<p>the payoff from play the trigger strategy will be:
$$ \sum_{i=0}^{\infty}(x+1) \delta^i=\frac{x+1}{1-\delta}$$ </p>

<p>if I deviate and I play $l$ or $d$ the payoff will be
$$ 2x + \sum_{i=1}^{\infty}x \delta^i = 2x+ x\frac{\delta}{1-\delta}=\frac{2x(1-\delta) + x\delta}{1-\delta}= \frac{x(2-\delta)}{1-\delta}$$</p>

<p>then, the condition is $$ \frac{x+1}{1-\delta} \geq \frac{x(2-\delta)}{1-\delta}$$</p>

<p>I hope that this will help you ;)</p>

CC BY-SA 3.0
2015-04-18T00:26:05.990Z
2015-04-18T00:41:39.450Z
2015-04-18T00:41:39.450Z
hllspwn
<p>I have done most of the legwork, but I have fallen short at the final hurdle.
Could you please correct my mistake(s)?</p>

<p><img src="https://i.sstatic.net/Wvla4.jpg" alt="enter image description here"></p>

<p>Given the condition $x&gt;0$ I need to consider the infinitely repeated game where the above strategic form is the stage game. The discount factor: $\delta=\frac{1}{2}$.</p>

<p>I need to find a further condition on the game, such that we have a subgame perfect equilibrium in which both players "cooperate" in each period.</p>

<p><strong>What I have done so far</strong></p>

<p>The profile of strictly dominated strategies $(u,r)$ constitutes the cooperative strategy where $x&gt;0$ and $x \ne 1$.</p>

<p>The prescription (cooperative strategy) will therefore be $(u,r)=(x+1,x+1)$</p>

<p>The grim trigger is $(d,l) = (x,x)$</p>

<p><strong>Payoff from obeying the prescription:</strong></p>

<p>$(x+1) + \delta(x+1) + \delta^2(x+1) +\delta^3(x+1)+... = \frac{x+1}{1-\delta}$</p>

<p><strong>Payoff from deviating</strong> (<em>here comes the mistake</em>)<strong>:</strong></p>

<p>$2x + \delta x + \delta^2 x +\delta^3 x +... =\color{red}{\frac{2x-\delta(x+1)}{1-\delta}}$</p>

<p>The part in red should apparently be:</p>

<p>$$\frac{(2-\delta)x}{1-\delta}$$</p>

<p>I don't understand why I am wrong, could you please explain?
My understanding was that if a player deviates in period 1 in order to obtain $2x$ then both players would have to play the grim trigger profile $(d,l)$ from period 2 to perpetuity. As a result, the players would be missing out on $\delta(x+1)$.  The payoff numerator would therefore be $\color{red}{2x - \delta(x+1)}$ instead of $\color{blue}{(2-\delta)x}$.</p>

<p>Given that the correct numerator is $\color{blue}{(2-\delta)x}$, I can easily finish the question:</p>

<p>$$\frac{x+1}{1-\delta} \ge \frac{(2-\delta)x}{1-\delta}$$</p>

<p>Therefore, we would have a subgame perfect equilibrium <strong>iff</strong> $x \le 2$.</p>

<p>The problem is that my mistake lies in:</p>

<blockquote>
  <p>$2x + \delta x + \delta^2 x +\delta^3 x +...$
  $=\color{red}{\frac{2x-\delta(x+1)}{1-\delta}}$</p>
</blockquote>

<p>But, I'm not sure why. I would like to know why I am wrong and why $\color{blue}{(2-\delta)x}$ is the correct numerator.</p>

<p>Thanks.</p>

<game-theory>
Infinitely repeated game: prisoners' dilemma
Five σ
# Infinitely repeated game: prisoners' dilemma

(Asked by: username_0 on 2015-04-17)

I have done most of the legwork, but I have fallen short at the final hurdle.
Could you please correct my mistake(s)?

![enter image description here](https://i.sstatic.net/Wvla4.jpg)

Given the condition $x>0$ I need to consider the infinitely repeated game where the above strategic form is the stage game. The discount factor: $\delta=\frac{1}{2}$.

I need to find a further condition on the game, such that we have a subgame perfect equilibrium in which both players "cooperate" in each period.

**What I have done so far**

The profile of strictly dominated strategies $(u,r)$ constitutes the cooperative strategy where $x>0$ and $x \ne 1$.

The prescription (cooperative strategy) will therefore be $(u,r)=(x+1,x+1)$

The grim trigger is $(d,l) = (x,x)$

**Payoff from obeying the prescription:**

$(x+1) + \delta(x+1) + \delta^2(x+1) +\delta^3(x+1)+... = \frac{x+1}{1-\delta}$

**Payoff from deviating** (*here comes the mistake*)**:**

$2x + \delta x + \delta^2 x +\delta^3 x +... =\color{red}{\frac{2x-\delta(x+1)}{1-\delta}}$

The part in red should apparently be:

$$\frac{(2-\delta)x}{1-\delta}$$

I don't understand why I am wrong, could you please explain?
My understanding was that if a player deviates in period 1 in order to obtain $2x$ then both players would have to play the grim trigger profile $(d,l)$ from period 2 to perpetuity. As a result, the players would be missing out on $\delta(x+1)$. The payoff numerator would therefore be $\color{red}{2x - \delta(x+1)}$ instead of $\color{blue}{(2-\delta)x}$.

Given that the correct numerator is $\color{blue}{(2-\delta)x}$, I can easily finish the question:

$$\frac{x+1}{1-\delta} \ge \frac{(2-\delta)x}{1-\delta}$$

Therefore, we would have a subgame perfect equilibrium **iff** $x \le 2$.

The problem is that my mistake lies in:

> $2x + \delta x + \delta^2 x +\delta^3 x +...$
> $=\color{red}{\frac{2x-\delta(x+1)}{1-\delta}}$

But, I'm not sure why. I would like to know why I am wrong and why $\color{blue}{(2-\delta)x}$ is the correct numerator.

Thanks.

**Tags:** game-theory


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## 1 answers

### Answer by username_1 on 2015-04-18. Score: 6 | Accepted answer

the payoff from play the trigger strategy will be:
$$ \sum\_{i=0}^{\infty}(x+1) \delta^i=\frac{x+1}{1-\delta}$$

if I deviate and I play $l$ or $d$ the payoff will be
$$ 2x + \sum\_{i=1}^{\infty}x \delta^i = 2x+ x\frac{\delta}{1-\delta}=\frac{2x(1-\delta) + x\delta}{1-\delta}= \frac{x(2-\delta)}{1-\delta}$$

then, the condition is $$ \frac{x+1}{1-\delta} \geq \frac{x(2-\delta)}{1-\delta}$$

I hope that this will help you ;)


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