VC Bargaining

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This page (and the discussion page) is for Ed and Ron to share their thoughts on VC Bargaining. Access is restricted to those with "Trusted" access.

A Basic Model

The players

The players are an Entrepreneur ([math]E\,[/math]) and a VC investor ([math]I\,[/math]), both are risk neutral.

The Value Function

[math]V_t=V_{t-1} + f(x_t) - k \,[/math]

with

[math]V_0=0, f(0)=0, f'\gt 0, f''\lt 0, k\gt 0 \,[/math]

Having [math]k\gt 0\,[/math] forces a finite number of rounds as the optimal solution providing there is a stopping constraint on [math]V_t\,[/math] (so players don't invest forever).

One possible stopping constraint is:

[math]V_t \ge \overline{V}\,[/math]

with

[math]\overline{V} \sim F(V)\,[/math]

where the distribution is known to both parties.

Bargaining

In each period there is Rubenstein finite bargaining, with potentially different patience, and one player designated as last. This will give a single period equilibrium outcome with the parties having different bargaining strength.

Simple First Steps

Address the question: How does the optimal policy compare to the current way of calculating shares and values?

Assume a fixed number of rounds: [math]t={1,2}\,[/math] Assume a fixed total investment: [math]\sum_t x_t = 1\,[/math] Assume a functional form for [math]f(x_t): f(x_t) = x_t^\frac{1}{2}\,[/math]

Again [math] V(0)=0 \,[/math].


[math] \therefore V_2 = x_1^\frac{1}{2} + x_2^\frac{1}{2}\,[/math]

Recalling that [math] x_2 = 1 - x_1 \,[/math]


[math] \frac{\partial V_2}{\partial x_1} =0 \implies x_1 = x_2 = \frac{1}{2}\,[/math]
[math]\therefore V_2 = \frac{1}{2}^\frac{1}{2} + \frac{1}{2}^\frac{1}{2} = 2\cdot\frac{1}{2}^\frac{1}{2} \approx 1.41\,[/math]

How much should be allocated to the investor?


Using Shapley values, Nash Bargaining and infinite Rubenstein bargaining will all imply each party gets :[math]\frac{1}{2}^\frac{1}{2}\approx 0.707\,[/math], assuming equal outside options of zero and equal bargaining power.


Proof using the Shapley value for a single stage of negotiation:

[math]v(\{\empty\}) = 0, \;v(\{I\}) = 0, \; v(\{E\}) = 0, \; v(\{I,E\}) = 2\cdot\frac{1}{2}^\frac{1}{2}\,[/math]


[math]\phi_i(v)=\sum_{S \subseteq N \setminus \{i\}} \frac{|S|!\; (n-|S|-1)!}{n!}(v(S\cup\{i\})-v(S))[/math]


[math]\therefore \phi_I(v)= \frac{1!0!}{2!}(2\cdot\frac{1}{2}^\frac{1}{2} - 0) + \frac{0!1!}{2!}(0 - 0) = \frac{1}{2}^\frac{1}{2} \approx 0.707\,[/math]


The entrepreneur gets the same (the efficient outcome is realised and the profits are fully distributed, so you know he must without doing the math).


For two stages of negotiation, the intermediate value of the firm is

[math]V_1 = x_1^\frac{1}{2} = \frac{1}{2}^\frac{1}{2} \approx 0.707\,[/math]


and the characteristic function is:

[math]v(\{\empty\}) = 0, \;v(\{I\}) = 0, \; v(\{E\}) = 0, \; v(\{I,E\}) = \frac{1}{2}^\frac{1}{2}\,[/math]


This gives:

[math]\phi_{I(1)}(v)= \frac{1!0!}{2!}(\frac{1}{2}^\frac{1}{2} - 0) + \frac{0!1!}{2!}(0 - 0) = \frac{1}{4}^\frac{1}{2} \approx 0.354\,[/math]


For the second stage, the characteristic function is:

[math]v(\{\empty\}) = 0, \;v(\{I\}) = \frac{1}{2}^\frac{1}{2}, \; v(\{E\}) = \frac{1}{2}^\frac{1}{2}, \; v(\{I,E\}) = 2\cdot\frac{1}{2}^\frac{1}{2}\,[/math]


This gives:

[math]\phi_{I(2)}(v)= \frac{1!0!}{2!}(2\cdot\frac{1}{2}^\frac{1}{2}-\frac{1}{2}^\frac{1}{2}) + \frac{0!1!}{2!}(\frac{1}{2}^\frac{1}{2} - 0) = \frac{1}{2}^\frac{1}{2} \approx 0.707\,[/math]


And we have the same result as the single negotiation version of the game.


Note that this assumes that value isn't created or destroyed by the presence of the investor or the entrepreneur alone after the first stage - the first stage value just sits there, waiting to be built upon by the combination of the investor and the entrepreneur together. This is a model where the outside options of both players are zero. If the entrepreneur doesn't turn up for both rounds the firm is worth zero, and likewise for the entrepreneur. Also, but differently, the bargaining strength is equal. To express different bargaining strengths we would use a weighted Shapley value. Note that this could still be used with zero outside options.