Maximizing a function in $\mathbb{R}^2$ when we can only control $x$.

Question

Consider the following problem:

Let $x \in [0,1]$ and $y \in [0,1]$. I control $x$ and my opponent controls $y$. Neither of us know the value that the other chose. Let $z$ be defined to be

\begin{align*} z=8xy-3x-3y+1 \end{align*}

I am seeking to maximize $z$ by choosing an optimal $x$, even though I don't know what $y$ is. My opponent is trying to minimize $z$.

My attempt

We start by taking partial derivatives and setting equal to $0$ to find local maxima/minima. $$\frac{\partial z}{\partial x} =0 \Rightarrow y = \frac{3}{8}$$ $$\frac{\partial z}{\partial y} =0 \Rightarrow x = \frac{3}{8}$$

However, when graphing this function, I notice that $\left(\frac{3}{8}, \frac{3}{8}\right)$ is neither a max nor a min.

Nevertheless, we try to evaluate $z\left(x,\frac{3}{8}\right)$, and we notice that $$z\left(x,\frac{3}{8}\right) = 8x*\frac{3}{8} - 3x - 3*\frac{3}{8} +1 = -\frac{1}{8}$$ This indicates that regardless of my choice of $x$, the value will be $-1/8$. But can I do better? Can my opponent do better? How do I pick an optimal $x$ if I don't know what what $y$ is?

P.S. I'm very ignorant with game theory concepts.

Answer 1

This is a zero-sum game: your win is your opponent's loss.

A convenient principle of optimality for zero-sum games is to play your maximin strategy; i.e., play the strategy that maximizes the worst payoff your opponent may force upon you. Intuitively, the maxmin strategy maximizes your payoff if your opponent is trying to maximise hers (and in so doing she's hurting yours).

To compute this strategy, rewrite the payoff function as $$z(x,y) = 8xy-3x-3y+1 = (8x-3)y-3x+1$$ and compute $$\min_{y \in [0,1]} (8x-3)y-3x+1$$ If you choose $x>3/8$, then $(8x-3)>0$ so your opponent minimises your payoff (and maximize hers) by choosing $y=0$. If you choose $x<3/8$, then $(8x-3)<0$ and your opponent minimises your payoff by choosing $y=1$. If you choose $x=3/8$, your payoff is $-1/8$ regardless of her choice (as you have noted).

Summarizing, we have found that $$\min_{y \in [0,1]} (8x-3)y-3x+1 = \left\{ \begin{array}{ll} 1 - 3x & \mbox{if } x > 3/8 \\ -\frac{1}{8} & \mbox{if } x = 3/8 \\ 5x-2 & \mbox{if } x < 3/8 \end{array} \right.$$ Maximizing this piece-wise linear function with respect to ~$x \in [01,]$, you find $x^*=\frac{3}{8}$. Choosing this value ensures that your payoff cannot be smaller than $-\frac{1}{8}$, regardless of what your opponent plays.

Incidentally, this solution does not rely on the notion of Nash equilibrium. However, it can be shown that any pair of maxmin strategies for a zero-sum game defines a Nash equilibrium.

Answer 2

If neither player knows anything about the other's strategy, there's not much to analyze. Note that for $z(x,y)=8xy-3x-3y+1$, the part you control is $8xy-3x$ or $x(8y-3)$. We split the possible values of $y$, then once $y$ is chosen, $z(x,y)$ becomes a function of one variable so we maximize that.

1. $y=3/8$: Then $z(x,3/8)=-1/8$.
2. $y<3/8$: Then $8y-3<0$ so $x(8y-3)\le 0$. You should choose $0$, then $z(0,y)=-3(y)+1>-3(3/8)+1=z(x,3/8)$.
3. $y>3/8$: Then $8y-3>0$ so $x(8y-3)\ge 0$. You should choose $1$, then $z(1,y)=8y-3-3y+1=5y-2$. But notice that $5\Big(\frac{3}{8}\Big)-2=-1/8$. Hence for $y>3/8$, $5y-2>-1/8$, i.e. $z(1,y)>z(x,3/8)$.

At this point you have to be specific about what you mean by "optimal".

Also, if they play $3/8$, you can't do better. Note that there is some symmetry in this game: The part they control is $y(8x-3$) so their strategy is almost equivalent to your strategy. Except for switching the $x$'s and $y$'s, they choose 1 if $x<3/8$ and $0$ if $x>3/8$.

Depending on your choice of $x$, they may have done better playing some other value $y$ but this is after the game has been played.