Respuesta :
We would bid $0 as the expected value of this game is always negative.
I’ll assume that cash means there will not be fractions of a dollar in the box. Let X be the payout of the box and Y be the amount that we pay for the box after bidding. Then we have
P(X)=1.5X−Y, X≤Y
as ur profit function. Since E[P(X)]=1.5E[X]−Y, we first need to find the expected value of X and Y. Each integer between 0 and 1000 is equally likely to be in the box
(P(X=x)={1}/{1001}
Since we only realize the payout of the box if X≤Y we only need to sum up to Y:
E[X]= x=0∑Y1/1001 = Y(Y+1)/2002
To find the expected value of Y, note that we will pay Y dollars for the box if X≤Y and 0 otherwise. Given a Y, the probability that X≤Y is {Y+1}/{1001}
since each potential value of X shy of Y has probability {1}/{1001}
Then{E}[Y] = {Y(Y+1)}/{1001}
It follows that
[tex]\begin{aligned} \mathbb{E}[P(X)] &= 1.5\mathbb{E}[X] - E[Y] = \frac{1.5Y(Y+1)}{2002} - \frac{Y(Y+1)}{1001} \\ &= \frac{-3Y(Y+1)}{4004} \end{aligned}[/tex]
E[P(X)]=1.5E[X]−E[Y]= 1.5Y(Y+1)/2002 − Y(Y+1)/1001 = −3Y(Y+1)/4004
which is always negative so we should not bid any amount for the box. Our optimal bid then is $0, with expected payoff $0. Note that my original assumption was the impetus for using a summation; should I have assumed that any value was possible in the box, then we would have to approach X as though it were a continuous variable.
So, We would bid $0 as the expected value of this game is always negative.
Learn more about PROBABILITY here https://brainly.com/question/24756209
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