On the optimal number of advertising slots in a generalized second-price auction

Abstract

In search advertising, a search engine uses a generalized second-price auction to sell advertising slots adjacent to search results on its webpage. In this paper, we study an interesting question related to the design of the generalized second-price auction: how should a search engine strategically decide on the number of advertising slots? To answer this question, we analyze the implication of varying the number of slots in a base model in which the click-through rates are assumed to be independent of the number of slots. When deciding the number of slots, we find that a search engine’s profit is based on two counteracting factors: the incremental clicks from an extra slot and the influence of the extra slot on advertisers’ payments per click. Our analysis characterizes the conditions for optimality of the number of slots and the implications of different distributions for advertiser valuations. We also extend the base model to allow for attraction and cannibalization of clicks from existing slots by new ad slots and show how such effects affect the optimal number of slots. Our overall results show that search engines need to optimize the number of ad slots offered for auction in order to maximize profit.

Appendix

Proof of Proposition 1

From Proposition 3 in Balachander et al. (2009), we have the total profit with K slots as¹²

$$ \pi_K^S = \sum\nolimits_{{k = 1}}^K {\sum\nolimits_{{j = k}}^k {{E_N}\left( {{v_{{j + 1}}}} \right)\left( {{\alpha_j} - {\alpha_{{j + 1}}}} \right) = \sum\limits_{{j = 1}}^K {j{E_N}\left( {{v_{{j + 1}}}} \right)\left( {{\alpha_j} - {\alpha_{{j + 1}}}} \right)} } } $$

(A.1)

Then,

$$ \pi_K^S = \sum\limits_{{j = 1}}^{{K - 1}} {j{E_N}\left( {{v_{{j + 1}}}} \right)\left( {{\alpha_j} - {\alpha_{{j + 1}}}} \right) + K{E_N}\left( {{v_{{K + 1}}}} \right){\alpha_K}} $$

(A.2)

$$ \pi_{{K + 1}}^S = \sum\limits_{{j = 1}}^{{K - 1}} {j{E_N}\left( {{v_{{j + 1}}}} \right)\left( {{\alpha_j} - {\alpha_{{j + 1}}}} \right) + K{E_N}\left( {{v_{{K + 1}}}} \right)\left( {{\alpha_K} - {\alpha_{{K + 1}}}} \right) + \left( {K + 1} \right){E_N}\left( {{v_{{K + 2}}}} \right)} $$

(A.3)

To compute the marginal profit we subtract (A.2) from (A.3).

$$ \pi_{{K + 1}}^S - \pi_K^S = - {\text{K}}{{\text{E}}_N}\left( {{v_{{K + 1}}}} \right){\alpha_{{K + 1}}} + \left( {K + 1} \right){E_N}\left( {{v_{{K + 2}}}} \right){\alpha_{{K + 1}}} $$

(A.4)

If Eq. (A.4) is positive, then a search engine has an incentive to increase the number of slot to K + 1. In other words, the marginal profit condition becomes

$$ \left( {K + 1} \right){E_N}\left( {{v_{{K + 2}}}} \right) - {\text{K}}{{\text{E}}_N}\left( {{v_{{K + 1}}}} \right) \geqslant 0 $$

(A.5)

Using property of order statistics, the expected valuation of K + 1^th and K + 2^th advertisers is given as follows:

$$ {E_N}\left( {{v_{{K + 1}}}} \right) = \int_0^{\infty } {\left\{ {v\frac{{N!}}{{\left( {N - K - 1} \right)!K!}}f(v){{\left[ {F(v)} \right]}^{{N - K - 1}}}{{\left[ {1 - F(v)} \right]}^K}} \right\}dv.} $$

(A.6)

$$ {E_N}\left( {{v_{{K + 2}}}} \right) = \int_0^{\infty } {\left\{ {v\frac{{N!}}{{\left( {N - K - 2} \right)!\left( {K + 1} \right)!}}f(v){{\left[ {F(v)} \right]}^{{N - K - 2}}}{{\left[ {1 - F(v)} \right]}^{{K + 1}}}} \right\}dv} $$

(A.7)

