Difference between revisions of "VC Bargaining"

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imported>Ed
imported>Ed
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===The players===
 
===The players===
  
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The players are an Entrepreneur and a VC, both are risk neutral.
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The players are an Entrepreneur (<math>E\,</math>) and a VC investor (<math>I\,</math>), both are risk neutral.
  
 
===The Value Function===
 
===The Value Function===
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:<math> \frac{\partial V_2}{\partial x_1} =0 \implies x_1 = x_2 = \frac{1}{2}\,</math>
 
:<math> \frac{\partial V_2}{\partial x_1} =0 \implies x_1 = x_2 = \frac{1}{2}\,</math>
  
−
\therefore V_2 = \frac{1}{2}^\frac{1}{2} + \frac{1}{2}^\frac{1}{2} = 1^\frac{1}{2} \approx 1.41
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:<math>\therefore V_2 = \frac{1}{2}^\frac{1}{2} + \frac{1}{2}^\frac{1}{2} = 1^\frac{1}{2} \approx 1.41\,</math>
  
 
How much should be allocated to the investor?
 
How much should be allocated to the investor?
  
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Using Shapley values, Nash Bargaining and infinite Rubenstein bargaining will all imply each party gets \frac{1}{2}^\frac{1}{2}\approx 0.707, assuming equal outside options of zero and equal bargaining power.
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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:
 
Proof using the Shapley value for a single stage of negotiation:
  
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V(C
+
<math>
 +
v(\empty) = 0 \\
 +
v(\{I\}) = 0 \\
 +
v(\{E\}) = 0 \\
 +
v(\{I,E\}) = 1^\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!}(1^\frac{1}{2} - 0) +  \frac{0!1!}{2!}(0 - 0) = \frac{1}{2}^\frac{1}{2} \approx 0.707\,</math>

Revision as of 22:03, 25 May 2011

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