# prisoner's dilemma bimatrix

I have a question about the following derivation

Consider the prisoner's dilemma with the following bimatrix:

$$(A, B) = \begin{pmatrix} (-5, -5) & (-1, -10) \\ (-10, -1) & (-2, -2) \end{pmatrix}$$

The payoff to the row player is given by:

$$\pi(x, y) = (x, 1 - x) \begin{pmatrix} -5 & -1 \\ -10 & -2 \end{pmatrix} \begin{pmatrix} y \\ 1 - y \end{pmatrix} = (x, 1 - x) \begin{pmatrix} -4y - 1 \\ -8y - 2 \end{pmatrix}$$

I don't understand the following:

Since $$-8y - 2 < -4y - 1$$ for any $$0 \leq y \leq 1$$, we have:

$$P = \{(1, y) : 0 \leq y \leq 1\}$$ (WHY?)

On the other hand:

$$\rho(x, y) = (x, 1 - x) \begin{pmatrix} -5 & -10 \\ -1 & -2 \end{pmatrix} \begin{pmatrix} y \\ 1 - y \end{pmatrix} = (-4x - 1, -8x - 2) \begin{pmatrix} y \\ 1 - y \end{pmatrix}$$

Since $$-8x - 2 < -4x - 1$$ for any $$0 \leq x \leq 1$$, we have:

$$Q = \{(x, 1) : 0 \leq x \leq 1\}$$

(AGAIN WHY?)

Now:

$$P \cap Q = \{(1, 1)\}$$

Therefore, the game has a unique Nash equilibrium $$(p, q) = ((1, 0), (1, 0))$$. $$\blacksquare$$

Edit:

$$π(x, y) = xAy^T$$

$$ρ(x, y) = xBy^T$$

P = {(x, y) : π(x, y) attains its maximum at x for fixed y.}

Q = {(x, y) : ρ(x, y) attains its maximum at y for fixed x.}

• How do you have 6 upvotes in 5 minutes? Also, what is P and Q? Commented May 19 at 4:52
• P = {(x, y) : π(x, y) attains its maximum at x for fixed y.} Q = {(x, y) : ρ(x, y) attains its maximum at y for fixed x.} I will edit my question as well
– user1324054
Commented May 19 at 5:04
• I think you need to give definition to the payoff as well right
– user1055322
Commented May 19 at 5:45
• Editted, thanks
– user1324054
Commented May 19 at 6:10
• In the first case, you want to maximise$$x(-4y - 1) + (1 - x)(-8y - 2)$$with respect to $x \in [0, 1]$, given $y \in [0, 1]$ is fixed. This is a weighted average of the two numbers $-4y - 1$ and $-8y - 2$. As observed, $-8y - 2 < -4y - 1$, since $y \ge 0$, so you want to give as little weight to $-8y - 2$ as possible, which you do by setting $x = 1$. You could also simplify:$$x(-4y - 1) + (1 - x)(-8y - 2) = x(-4y - 1) + (2 - 2x)(-4y - 2) = (2 - x)(-4y - 2),$$which is a product of the negative number $(-4y - 2)$ and $2 - x$. The maximum occurs when $2 - x$ is at minimum, i.e. when $x = 1$. Commented May 19 at 6:18

In the first case, you want to maximise $$x(−4y−1)+(1−x)(−8y−2)$$ with respect to $$x\in[0,1]$$, given $$y\in[0,1]$$ is fixed. This is a weighted average of the two numbers $$−4y−1$$ and $$−8y−2$$. As observed, $$−8y−2<−4y−1$$, since $$y\ge0$$, so you want to give as little weight to $$−8y−2$$ as possible, which you do by setting $$x=1$$. You could also simplify: $$x(−4y−1)+(1−x)(−8y−2)=x(−4y−1)+(2−2x)(−4y−2)=(2−x)(−4y−2),$$ which is a product of the negative number $$−4y−2$$ (which is constant with respect to $$x$$) and $$2−x$$. Since $$2 - x$$ is a decreasing function of $$x$$, $$(2 - x)(-4y - 2)$$ an increasing function of $$x$$. Thus, the maximum occurs when $$x$$ is at its maximum, i.e. $$x = 1$$.
The second case is identical to the first, except the $$x$$ and $$y$$ are reversed.