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FRAND Licensing of Standard-Essential Patents: Comparing Realistic Ex-Ante and Ex-Post Contracts

  • Pier Luigi Parcu , Maria Alessandra Rossi EMAIL logo and David Silei
Published/Copyright: January 27, 2025

Abstract

The paper presents a bargaining model that compares two forms of licensing of Standard Essential Patents (SEPs) in terms of their ability to deliver Fair, Reasonable and Non-Discriminatory (FRAND) royalties. The first is a profit-based or ad valorem contract signed ex-ante, before the product market entry of the implementer. The second is a lump sum contract signed ex-post, after implementers’ profits have been realized. Both contracts are realistic, i.e. compatible with the typical timing of standardization activities. We find that the ex-ante contract dominates the ex-post one in many but not all situations. It does when the imbalance in bargaining power between the parties is not too strong, independently of how the latter is distributed. In more extreme situations of bargaining power imbalance (typical hold-up and hold-out situations), ex-ante contracts are again better when the level of litigation costs relative to profits is very high or very low. However, for intermediate levels of litigation costs relative to profits, the threat of litigation may be useful to set royalties that are closer to FRAND. This finding is relevant in light of current regulatory initiatives over-emphasizing ex-ante licensing, like the European SEP Regulation.

JEL Classification: O32; O34; K12

Corresponding author: Maria Alessandra Rossi, Robert Schuman Centre for Advanced Studie s, European University Institute, Firenze, Italy; and Department of Economic Studies, University “G. d’Annunzio” of Chieti-Pescara and Robert Schuman Centre for Advanced Studies, European University Institute, Firenze, Italy, E-mail:

Acknowledgements

The authors would like to thank Justus Baron, Benno Buehler, Lapo Filistrucchi, Gregor Langus and other participants to the Florence Seminar on Standard-Essential Patents held on 6-7 October 2022 at the EUI for their helpful comments. Usual disclaimers apply.

  1. Disclosure statement: This research was developed in the context of the project Innovation and Intellectual Property in the digital age carried out at the Centre for a Digital Society of the European University Institute. In addition to public funding, the Centre benefits from the financial support of private companies that may have an interest in intellectual property and standards policy, including Deutsche Telekom, Meta, Qualcomm Inc., Telecom Italia, and others. The views expressed in the paper are, however, the authors’ only.

Appendix A: Derivation of Condition 9

Let us start from conditions 8. Let us set m β β * ρ and n ( 1 2 β ) ( 1 ρ ) ρ . Substituting for equations (3) and (7), which define R and R*, we obtain:

m Π + n L > L m Π + n L < L

and thus:

m Π + ( n 1 ) L > 0 m Π + ( n + 1 ) L < 0

We have three possible settings for parameter n:

  1. n < −1

  2. n > 1

  3. −1 ≤ n ≤ 1

Working out n < −1, we have ρ < 1 1 2 β . However, ρ is a probability and thus belongs to the unit interval. For ρ to be smaller than 1 1 2 β , we need to impose 1 1 2 β > 0 , which requires necessarily β > 1 2 . Since there are values of β that make ρ < 1 1 2 β fail, n < −1 should be excluded from parameter region 9.

Same goes for n > 1, which is equivalent to impose ρ < 1 2 β 2 2 β . Since ρ is a probability, we must have 1 2 β 2 2 β > 0 , which is true only for β < 1 2 . Since there are values of β that make ρ < 1 2 β 2 2 β fail, n > 1 should be excluded from parameter region 9.

On the other hand, −1 ≤ n ≤ 1 is satisfied for every value of β and is the only one that can be accepted. In fact −1 ≤ n implies ρ > 1 1 2 β . Since ρ is a probability, we must impose 1 1 2 β < 1 , which is always true for every β ∈ [0, 1]. Moreover n ≤ 1 implies ρ > 1 2 β 2 2 β , which requires 1 2 β 2 2 β < 1 and this is always true too.

Saying that −1 ≤ n ≤ 1 is the only setting that can be accepted is equivalent to state part (i) of condition 9, which was thus proved.

As for part (ii) of condition 9, it comes from solving system:

m Π + ( n 1 ) L > 0 m Π + ( n + 1 ) L < 0

in the case −1 ≤ n ≤ 1 and β ∈ [0, 1].

If m ≥ 0, that is ββ*, we have:

L Π > m n + 1 L Π > m n 1

The first inequality is always true because m n + 1 is nonpositive whereas L and Π are positive; while the second inequality is true when, substituting for m and n, L Π > β β * 2 ρ + 2 β 2 β ρ 1 . Thus the solution of the system is L Π > β β * 2 ρ + 2 β 2 β ρ 1 , as we wanted to show.

If m < 0, that is β < β*, the first inequality is true when, substituting for m and n, L Π > β β * 1 2 β + 2 β ρ ; whereas the second inequality is always true because m n 1 is nonpositive while L and Π are positive. Thus the solution of the system is L Π > β β * 1 2 β + 2 β ρ , as we wanted to show.

Appendix B: Proof of Proposition 1

Let us consider condition 9 and equations (14) and (15) to prove the proposition.

As a first step, let us consider the special case β = β*. In this case:

R expost R * = ( 1 2 β ) ( 1 ρ ) ρ L R exante R * = 0

thus the ex-ante royalty is always equal to the FRAND level whereas the ex-post royalty generally differs from it (unless β = 1/2), as we wanted to show.

Second, let us set ββ*. Thus δ = 1 ρ + ( 1 2 β ) ( 1 ρ ) L ( β β * ) ρ Π . For convenience, let us impose m β β * ρ and n ( 1 2 β ) ( 1 ρ ) ρ . Thus δ = m Π + n L ρ m Π . Let us impose δ > 1 to see when the ex-ante negotiation is more FRAND than the ex-post one. We shall consider each of the 4 possible cases.

Case a) with m, n ≥ 0. Then we have L Π > m ( 1 ρ ) n , which is always true since L Π is by definition larger than zero. The same result is obtained in case b) with m, n < 0. This shows that the ex-ante negotiation is better than the ex-post for both m, n ≥ 0 and m, n < 0. However, saying m, n ≥ 0 is equivalent to state ββ* and β ≤ 1/2, whereas m, n < 0 is equivalent to β < β* and β > 1/2. This proves the first part of proposition 1.

Case c) with m ≥ 0 (that is β ≥ 0) and n<0 (β > 1/2). In this case δ > 1 implies solution L Π < m ( 1 ρ ) n L Π > m ( 1 + ρ ) n , which is consistent with parameter region 9 and, after substituting for m and n, is equivalent to L Π < β β * 1 2 β L Π > 1 + ρ 1 ρ β β * 1 2 β , as we wanted to show. Analogous result in case d) with m < 0 (that is β < 0) and n ≥ 0 (β ≤ 1/2). This ends the proof of the proposition.

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Received: 2023-08-02
Accepted: 2024-11-04
Published Online: 2025-01-27

© 2025 Walter de Gruyter GmbH, Berlin/Boston

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