Then, integrating E _N (v _K + 2) by parts we have

$$ \matrix{ {{E_N}\left( {{v_{{K + 2}}}} \right)} \hfill \\ { = \frac{{N!}}{{\left( {N - K - 1} \right)!\left( {K + 1} \right)!}}\left\{ { - \int_0^{\infty } {{{\left[ {F(v)} \right]}^{{N - K - 1}}}{{\left[ {1 - F(v)} \right]}^{{K + 1}}}dv - \left( {K + 1} \right)\int_0^{\infty } {vf(v)F{{(v)}^{{N - K - 1}}}{{\left[ {1 - F(v)} \right]}^K}dv} } } \right\}} \hfill \\ { = - \frac{{N!}}{{\left( {N - K - 1} \right)!\left( {K + 1} \right)!}}\int_0^{\infty } {{{\left[ {F(v)} \right]}^{{N - K - 1}}}{{\left[ {1 - F(v)} \right]}^{{K + 1}}}dv + {E_N}\left( {{v_{{K + 1}}}} \right)} } \hfill \\ }<!end array> $$

(A.8)

Plugging (A.8) and (A.6) into (A.5) gives (A.9).

$$ \int_0^{\infty } {\left\{ {\frac{{N!}}{{\left( {N - K - 1} \right)!K!}}f(v){{\left[ {F(v)} \right]}^{{N - K - 1}}}{{\left[ {1 - F(v)} \right]}^K}\left[ {v - \frac{{1 - F(v)}}{{f(v)}}} \right]} \right\}dv \geqslant 0} $$

(A.9)

Let $J(v) = v - \frac{{1 - F(v)}}{{f(v)}}$, then (A.9) becomes E _N (J(v _K + 1)).

Proof of Proposition 3(a)

For the uniform valuation distribution, Eq. (A.5) reduces to the following:

$$ \frac{{N!}}{{\left( {N - K - 2} \right)!K!}}\int_0^{{{v^{{\max }}}}} {\left[ {{v^{{N - K - 1}}} - {{\left( {1 - v} \right)}^{{K + 1}}}} \right]dv - \frac{{N!}}{{\left( {N - K - 1} \right)!\left( {K - 1} \right)!}}\int_0^{{{v^{{\max }}}}} {\left[ {{v^{{N - K}}} - {{\left( {1 - v} \right)}^K}} \right]dv \geqslant 0} } $$

(A.10)

By letting $\int_0^{{{v^{{\max }}}}} {\left[ {{v^{{N - K - 1}}} - {{\left( {1 - v} \right)}^{{K + 1}}}} \right]dv}$ be A and $\int_0^{{{v^{{\max }}}}} {\left[ {{v^{{N - K}}} - {{\left( {1 - v} \right)}^K}} \right]dv}$ be B, (A.10) reduces to

$$ \frac{{N!}}{{\left( {N - K - 2} \right)!K!}}\left\{ A \right\} - \frac{{N!}}{{\left( {N - K - 1} \right)!\left( {K - 1} \right)!}}\left\{ B \right\} \geqslant 0 $$

Through integration by parts, B becomes

$$ \int_0^{{{v^{{\max }}}}} {\frac{{{v^{{N - K}}}{{\left( {1 - v} \right)}^{{K + 1}}}}}{{K + 1}}dv + \frac{{N - K}}{{K + 1}}\int_0^{{{v^{{\max }}}}} {\left[ {{v^{{N - K - 1}}} - {{\left( {1 - v} \right)}^{{K + 1}}}} \right]dv} } $$

(A.11)

Since the first term in (A.11) reduces to zero, (A.11) reduces to the following:

$$ \frac{{N - K}}{{K + 1}}\int_0^{{{v^{{\max }}}}} {\left[ {{v^{{N - K - 1}}} - {{\left( {1 - v} \right)}^{{K + 1}}}} \right]dv} $$

(A.12)

Plugging (A.12) into (A.10), (A.10) becomes

$$ \left[ {\frac{{N!}}{{\left( {N - K - 2} \right)!K!}} - \frac{{N!\left( {N - K} \right)}}{{\left( {N - K - 1} \right)!\left( {K - 1} \right)!\left( {K + 1} \right)}}} \right]\int_0^{{{v^{{\max }}}}} {\left[ {{v^{{N - K - 1}}} - {{\left( {1 - v} \right)}^{{K + 1}}}} \right]dv \geqslant 0} $$

