $b < v$.

Proof

Suppose an advertiser with value v changes his bid b from some $b = b^{\prime } < v$ to $b = v$. Note that this change in an advertiser’s bidding strategy only affects the advertiser’s payoff if the advertiser would have the highest bid in the auction when $b = v$ but not when $b = b^{\prime } < v$. In this case, the advertiser’s payoff changes from zero to $v - p$ for some $p \le v$ that reflects the amount the advertiser has to pay for the advertising opportunity. Since the advertiser’s payoff increases from such a change, an advertiser can always achieve at least as high a payoff from making a bid $b = v$ than by making a bid $b < v$. $\square$

Proof of Proposition 3

First we derive necessary and sufficient conditions for there to exist an equilibrium in which all advertisers follow the same strictly monotonic bidding strategy b(v). Suppose there are m advertisers in each advertiser network. Note that if all other advertisers follow the same strictly monotonic bidding strategy b(v) and an advertiser with value v makes a bid of $b(\hat{v})$, then this advertiser obtains an expected payoff of $F(\hat{v})^{m-1} \int _{0}^{\hat{v}} (v - b(y)) (nm - m) f(y) F(y)^{nm-m-1} \; dy$. In order for it to be an equilibrium for all advertisers to follow the bidding strategy b(v), this expected payoff must be maximized when $\hat{v} = v$.

We first show that a sufficient condition for this expected payoff to be maximized when $\hat{v} = v$ is that the derivative of this expected payoff with respect to $\hat{v}$ is zero when $\hat{v} = v$. To see this, note that we can rewrite an advertiser’s expected payoff from making a bid of $b(\hat{v})$ when the advertiser’s value is v as $U(v,\hat{v}) = v q(\hat{v}) - p(\hat{v})$, where $q(\hat{v})$ represents the probability that an advertiser who makes a bid of $b(\hat{v})$ wins the object and $p(\hat{v})$ represents the expected payment for an advertiser who makes a bid of $b(\hat{v})$. $q(\hat{v})$ and $p(\hat{v})$ are both continuously differentiable and strictly increasing in $\hat{v}$. Thus if the derivative of the advertiser’s expected payoff with respect to $\hat{v}$ is zero when $\hat{v} = v$, then $v q^{\prime }(v) - p^{\prime }(v) = 0$. Now $\frac{\partial U(v,\hat{v})}{\partial \hat{v}} = v q^{\prime }(\hat{v}) - p^{\prime }(\hat{v})$ so if $v q^{\prime }(v) - p^{\prime }(v) = 0$ for all v, then $\frac{\partial U(v,\hat{v})}{\partial \hat{v}} > 0$ if $v > \hat{v}$ and $\frac{\partial U(v,\hat{v})}{\partial \hat{v}} < 0$ if $v < \hat{v}$. From this it follows that $U(v,\hat{v})$ is increasing in $\hat{v}$ for $\hat{v} < v$ and decreasing in $\hat{v}$ for $\hat{v} > v$. Thus if the derivative of this expected payoff with respect to $\hat{v}$ is zero when $\hat{v} = v$, then this expected payoff is maximized when $\hat{v} = v$.

Since the first-order conditions are sufficient for a global maximum, it is an equilibrium for all advertisers to follow the bidding strategy b(v) if and only if the derivative of $U(v,\hat{v}) = F(\hat{v})^{m-1} \int _{0}^{\hat{v}} (v - b(y)) (nm - m) f(y) F(y)^{nm-m-1} \; dy$ with respect to $\hat{v}$ is zero when $\hat{v} = v$ or if and only if

$$\begin{aligned}&(m-1) f(v) F(v)^{m-2} \int _{0}^{v} (v - b(y)) f(y) F(y)^{nm-m-1} \; dy\\&\quad +\, (v - b(v)) f(v) F(v)^{nm-2} = 0 \\&\quad \Leftrightarrow (m-1) \int _{0}^{v} (v - b(y)) f(y) F(y)^{nm-m-1} \; dy + (v - b(v)) F(v)^{nm-m} = 0 \end{aligned}$$

By differentiating this expression with respect to v, we further see that it is an equilibrium for all advertisers to follow the bidding strategy b(v) if and only if

$$\begin{aligned}&(m-1) (v - b(v)) f(v) F(v)^{nm-m-1} + (m-1) \int _{0}^{v} f(y) F(y)^{nm-m-1} \; dy \\&\quad +\, (1 - b^{\prime }(v)) F(v)^{nm-m} + (nm-m) (v - b(v)) f(v) F(v)^{nm-m-1} = 0 \\&\quad \Leftrightarrow (nm-1) (v - b(v)) f(v) F(v)^{nm-m-1} + \frac{m-1}{nm-m}F(v)^{nm-m}\\&\quad +\, (1 - b^{\prime }(v)) F(v)^{nm-m} = 0 \\&\quad \Leftrightarrow b^{\prime }(v) = (nm-1) (v - b(v)) \frac{f(v)}{F(v)} + \frac{nm-1}{nm-m}. \end{aligned}$$

Thus there exists an equilibrium in which all bidders follow the same strictly monotonic bidding strategy b(v) if and only if there exists a solution to the differential equation given above in which $b(0) = 0$ and b(v) is strictly increasing in v. Standard results on differential equations guarantee that there exists a solution to this differential equation in which $b(0) = 0$, so it only remains to prove that any such function b(v) that solves this differential equation is strictly increasing in v.

