Symmetrical nash equilibrium

I read this paper on placement public auctions for internet advertisements. Primarily, there are N ports each with an anticipated variety of clicks (in a certain amount of time) $x$. Each representative tries $B_i$ of just how much they agree to ppc. The proposals are placed in lowering order and also representative that makes the $i^\text{th}$ highest possible proposal obtains obtains the port with the $i^\text{th}$ highest possible click via price for a rate $P_i$ equivalent to $B_{i+1}$ (with the exception of the last representative that pays absolutely nothing).

To get the least rate to win the $i^\text{th}$ port, we keep in mind that we need to defeat the $i^\text{th}$ representative's proposal if we are going up, yet that we just need to defeat the $i^\text{th}$ representatives rate if we are relocating down. We conveniently get the adhering to formulas for Nash Equilibria:

$(v_s-p_s)x_s\ge(v_s-p_t)x_t$ for $t>s$

$(v_s-p_s)x_s\ge(v_s-P_{t-1})x_t$ for $t < s$

The paper after that specifies the symmetrical Nash equilibrium to be a set of rates with:

$(v_s-p_s)x_s\ge(v_s-p_t)x_t$ for all $t$ and also $s$

Primarily, as opposed to having the 2nd component of the previous problems, the first component of the previous formulas stands anywhere.

Is the symmetrical Nash equilibrium specified extra usually? Specifically, is it the like the symmetric equilibrium in this Wikipedia article

0
2019-05-05 16:35:18
Source Share
Answers: 1

The idea "symmetrical stability" (the one from Wikipedia write-up) is not relevant below, due to the fact that the video game is not symmetrical (various gamers have various "earnings per click").

I've an appearance at the paper and also I assume, that "symmetrical Nash equilibrium" in your instance is just practically hassle-free instance of Nash equilibrium (the latter is not one-of-a-kind in your instance).

Additionally I've kept in mind a potential misprint in the evidence of the "Fact 1". It needs to be $(v_s-p_s)x_s \geq (v_s-p_{S+1})x_{S+1}$ as opposed to $(v_s-p_s)x_s \geq (v_{S+1}-p_{S+1})x_{S+1}$.

0
2019-05-08 21:20:50
Source