Changes

*This page is referenced in [[PHDBA602 (Theory of the Firm)]]

*If both firms enter they each get <math>\mu\,</math>, for a payoff of <math>\mu - S\,</math>

*<math>S \sim F(\cdot)\,</math>, with support such that <math>\mu < S < \lambda\,</math> for all realizations

*The game ends when one or more firms <math>Enter\,</math>

===The General Result===

Then it must be the case that <math>t_1 < t_2\,</math>

To see this assume that <math>A\,</math> believes that there is:

*a probability that <math>B\,</math> will not have entered prior to <math>t\,</math> of <math>\alpha(t)\,</math>, where we denote <math>a(t) = \delta^t \alpha(t)\,</math>.

For <math>t_1\,</math> to be prefered with <math>S_A^1\,</math>, and <math>t_2\,</math> to be prefered with <math>S_A^2\,</math>, it must be true that:

and

So:

:<math>\therefore (a(t_1) - a(t_2))(S_A^1 - S_A^2) \le 0\,</math>

As <math>a(t)\,</math> is decreasing in <math>t\,</math>, and <math>S_A^1 < S_A^2\,</math> by assumption, it must be that <math>t_1 < t_2\,</math>.

Th paper then sets up the '''Fundamental Difference Equation''', where is supposes cutoffs <math>S_1\,</math>, <math>S_2\,</math> and so forth such that firms with costs between <math>S_{t-1}\,</math> and <math>S_t\,</math> will enter in <math>t\,</math> (providing no previous entry has occurred). Using the indifference of such a firm between periods <math>t\,</math> and <math>t+1\,</math> we have:

One Bayesian equilibrium is then:

(Note that this equation is not the same as in the paper, but does give the correct <math>v\,</math>).

:<math>q_t = (1-p)^{2(t-1)}\cdot 2p(1-p)\,</math>

:<math>r_t = (1-p)^{2(t-1)}\cdot p^2\,</math>

:<math>W^G = \sum_{t=1}^{\infty} \delta^{t-1}(q_t(1-S) + r_t(1-2S)\,</math>

There is an inherent trade-off between delay and duplication - both can be expressed in terms of <math>p\,</math>, or just in terms of each other.

:<math>y = \frac{(x-1)^2}{4x}\,</math>

By changing the parameters <math>\lambda\,</math>, <math>\mu\,</math>, or <math>S\,</math>, a planner could make this trade-off, allowing less delay at the expense of more duplication, or vice versa.

This can be written as the difference from the first-best:

:<math>W^R = 1 -(qS_L + (1-q)S_H)\,</math>

Which gives:

:<math>W^D (\delta=0) = q^2(1-2S_L) + 2q(1-q)(1-S_L)\,</math>

or

:<math>W^* - W^D (\delta=0) = q^2S_L + (1-q)^2(1-S_H)\,</math>

Then decentraliazation is better than random choice iff:

<math>W^* - W^D (\delta=0) < W^* - W^R\,</math>

<math>q^2S_L + (1-q)^2(1-S_H) < q(1-q)(S_H - S_L)\,</math>

Which is less likely to hold when <math>S_H\,</math> and <math>S_L\,</math> are close (private info is almost unimportant), and more likely to hold when <math>1-S_H\,</math> is small (delay is not costly - not this is opposite of what the paper says) and <math>S_L\,</math> is large (duplication is costly).

'''For non-urgent problems''', when <math>\delta = 1\,</math>, decentralization gives:

<math>W^* - W^D (\delta=1) = q^2 S_L + \left (1-q)^2 \cdot(1 - \frac{\lambda - S_H}{\lambda + S_H - 2 \mu}\right)(1-S_H)\,</math>

Then decentralization is better than random chocie iff:

:<math>W^* - W^D (\delta=1) < W^* - W^R\,</math>

:<math>q^2 S_L + \left (1-q)^2 \cdot(1 - \frac{\lambda - S_H}{\lambda + S_H - 2 \mu}\right)(1-S_H) < q(1-q)(S_H - S_L)\,</math>