1
$\begingroup$

Given $2$ random variables $X, Y$ that take integer values with uniform distribution from $0$ to $100$. You play a game in which a random value of $x$ comes first & you have to decide if the random value of $Y$ will be greater than $x$ or less than $x$. You will get reward of $y$ if your guess is correct and else $0$. Find the value you may like to pay for the game.(That is maximize the expectation payoff by finding the value of threshold $c$ if $x\ge c$ --> you guess $y < x$ and if $x < c$ --> you guess $y > x$)

$\endgroup$
  • $\begingroup$ Equality of X and Y can occur with positive probability. So, I think that your decision should include greater or equal than x or less than x or vise versa. $\endgroup$ – Jimmy R. Nov 20 '14 at 10:13
1
$\begingroup$

You have two strategies say $l$ for guessing that $Y$ will be less than the given $x$ and $g$ for guessing that $Y$ will be greater than or equal to the given $x$ (since the case of equal is missing from the exercise, I make the assumption that you can put it with greater than). Now the expected reward $R$ of each choice is equal to $$R_l=\sum_{y=0}^{x-1}P(Y=y)\cdot y+P(Y\ge x)\cdot 0=\sum_{y=0}^{x-1}\frac{1}{101}\cdot y=\frac{1}{101}\cdot\frac{x(x-1)}{2}$$ and similarly $$\begin{align*}R_g&=P(Y< x)\cdot 0+\sum_{y=x}^{100}P(Y=y)\cdot y=\sum_{y=x}^{100}\frac{1}{101}\cdot y=\frac{1}{101}\cdot\frac{(x+100)(100-(x-1))}{2}\\\\&=\frac{1}{101}\cdot\frac{(100+x)(101-x)}{2}\end{align*}$$ Now choice $g$ has a greater (or equal) expected reward than choice $l$ iff $$\frac{1}{101}\cdot\frac{(100+x)(101-x)}{2}\ge \frac{1}{101}\cdot\frac{x(x-1)}{2}$$ which is equivalent to $$-2x^2+2x+10100\ge0 \implies -(x+70.565)(x-71.565)\ge 0$$ Thus for every $x$ in $\{0,1,\ldots 71\}$ $g$ has a greater expected payoff and the opposite for $x \in \{72, 73, \ldots, 100\}$. In total the optimal strategy of the player is $$\text{Optimal choice }=\begin{cases}\text{greater or equal than }, & 0\le x\le 71 \\ \text{less }, & 72\le x \le 100 \end{cases}$$ (The threshold $c$ is equal to $71.565$).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.