In a Nash equilibrium, no player has incentive to change their action, holding fixed the actions of the others. Here, actions are bids.
Take the action profile proposed by Osborne and Rubinstein. Does player one have incentive to increase bid? No, he will still win but pay more. Does he have incentive to lower it? No, he will lose the auction, and give up the surplus (at least zero) he is current receiving.
Do other's have incentive to lower their bids? No, they will continue to lose the auction. Do others have incentive to increase their bids? No, either they will continue to lose the auction, or if they raise it enough, they will win, but at a price that is at least as high as their valuation since $b_1 \geq v_2$.
Now, we consider your proposed strategy: player 1 pays $v_2$ and every other players bid whatever they want $b_i ≤ v_2$. This may not be a Nash equilibrium. If other's all bid zero for instance, than player 1 has incentive to lower his bid to zero, he will continue to win and pay nothing! In fact, if the other's leave player 1 and room to lower his bid, he will. That's why, in equilibrium we need someone else to bid what player one bids in equilibrium.
But, why are these the only equilibrium? Well, suppose someone else wins. If it is an equilibrium, they must have bid no more than their valuation. Otherwise, they would be better off losing the auction and so they would have incentive to bid zero. But, if the winning bidder isn't player 1 and is bidding a value no higher than their own valuation, player 1 has incentive to raise his bid to just above the currently winning bid and win the auction at a price that gives him a positive surplus. Thus, in equilibrium, player 1 has to win.
In (a) you are not asked whether bidding the valuation is a dominant strategy. Instead, you need to show that $(b_1,b_2,b_3)=(v_1,v_2,v_3)$ is not a Nash equilibrium. To do this, you just need to show that a player has a strictly profitable deviation. So, suppose $(b_1,b_2,b_3)=(v_1,v_2,v_3)$. Player 2's pay off is zero. If they increase their bid to something more than $v_1$, then they win the auction, pay the third highest bid $v_3$, and get a payoff of $v_2-v_3>0$. This is a strictly profitable deviation, so all players bidding their valuation is not a Nash equilibrium.
For (b), to show $(b_1,b_2,b_3)=(v_1,v_1,v_1)$ is a Nash equilibrium, you need to show that no player has a strictly profitable deviation from bidding $v_1$ (when the other players are bidding $v_1$). When players all bid $v_1$, they all get zero payoff (players 2 and 3 because they do not win the auction, and player 1 because they win the auction and pay $v_1$).
- If player 1 raises their bid, they still win the auction and pay $v_1$ and get zero payoff. If player 1 lowers their bid, they lose the auction and get zero payoff.
- If player 2 or 3 raise their bid, they win the auction and pay $v_1(>v_2>v_3)$ so make a loss. If player 2 or 3 lowers their bid, they lose the auction and get zero payoff.
Thus, no player has a strictly profitable deviation, and hence it is a Nash equilibrium for all players to bid the highest valuation $v_1$.
Best Answer
Truth-telling is not a dominant strategy with the third-price auction. Suppose I value an item at £100 and there are two other bids, £200 and £10. I should bid £205 and pay £10 for the item, even though this is more than my private valuation.
Similarly, truth-telling is not dominant in the average-price auction. Suppose I value the item at $100 and the bid is £10. If I bid my true value I will pay £55, but if I bid £12 I will pay only £11.
Both examples show that truth-telling is not a Nash equilibrium, because you could suppose that the other bids are at their truthful values.