Technological endowments in entrepreneurial partnerships

Abstract

This paper discusses a novel argument to interpret the importance of thinking of collaborative partnerships in pre-competitive agreements. To do so, we adopt a dynamic iterative process to model technology diffusion between the partners of an agreement. We find that the success of an agreement of a given length hinges around identifying the suitable efficient combinations of the initial technological endowments of partners. As the time horizon of the agreement expands, the probability of identifying a suitable partner decreases, thus justifying the prevalence of short-horizon R&D agreements.

Appendices

Appendix A: Proof of Proposition 2

Here, firms compete à la Bertrand in the market. The equilibrium prices, quantities, and profits are,

(16)

(17)

where $\overline{a}_{1}=a_{1}-\mu \lambda_{0}(1-\lambda_{0})[b_{1}-c\xi ]$ and $\overline{a}_{2}=a_{2}-\mu \lambda_{0}(1-\lambda_{0})[b_{2}\xi -c]$.

As before, given the symmetry of the problem, we concentrate on the behavior of firm 1. Firm 1 evaluates the benefits it can get from the agreement and compares the level of profits with and without the agreement. That is, it compares profits in (6) and (16). Participating in an agreement will be profitable if and only if,

$$ b_{1} \biggl( \frac{2b_{2}\overline{a}_{1}+ c\overline{a}_{2}}{4b_{1}b_{2}-c^{2} } \biggr)^{2}> b_{1} \biggl( \frac{2b_{2}\hat{a}_{1}+c\hat{a}_{2}}{ 4b_{1}b_{2}-c^{2}} \biggr)^{2}. $$

After some algebraic computations, the previous inequality reduces to,

$$ b_1\bigl[ 1-\mu \lambda_{0}(1-\lambda_{0})\bigr] \bigl(2b_{1}b_{2}-b_{2}c\xi -c^{2} \bigr)>0. $$

(18)

Note that (18) differs from (12) in the second term in (round) brackets. Note also that this term is independent of λ ₀. The second term in round brackets is concave in c, and has one positive and one negative root. Therefore, for positive values of c smaller than the positive root, the term (2b ₁ b ₂−b ₂ cξ−c ²) is positive, and inequality (18) behaves as (12). Thus, we obtain the same result as in the monopoly case. In contrast, for large enough values of c (beyond the positive root), the term (2b ₁ b ₂−b ₂ cξ−c ²) is negative, so that the inequality is fulfilled when [1−μλ ₀(1−λ ₀)]<0 that is, for $\lambda_{0}\in (\frac{1-\overline{\mu }}{2},\frac{1+\overline{\mu }}{2})$, where $\overline{\mu }= (\frac{\mu -4}{\mu } )^{1/2}\in (0,1)$, for μ>4.

When 0<μ≤4 the first term of (18) in square brackets is always positive. Hence, a solution exists only if the second term in round brackets is positive, i.e. for positive values of c smaller than the positive root.

Appendix B: Proof of Proposition 3

We need to prove that Λ is a Cantor set, namely, that it is a closed, perfect and totally disconnected subset of I. Following Devaney (1985), we structure the proof in three steps.

• 1. Λ is a closed set. Let us define G(λ ₀)=1−F(λ ₀) and re-write it as G≡1−F. By construction A _i is an open interval centered around 1/2 (see Fig. 4). Let us concentrate on A ₀. In that case, the function G maps both the intervals I ₀=[0,λ ₁] and I ₁=[λ ₂,1] monotonically onto I. Moreover, G is decreasing on the first interval and increasing on the second. Since G(I ₀)=G(I ₁)=I there is a pair of intervals (one in I ₀ and the other in I ₁) which are mapped onto A ₀ by G. These intervals define the set A ₁. Next, let us consider Λ ₁=I−(A ₀∪A ₁). This set consists of four closed intervals (see Fig. 4) and G maps them monotonically onto either I ₀ or I ₁, but as before, each of the four intervals contains an open subinterval which is mapped by G ₂ onto A ₀, i.e., the points of this interval escape from I after the third iteration of G. By applying this iterative process, we note that A _t consists of 2^t disjoint open intervals and Λ _t =I−(A ₀∪…∪A _t ) consists of 2^t+1 closed intervals. Hence, Λ is a nested intersection of closed intervals, and is thus a closed set.

• 2. Λ is a perfect set. Note that all endpoints of A _t , (t=1,…) are contained in Λ. Such points are eventually mapped to the fixed point of G at 1, and they stay in I under iteration. If a point x∈Λ, were isolated, each nearby point must leave I under iteration, and therefore these points must belong to some A _t . Two possibilities arise. We can think of a sequence of endpoints of A _t converging to x. In this case the endpoints of A _t map to 1 and so, they are in Λ. Alternatively, all points in a deleted area nearby x are mapped out of I by some iteration of G. In this case, we may assume that G _τ maps x to 1 and all the other nearby points are mapped in the positive axis above 1. Then G _τ has a minimum at x, i.e., $G_{\tau }^{\prime }(x)=0$. This iterative process ensures that it must be so for some t<τ. Hence, G _t (x)=1/2, but then G _t+1(x)∉I and G _τ (x)→−∞, contradicting the fact that G _τ (x)=1.

• 3. Λ is a totally disconnected set. Let us focus in the first iteration and assume μ is large enough so that |G′(x)|>1 for all x∈I ₀∪I ₁. For those values of μ, there exists γ>1 such that |G′(x)|>γ for all x∈Λ. Our iterative process yields $\vert G_{\tau }^{\prime}(x)\vert >\gamma_{\tau }$. We want to prove that Λ does not contain any interval. Let us proceed by contradiction and assume that there is a closed interval [x,y]∈Λ, x,y∈I ₀∪I ₁, x≠y. In this case, $\vert G_{\tau }^{\prime }(z)\vert>\gamma_{\tau }$, for all z∈[x,y]. Choose τ so that λ _τ |y−x|>1. Applying the Mean Value Theorem, it follows that |G _τ (y)−G _τ (x)|≥γ _τ |y−x|>1, implying that either G _τ (y) or G _τ (x) lies outside of I. This contradicts with our main hypothesis, and thus Λ does not contain intervals. It remains to be determined the μ-values for which the previous argument holds. Finding the values of μ allowing |G′(x)|>1 means to identify μ-values for which [−μ(1−2x)]²>1. When G=0, this inequality holds for $\mu >2+\sqrt{5}$. Thus, we have proven that Λ is totally disconnected for $\mu >2+\sqrt{5}$. Recall from Lemma 1 that we already know that μ>4. Hence, we need to verify whether Λ is also totally disconnected for $\mu\in (4, 2+\sqrt{5}]$. We appeal to Kraft (1999) who establishes that Λ is a Cantor set for μ>4. The idea behind the proof is that for $\mu\in (4, 2+\sqrt{5}]$ it turns out that |G′(x)|⪋1. Kraft argues that the iteration process shrinks some components of I, and stretches some others. His proof thus consists in showing that in the interval $(4, 2+\sqrt{5})$ the stretching is dominated by the shrinking. To this end, he proves that Λ is an hyperbolic set, namely that $\vert G_{\tau }^{\prime}(x)\vert>k\delta_{\tau }>1$ for x∈Λ, k>0, δ>1.