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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.}

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  • $\begingroup$ How do you have 6 upvotes in 5 minutes? Also, what is P and Q? $\endgroup$ Commented May 19 at 4:52
  • $\begingroup$ 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 $\endgroup$
    – user1324054
    Commented May 19 at 5:04
  • $\begingroup$ I think you need to give definition to the payoff as well right $\endgroup$
    – user1055322
    Commented May 19 at 5:45
  • $\begingroup$ Editted, thanks $\endgroup$
    – user1324054
    Commented May 19 at 6:10
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    $\begingroup$ 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 $x = 1$. $\endgroup$ Commented May 19 at 6:18

1 Answer 1

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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.

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