Difference between revisions of "VC Bargaining"

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Address the question: How does the optimal policy compare to the current way of calculating shares and values?
 
Address the question: How does the optimal policy compare to the current way of calculating shares and values?
  
Assume a fixed number of rounds: t={1,2}
+
Assume a fixed number of rounds: <math>t={1,2}\,</math>
Assume a fixed total investment: \sum_t x_t = 1
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Assume a fixed total investment: <math>\sum_t x_t = 1\,</math>
Assume a functional form for f(x_t): f(x_t) = ln (x_t)
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Assume a functional form for <math>f(x_t): f(x_t) = ln (x_t)\,</math>
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Again <math> V(0)=0 \,</math>.
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:<math> \therefore V_2 = ln(x_1) + ln(x_2)\,</math>

Revision as of 21:07, 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 and a VC, 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) = ln (x_t)\,[/math]

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

[math] \therefore V_2 = ln(x_1) + ln(x_2)\,[/math]