To see this, first note that b(v) is increasing in v for values of v close to zero since $b^{\prime }(0) > 0$. Thus it suffices to show that there cannot exist some $\hat{v} > 0$ for which b(v) is increasing in v for values of $v < \hat{v}$ but non-increasing in v in some interval $[\hat{v}, \tilde{v}]$ with $\tilde{v} > \hat{v}$. Suppose by means of contradiction that such a $\hat{v}$ exists. This implies that in the limit as v approaches $\hat{v}$ from below, $b^{\prime }(v)$ becomes close to zero, which in turn implies that $(nm-1) (v - b(v)) \frac{f(v)}{F(v)}$ will be becoming less negative as v increases for values of v just below $\hat{v}$ and thus $(nm-1) (v - b(v)) \frac{f(v)}{F(v)} + \frac{nm-1}{nm-m}$ will be increasing in v for values of v just below $\hat{v}$. This contradicts the fact that $b^{\prime }(v) > 0$ for values of v just below $\hat{v}$ but $b^{\prime }(v) \le 0$ for values of v just above $\hat{v}$, and proves that b(v) must be strictly increasing in v. Thus there is an equilibrium in which each advertiser with value v follows a strategy of bidding some amount b(v), where b(v) is a strictly increasing function of v. $\square$

Proof of Proposition 4

First we derive necessary and sufficient conditions for there to exist an equilibrium in which all advertisers in the first advertiser network follow the same strictly monotonic bidding strategy b(v). Let $f(\cdot )$ denote the probability density function corresponding to $F(\cdot )$. Note that if all the advertisers in the first advertiser network follow the same strictly monotonic bidding strategy b(v) and an advertiser with value v makes a bid of $b(\hat{v})$, then this advertiser obtains an expected payoff of $F(\hat{v})^{m-1} \int _{0}^{b(\hat{v})} (v - y) f(y) \; dy$. In order for it to be an equilibrium for these advertisers to follow the bidding strategy b(v), this expected payoff must be maximized when $\hat{v} = v$, meaning it is an equilibrium for these advertisers to follow the bidding strategy b(v) if and only if

$$\begin{aligned}&(m\!-\!1) f(v) F(v)^{m-2} \int _{0}^{b(v)} (v \!-\! y) f(y) \; dy \!+\! F(v)^{m-1} b^{\prime }(v) (v - b(v)) f(b(v)) = 0 \nonumber \\&\quad \Leftrightarrow (m-1) f(v) \int _{0}^{b(v)} (v - y) f(y) \; dy + F(v) b^{\prime }(v) (v - b(v)) f(b(v)) = 0. \end{aligned}$$

(1)

For the exponential distribution, Eq. (1) holds if and only if

$$\begin{aligned}&\Leftrightarrow (m-1) e^{-v} \int _{0}^{b(v)} (v - y) e^{-y} \; dy + (1 - e^{-v}) b^{\prime }(v) (v - b(v)) e^{-b(v)} = 0 \nonumber \\&\Leftrightarrow (m-1) e^{-v} [v (1 - e^{-b(v)}) - 1 + (1 + b(v))e^{-b(v)}] \nonumber \\&\quad \, +\, (1 - e^{-v}) b^{\prime }(v) (v - b(v)) e^{-b(v)} = 0, \end{aligned}$$

(2)

which reduces to the condition given in the statement of the proposition for $v > 0$.

For the generalized Pareto distribution in which $F(v) = 1 - \frac{1}{(1+v)^2}$ and $f(v) = \frac{2}{(1+v)^3}$, Eq. (1) holds if and only if