(A.13)

Inequality (A.13) further reduces to $\left[ {\frac{{N!\left( {N - 2K - 1} \right)}}{{\left( {N - K - 1} \right)!\left( {K + 1} \right)!}}} \right]\int_0^{{{v^{{\max }}}}} {\left[ {{v^{{N - K - 1}}} - {{\left( {1 - v} \right)}^{{K + 1}}}} \right]dv \geqslant 0}$, and by simplifying this we can show that this inequality holds if $K \leqslant \frac{{N - 1}}{2}$. Since the number of slots is an integer, we assume that a search engine increases K until the marginal profit becomes zero or negative. This implies ${K^{ * }} = \left\lceil {\frac{{N - 1}}{2}} \right\rceil$ .

Proof of Proposition 3(b)

Let the CDF of the Pareto distribution be $F\left( {v;a,\beta } \right) = 1 - {\left( {\frac{a}{v}} \right)^{\beta }},v > a$, where a is the lower bound of v, a > 0, and β > 0, which is the shape parameter.

According to the well-known representation theorem of Renyi (1953) and Huang (1975),

$$ {v_K} = a\prod\nolimits_{{i = k}}^N {{\eta_i}} $$

(A.14)

Where η _i ,…,η _N are independent Pareto variables, with η _i  ∼ F(1,iβ). Because the mean of F(1,iβ) is $\frac{{a\beta }}{{\beta - 1}}$ for β > 1and η _i  ∼ F(1,iβ), we can easily obtain that (Arnold et al. 1992)

$$ {E_N}\left( {{v_K}} \right) = \frac{{a\Gamma \left( {N + 1} \right)\Gamma \left( {K - 1/\beta } \right)}}{{\Gamma (K)\Gamma \left( {N + 1 - 1/\beta } \right)}} \;{\text{for}}\;\beta \; > \;1 $$

(A.15)

Using (A.15), the marginal profit condition, (A.5) reduces to

$$ \Gamma \left( {K + 2 - 1/\beta } \right) - K\Gamma \left( {K + 1 - 1/\beta } \right) = \left[ {K + 1 - 1/\beta } \right]\Gamma \left( {K + 1 - 1/\beta } \right) \geqslant 0 $$

(A.16)

Because 1 − 1/β > 0, we have

$$ \matrix{ {\Gamma \left( {K + 1 - 1/\beta } \right) = \left( {K - 1/\beta } \right)\left( {K - 1/\beta - 1} \right) \ldots \left( {1 - 1/\beta } \right)\Gamma \left( {1 - 1/\beta } \right)} \hfill \\ { = \left( {1 - 1/\beta } \right)\Gamma \left( {1 - 1/\beta } \right)\sum\nolimits_{{r = 1}}^{{K - 1}} {\left( {K + 1 - 1/\beta - r} \right)} = \left( {K \frac{{ - 1}}{2} } \right)\left( {\beta \left( {K + 2} \right) - 2} \right) \geqslant 0} \hfill \\ }<!end array> $$

(A.17)

(A.17) implies that the marginal profit condition is positive for all K because β > 1. Therefore, K* = N.

Proof of Proposition 3(c)

According to Arnold et al. (1992), the (ξ − 1)th moment of v _i:N , the ith order statistic of N valuations, is given by

$$ \mu_{{i:N}}^{{\xi - 1}} = B{\left( {i,N - i + 1} \right)^{{ - 1}}}\int_0^{\infty } {{v^{{\xi - 1}}}{{\left( {1 - {e^{{ - v}}}} \right)}^{{N - i}}}{{\left( {{e^{{ - v}}}} \right)}^i}dv} $$

(A.18)

