Changes

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

*There are infinite periods (<math>t\,</math>), discounted by <math>\delta\,</math>, and the strategy space is <math>\{Enter,Wait\}\,</math>

*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 hazard-rate probability that <math>B\,</math> will enter at exactly <math>t\,</math> of <math>h(t)\,</math>

*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>.

Then <math>A\,</math>'s expected payoff from entering at <math>t\,</math> is:

:<math>a(t)(\lambda - S) + a(t)(h(t)(\mu - \lambda)) = a(t)(\lambda - S - h(t)(\lambda -\mu))\,</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:

:<math>a(t_1)(\lambda - S_A^1 - h(t)(\lambda -\mu)) \ge a(t_2)(\lambda - S_A^1 - h(t)(\lambda -\mu))\,</math>

and

:<math>a(t_1)(\lambda - S_A^1 - h(t)(\lambda -\mu)) + a(t_2)(\lambda - S_A^2 - h(t)(\lambda -\mu)) \ge a(t_2)(\lambda - S_A^1 - h(t)(\lambda -\mu)) + a(t_1)(\lambda - S_A^2 - h(t)(\lambda -\mu))\,</math>

:<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>.

(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>

:<math>\therefore W^G = (1-S) - \underbrace{\left (1 - \frac{p(2-p)}{1-\delta(1-p)^2}\right)(1-S)}_{\mbox{Delay Loss}} - \underbrace{\frac{p^2}{1-\delta(1-p)^2}S}_{\mbox{Duplication Loss}}\,</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).

<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>