$$\begin{aligned}&\frac{2(m-1)}{(1+v)^3} \int _{0}^{b(v)} \frac{2(v-y)}{(1+y)^3} \; dy + \left( 1 - \frac{1}{(1+v)^2}\right) b^{\prime }(v) (v - b(v)) \frac{2}{(1 + b(v))^3} = 0 \nonumber \\&\quad \Leftrightarrow (m\!-\!1)(1 \!+\! b(v))^3 \int _{0}^{b(v)} \frac{2(v-y)}{(1+y)^3} \; dy \!+\! (1 + v)(2v + v^2) b^{\prime }(v) (v - b(v)) = 0 \nonumber \\&\quad \Leftrightarrow (m-1)(1 + b(v))^3 \left[ \frac{(v+1)(2b(v) + b(v)^2)}{(1 + b(v))^2} - \frac{2b(v)}{1+b(v)} \right] \nonumber \\&\qquad +\, v(1 + v)(2 + v) b^{\prime }(v) (v - b(v)) = 0 \nonumber \\&\quad \Leftrightarrow (m-1)b(v)(1 + b(v)) [(v+1)(2 + b(v)) - 2(1+b(v))]\nonumber \\&\qquad +\, v(1 + v)(2 + v) b^{\prime }(v) (v - b(v)) = 0 \nonumber \\&\quad \Leftrightarrow (m-1)b(v)(1 + b(v)) [v(2 + b(v)) - b(v)] \nonumber \\&\qquad +\, v(1 + v)(2 + v) b^{\prime }(v) (v - b(v)) = 0, \end{aligned}$$

(3)

which also reduces to the condition given in the statement of the proposition for $v > 0$.

To prove the result, it only remains to show that $b^{\prime }(0) = \frac{2m}{m+1}$. To see this, first note that $b(v) = \Theta (v)$ when v is arbitrarily close to 0. We already know from Lemma 1 that $b(v) \ge v$ for all v, so it suffices to show that there cannot be some sequence of values of v with limit 0 for which $b(v) = \omega (v)$ in this sequence. To see this, note that since $e^{-b(v)} = 1 - b(v) + \frac{b(v)^2}{2} + o(b(v)^2)$, the left-hand side of Eq. (2) reduces to

$$\begin{aligned}&(m-1) e^{-v} \left[ v b(v) - 1 + (1 + b(v)) \left( 1 - b(v) + \frac{b(v)^2}{2}\right) \right] \\&\quad +\, (1 - e^{-v}) b^{\prime }(v) (v - b(v)) e^{-b(v)} + o(b(v)^2) \\&\qquad = (m-1) \left[ v b(v) - \frac{b(v)^2}{2} \right] + (1 - e^{-v}) b^{\prime }(v) (v - b(v)) e^{-b(v)} + o(b(v)^2) \end{aligned}$$

But if $b(v) = \omega (v)$ in some sequence, then $(m-1) \left[ v b(v) - \frac{b(v)^2}{2} \right] < 0$ for values of v close to 0 in this sequence. And we know from Lemma 1 that $b(v) \ge v$, so it follows that $(1 - e^{-v}) b^{\prime }(v) (v - b(v)) e^{-b(v)} \le 0$ for all v. Thus if $b(v) = \omega (v)$ along some sequence, then the left-hand side of Eq. (2) is negative for sufficiently small values of v in this sequence. Thus $b(v) = \Theta (v)$ must hold for the exponential distribution.

Similarly, if there is some sequence of values of v with limit 0 for which $b(v) = \omega (v)$ in this sequence, then the left-hand side of Eq. (3) reduces to $-(m-1)b(v)^2 + v(1 + v)(2 + v) b^{\prime }(v) (v - b(v)) + o(b(v)^2)$, which is negative for sufficiently small values of v in this sequence. Thus $b(v) = \Theta (v)$ must hold for the generalized Pareto distribution.

Since $b(v) = \Theta (v)$, in the limit as $v \rightarrow 0$, we have $e^{-b(v)} = 1 - b(v) + \frac{b(v)^2}{2} + o(b(v)^2) = 1 - b(v) + \frac{b(v)^2}{2} + o(v^2)$. Thus the left-hand side of equation (2) reduces to

$$\begin{aligned}&(m-1) e^{-v} \left[ v b(v) - 1 + (1 + b(v)) \left( 1 - b(v) + \frac{b(v)^2}{2}\right) \right] \nonumber \\&\quad +\, v b^{\prime }(v) (v - b(v)) e^{-b(v)} + o(v^2) \nonumber \\&\qquad = (m-1) \left[ v b(v) - \frac{b(v)^2}{2} \right] + v b^{\prime }(v) (v - b(v)) + o(v^2) \end{aligned}$$

(4)

Now if $b^{\prime }(0) = k$, then the expression in Eq. (4) will equal zero for small v if and only if $(m-1)(k - \frac{k^2}{2}) + k(1-k) = 0 \Leftrightarrow (m-1)(1 - \frac{k}{2}) + 1 - k = 0 \Leftrightarrow k = \frac{2m}{m+1}$. Thus $b^{\prime }(0) = \frac{2m}{m+1}$ for the exponential distribution.