Integrating this by parts with v ^ξ−1 dv and (1 − e ^−v)^N−i (e ^−v)ⁱ, we have

$$ \matrix{ {\mu_{{i:N}}^{{\xi - 1}}} \hfill \\ { = B{{\left( {i,N - i + 1} \right)}^{{ - 1}}}\left[ { - \int_0^{\infty } {\frac{{{v^{\xi }}}}{\xi }\left( {N - i} \right){{\left( {1 - {e^{{ - v}}}} \right)}^{{N - i - 1}}}{{\left( {{e^{{ - v}}}} \right)}^{{i + 1}}}dv + \int_0^{\infty } {\frac{{{v^{\xi }}}}{\xi }\left( {N - i} \right){{\left( {1 - {e^{{ - v}}}} \right)}^{{N - i}}}{{\left( {{e^{{ - v}}}} \right)}^i}(i)dv} } } \right]} \hfill \\ { = - \frac{{N!(i)}}{{\left( {N - i - 1} \right)!(i)!\xi }}\int_0^{\infty } {{v^{\xi }}{{\left( {1 - {e^{{ - v}}}} \right)}^{{N - i - 1}}}{{\left( {{e^{{ - v}}}} \right)}^{{i + 1}}}dv} + \frac{{N!(i)}}{{\left( {N - 1} \right)!\left( {i - 1} \right)!\xi }}\int_0^{\infty } {{v^{\xi }}{{\left( {1 - {e^{{ - v}}}} \right)}^{{N - i}}}{{\left( {{e^{{ - v}}}} \right)}^i}dv} } \hfill \\ { = - \mu_{{i + 1:N}}^{\xi }\frac{i}{\xi } + \mu_{{i + 1:N}}^{\xi }\frac{i}{\xi }} \hfill \\ }<!end array> $$

(A.19)

Thus, $\mu_{{i:N}}^{\xi } = - \mu_{{i + 1:N}}^{\xi } + \frac{{\xi \mu_{{i:N}}^{{\xi - 1}}}}{i}$. Using this last identity and recursively plugging the E _N (v _i ), i = 1,…,N, we obtain

$$ {E_N}\left( {{v_K}} \right) = \sum\nolimits_{{r = K}}^N {K\frac{1}{r}} $$

(A.20)

Thus, the marginal profit condition becomes

$$ \left( {K + 1} \right){E_N}\left( {{v_{{K + 2}}}} \right) - K{E_N}\left( {{v_{{K + 1}}}} \right) = \left( {K + 1} \right)\sum\nolimits_{{r = K + 2}}^N {\frac{1}{r} - K} \sum\nolimits_{{r = K + 1}}^N {\frac{1}{r} = K} \left\lceil { - \frac{1}{{K + 1}}} \right\rceil + \sum\nolimits_{{r = K + 2}}^N {\frac{1}{r} \geqslant 0} $$

(A.21)

Proof of Proposition 4(a)

From Proposition 3 in Balachander et al. (2009), we have the total profit with K slots as

$$ \pi_K^S = \sum\nolimits_{{j = 1}}^K {j{E_N}\left( {{v_{{j + 1}}}} \right)\left( {\alpha_j^K - \alpha_{{j + 1}}^K} \right) = \sum\nolimits_{{j = 1}}^K {\alpha_j^K\left[ {j{E_N}\left( {{v_{{j + 1}}}} \right) - \left( {j - 1} \right){E_N}\left( {{v_j}} \right)} \right]} } . $$

(A.22)

Then, we have

$$ \matrix{ {\pi_{{K + 1}}^S - \pi_K^S = \sum\nolimits_{{j = 1}}^K {\alpha_j^{{K + 1}}\left[ {j{E_N}\left( {{v_{{j + 1}}}} \right) - \left( {j - 1} \right){E_N}\left( {{v_j}} \right)} \right]} - \sum\nolimits_{{j = 1}}^K {\alpha_j^K\left[ {j{E_N}\left( {{v_{{j + 1}}}} \right) - \left( {j - 1} \right){E_N}\left( {{v_j}} \right)} \right]} } \hfill \\ { + \alpha_{{K + 1}}^{{K + 1}}\left[ {\left( {K + 1} \right){E_N}\left( {{v_{{K + 2}}}} \right) - (K){E_N}\left( {{v_{{K + 1}}}} \right)} \right]} \hfill \\ { = \sum\nolimits_{{j = 1}}^K {\left( {\alpha_j^{{K + 1}} - \alpha_j^K} \right)\left[ {j{E_N}\left( {{v_{{j + 1}}}} \right) - \left( {j - 1} \right){E_N}\left( {{v_j}} \right)} \right] + \alpha_{{K + 1}}^{{K + 1}}\left[ {\left( {K + 1} \right){E_N}\left( {{v_{{K + 2}}}} \right) - (K){E_N}\left( {{v_{{K + 1}}}} \right)} \right]} } \hfill \\ }<!end array> $$

