<p>The Nash bargaining solution DOES maximize the Nash product. You have to separate the playing of the game from the bargaining problem. If the players negotiate a binding agreement they will realize that their maximal total payoff from playing the game is <span class="math-container">$3$</span>. This can be achieved by playing <span class="math-container">$(T,L)$</span> or <span class="math-container">$(T,C)$</span>, or any mixture of those two profiles.</p>

<p>The bargaining solution just asks: Given a total payoff of <span class="math-container">$3$</span>, how will the players divide this payoff among them, given a threat point of <span class="math-container">$(0,0)$</span>? So the efficient frontier is (a part of) the line <span class="math-container">$u_1+u_2=3$</span>, and by symmetry (or by maximizing the Nash product <span class="math-container">$u_1(3-u_1)$</span>) you find <span class="math-container">$u_1=u_2=3/2$</span>. The only remaining question is how to play the game (under a binding agreement) in such a way that these are the expected payoffs. But you already solved this part.</p>

CC BY-SA 4.0
2020-03-18T10:42:27.347Z
Oh I think I start to get my head around it, so in a way we think of the Nash product maximization as being abstracted from the game?
Jhonny
CC BY-SA 4.0
2020-03-19T00:14:51.233Z
@Jhonny: Yes, I would say so.
VARulle
CC BY-SA 4.0
2020-03-12T14:14:26.487Z
2020-03-12T14:14:26.487Z
VARulle
<p>In my game theory class this term, we studied Nash bargaining. It is only now when starting to prepare for the exam that I have come to realise there is something I fundamentally don't understand, and hope that somebody here might be able to help. :)
This is a question highlighting my issue:</p>

<p><a href="https://i.sstatic.net/t4EdJ.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/t4EdJ.png" alt="enter image description here"></a></p>

<p>The Nash bargaining solution to this is (confirmed with the lecturer) the payoff pair (3/2,3/2). This is achieved by player 1 playing T and player 2 mixing L and C with 1/4, 3/4 probability. The perhaps simplest way to find this solution is to find the midpoint of a tangent to the convex set of possible utilities.</p>

<p><strong>My issue is, that I am not convinced that this maximises the Nash product s_1*s_2</strong> (the disagreement pair is (0,0)).</p>

<p><strong>IF</strong> one assumes that the payoff of player 1 and player 2 vary <em>independently</em>, then the above is indeed the solution because E[s_1] = 1/4*3 + 3/4*1 = 3/2 and E[s_2] = 1/4*0 + 3/4*2 = 3/2, which gives E[s_1*s_2] = 9/4 (which can be shown to be maximum).</p>

<p>However, the payoffs do <strong>not</strong> vary independently! (T,L) corresponds to Nash product 0 and (T,C) corresponds to 2. Hence the true expected value for the Nash product under this randomisation is 1/4*0+3/4*2=3/2. This is not optimal, as purely playing (T,C) gives a higher Nash product.</p>

<p>Hence, <strong>how comes that the Nash bargaining solution isn't actually maximising the Nash product? Where do I go wrong in my thinking?</strong></p>

<p>Very many thanks in advance! :) </p>

<game-theory><nash-equilibrium><cooperative-game-theory><bargaining>
Is the Nash product really maximised ex post?
Jhonny
# Is the Nash product really maximised ex post?

(Asked by: username_0 on 2020-03-12)

In my game theory class this term, we studied Nash bargaining. It is only now when starting to prepare for the exam that I have come to realise there is something I fundamentally don't understand, and hope that somebody here might be able to help. :)
This is a question highlighting my issue:

[![enter image description here](https://i.sstatic.net/t4EdJ.png)](https://i.sstatic.net/t4EdJ.png)

The Nash bargaining solution to this is (confirmed with the lecturer) the payoff pair (3/2,3/2). This is achieved by player 1 playing T and player 2 mixing L and C with 1/4, 3/4 probability. The perhaps simplest way to find this solution is to find the midpoint of a tangent to the convex set of possible utilities.

**My issue is, that I am not convinced that this maximises the Nash product s\_1\*s\_2** (the disagreement pair is (0,0)).

**IF** one assumes that the payoff of player 1 and player 2 vary *independently*, then the above is indeed the solution because E[s\_1] = 1/4\*3 + 3/4\*1 = 3/2 and E[s\_2] = 1/4\*0 + 3/4\*2 = 3/2, which gives E[s\_1\*s\_2] = 9/4 (which can be shown to be maximum).

However, the payoffs do **not** vary independently! (T,L) corresponds to Nash product 0 and (T,C) corresponds to 2. Hence the true expected value for the Nash product under this randomisation is 1/4\*0+3/4\*2=3/2. This is not optimal, as purely playing (T,C) gives a higher Nash product.

Hence, **how comes that the Nash bargaining solution isn't actually maximising the Nash product? Where do I go wrong in my thinking?**

Very many thanks in advance! :)

**Tags:** game-theory, nash-equilibrium, cooperative-game-theory, bargaining

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## 1 answers

### Answer by username_1 on 2020-03-12. Score: 4 | Accepted answer

The Nash bargaining solution DOES maximize the Nash product. You have to separate the playing of the game from the bargaining problem. If the players negotiate a binding agreement they will realize that their maximal total payoff from playing the game is $3$. This can be achieved by playing $(T,L)$ or $(T,C)$, or any mixture of those two profiles.

The bargaining solution just asks: Given a total payoff of $3$, how will the players divide this payoff among them, given a threat point of $(0,0)$? So the efficient frontier is (a part of) the line $u\_1+u\_2=3$, and by symmetry (or by maximizing the Nash product $u\_1(3-u\_1)$) you find $u\_1=u\_2=3/2$. The only remaining question is how to play the game (under a binding agreement) in such a way that these are the expected payoffs. But you already solved this part.

> **Comments:**
> **username_0**: Oh I think I start to get my head around it, so in a way we think of the Nash product maximization as being abstracted from the game?  
> **username_1**: @Jhonny: Yes, I would say so.  

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