Similarly, if $b^{\prime }(0) = k$, the left-hand side of Eq. (3) reduces to

$$\begin{aligned}&(m-1)kv(1 + kv) [v(2 + kv) - kv] + v(1 + v)(2 + v)k(v - kv) + o(v^2) \nonumber \\&\quad = (m-1)kv^2 (2 - k) + v^2 2k(1 - k) + o(v^2), \end{aligned}$$

(5)

so the expression in Eq. (5) will equal zero for small v if and only if $(m-1)k(2-k) + 2k(1-k) = 0 \Leftrightarrow (m-1)(2-k) + 2(1-k) = 0 \Leftrightarrow k = \frac{2m}{m+1}$. Thus $b^{\prime }(0) = \frac{2m}{m+1}$ for the Pareto distribution as well. $\square$

Proposition 5

Suppose any bid submitted by an advertiser network is rejected with probability p and no side payments are made to the advertiser networks. Then if $p \in [\frac{k-1}{k}, \frac{k}{k+1})$ for some positive integer k, each advertiser network i has a dominant strategy of submitting exactly $\min \{k, a_{i}\}$ bids to the advertising exchange.

Proof

To prove this result it suffices to prove that an advertiser network would prefer to submit the top $k+1$ bids the advertiser network has received rather than only submitting the top k bids if and only if $p \ge \frac{k}{k+1}$. Consider the incentives faced by a generic advertiser network i, let $b_j$ denote the highest bid submitted by some other advertiser network $j \ne i$ that is not rejected, and let $b_{(k)}$ denote the kth-highest bid that advertiser network i has received.

Note that whether advertiser network i submits the $k+1$th-highest bid that this network has received can only affect this network’s payoff if $b_{(k+1)} > b_j$ because if this inequality does not hold, then the advertiser with bid $b_{(k+1)}$ will never win the auction and will also never affect the price paid by any of the advertisers in network i if one of these advertisers wins the auction. Also note that whether advertiser network i submits the $k+1$th-highest bid that this network has received can only affect this advertiser network’s payoff if this bid is not rejected since if this bid is rejected, the advertiser network is in the same situation that would arise if this network never submitted this bid in the first place. Finally, note that whether advertiser network i submits the $k+1$th-highest bid that this network has received can only affect this advertiser network’s payoff if no more than one of the top k bids that this advertiser network submitted are accepted because if at least two of the other k bids that this advertiser network submitted are accepted, then submitting the $k+1$th-highest bid will not affect either who wins the auction or the price paid by this advertiser.

Thus in analyzing whether advertiser network i prefers to submit the $k+1$th-highest bid that this network has received, we can condition on the possibility that $b_{(k+1)} > b_j$, the $k+1$th-highest bid will not be rejected if it is submitted, and no more than one of the top k bids that this advertiser network submits will be accepted. Note that the probability the rth-highest bid that this advertiser network submits will be accepted and all of the other top k bids that this network submits will be rejected is $(1-p) p^{k-1}$ for all $r \le k$. And the probability that all of the top k bids that this network submits will be rejected is $p^k$.

If the rth-highest bid that this advertiser network submits is accepted and all of the other top k bids that this network submits are rejected, then the advertiser network’s payoff is $\beta (b_{(r)} - b_{(k+1)})$ if this network submits the $k+1$th-highest bid to the advertising exchange and $\beta (b_{(r)} - b_{j})$ otherwise, meaning the advertiser network’s payoff decreases by $\beta (b_{(k+1)} - b_{j})$ as a result of submitting the $k+1$th-highest bid to the advertising exchange. And if all of the top k bids that the network submits are rejected, then the advertiser network’s payoff is $\beta (b_{(k+1)} - b_{j})$ if this network submits the $k+1$th-highest bid to the advertising exchange and 0 otherwise, meaning the advertiser network’s payoff increases by $\beta (b_{(k+1)} - b_{j})$ as a result of submitting the $k+1$th-highest bid to the advertising exchange.

By combining the results in the previous two paragraphs, it follows that if the advertiser network submits the $k+1$th-highest bid that this network has received to the advertising exchange, then conditional on $b_{(k+1)} > b_j$ and the $k+1$th-highest bid not being rejected, this advertiser network’s payoff changes by

$$\begin{aligned}&p^k \beta (b_{(k+1)} - b_{j}) - \sum _{r=1}^{k} (1-p) p^{k-1} \beta (b_{(k+1)} - b_{j})\\&\quad = p^{k-1} \beta (b_{(k+1)} - b_{j}) (p - k(1-p)), \end{aligned}$$

which is non-negative if and only if $p - k(1-p) \ge 0$ or $p \ge \frac{k}{k+1}$. Thus advertiser networks prefer to submit the top $k+1$ bids they have received to the advertising exchange rather than submitting only the top k bids if and only if $p \ge \frac{k}{k+1}$. The result then follows. $\square$

Corollary 1

Suppose any bid submitted by an advertiser network is rejected with probability $p < \frac{1}{2}$ and no side payments are made to the advertiser networks. Then each advertiser network has a dominant strategy of submitting exactly one bid to the advertising exchange.