(A.23)

Using the relation in Proposition 1 that $\left( {K + 1} \right){E_N}\left( {{v_{{K + 2}}}} \right) - K{E_N}\left( {{v_{{K + 1}}}} \right) = {E_N}\left( {J\left( {{v_{{K + 1}}}} \right)} \right)$, Eq. (A.23) becomes

$$ \matrix{ {\pi_{{K + 1}}^S - \pi_K^S = \sum\nolimits_{{j = 1}}^K {\left\{ {\left[ {\alpha_j^{{K + 1}} - \alpha_j^K} \right]{E_N}\left( {J\left( {{v_j}} \right)} \right)} \right\} + \alpha_{{K + 1}}^{{K + 1}}\left[ {{E_N}\left( {J\left( {{v_{{k + 1}}}} \right)} \right)} \right]} } \hfill \\ { = \sum\nolimits_{{j = 1}}^K {\left( {\alpha_j^{{K + 1}} - \alpha_j^K} \right){E_N}\left( {J\left( {{v_j}} \right)} \right) + \left[ {{Z^{{K + 1}}} - {Z^K} + \sum\nolimits_{{j = 1}}^K {\left( {\alpha_j^K - \alpha_j^{{K + 1}}} \right)} } \right]{E_N}\left( {J\left( {{v_{{K + 1}}}} \right)} \right)} } \hfill \\ { = - \sum\nolimits_{{j = 1}}^K {\left( {\alpha_j^{{K + 1}} - \alpha_j^K} \right)\left[ {{E_N}\left( {J\left( {{v_j}} \right)} \right) - {E_N}\left( {J\left( {{v_{{K + 1}}}} \right)} \right)} \right] + \left( {{Z^{{K + 1}}} - {Z^K}} \right){E_N}\left( {J\left( {{v_{{K + 1}}}} \right)} \right)} } \hfill \\ }<!end array> $$

(A.24)

Since the search engine would add slots when (A.24) is non-negative, the condition to profitably add a slot is

$$ \left( {{Z^{{K + 1}}} - {Z^K}} \right){E_N}\left( {J\left( {{v_{{K + 1}}}} \right)} \right) \geqslant \sum\nolimits_{{j = 1}}^K {\left( {\alpha_j^K - \alpha_j^{{K + 1}}} \right)\left[ {{E_N}\left( {J\left( {{v_i}} \right)} \right) - {E_N}\left( {J\left( {{v_{{K + 1}}}} \right)} \right)} \right]} . $$

(A.25)

Proof of Proposition 4(b-2)

When ${Z^{{K + 1}}} = {Z^K},\forall K$, the left-hand side term in (A.25) is zero. However, the right-hand side term is always non-negative because, ${v_j} > {v_{{j + 1}}},\forall j$ in equilibrium, J(v) is monotonically increasing in v, and $\alpha_j^K > \alpha_j^{{K + 1}}$. Therefore K* = 1. When ${Z^K} > {Z^{{K + 1}}},\forall K$, the optimal number of slots is one, because the left-hand side of (A.25) is negative for all K ≥ 1 while the right-hand side is non-negative.

Proof of Proposition 4(b-3)

Note that $\alpha_j^K < \alpha_j^{{K + 1}}$ implies ${Z^K} < {Z^{{K + 1}}}$. Thus, the left-hand side of (A.25) is always positive. However, the right-hand side is always negative because ${E_N}\left( {J\left( {{v_j}} \right)} \right) \geqslant {E_N}\left( {J\left( {{v_{{K + 1}}}} \right)} \right)$ given that ${v_j} > {v_{{j + 1}}},\forall j$ in equilibrium and J(v) is monotonically increasing in v. Thus, K* = N.