# Min/Max Expectation problem (very difficult)

Problem: The manager of a fish market speculates that the number of requests for salmon on any given day is a random variable $X$ with the probability function

$$f(x)=\begin{cases} \frac14,&\text{if }x=0\\\\ \frac12,&\text{if }x=1\\\\ \frac14,&\text{if }x=2\;. \end{cases}$$

There is a profit of $\$2$on each salmon he sells and a loss of$\$1$ on each salmon he does not sell. Assuming that each salmon can be sold only on the day it is up for sale, that each request is for a single salmon, and that the managers's speculation is correct, find the number of salmons the market should have per day to maximize profit.

Attempt at a solution: So, what I think this is saying is that there is a $.25$ chance that no salmon will be requested, a $.50$ chance that $1$ salmon will be requested, and a $.25$ chance that $2$ salmon will be requested. This adds up to $1$, so it makes sense. In the second part, it states that each request is for a single salmon, but each the probability function of that only gives $.50$ (not sure if this is right)

Anyway, I attempt trial and error. So if we start with $10$ salmon, then he sells only $5$, and the other $5$ go to waste. So he loses $\$5$. If he starts with$8$salmon, he loses$\$4$ ($8 *.5 *2 - 8*.5 * \$1$), gives us$4$... Continue this to$0$, he loses none, but has no salmon. I'm lost I think I need help modeling and solving , thanks in advance - ## 2 Answers What is the range of salmon he can sell? It appears from the question that x can only equal 0, 1, or 2. - I don't think there is a limit - at least I believe this is what the question is asking, i.e. to find the number of salmon he can sell that maximizes his profit. – EulerChild May 1 '12 at 2:48 The probability that he will sell 0, 1, or 2 is .25, .50, and .25 respectively. – EulerChild May 1 '12 at 2:49 This suggests that there is a probability of 0 that a request for 3 or more salmon will be made. As such the only requests that can be made are for 0, 1, or 2 salmon. There is a 25% chance that the request will be 0, a 50% chance the request will be 1 and a 25% chance the request will be 2. This means that there is a 75% chance that the request will be for at least one salmon. The expected value of stocking one salmon is$1.50 (0.75x2). This is higher than the expected value of stocking 2 salmon (0.25x4), and so the profit maximizing amount should be x=1. –  user30343 May 1 '12 at 2:58
Interesting! I think he can either sell 0, 2 (1 and 1), 5 (1 and 2 and 2), or 8 (2*.25 * 4) . Not really sure, this problem is kind of crazy –  EulerChild May 1 '12 at 2:59
haha nice! Got it! Thank you!!!!! :) –  EulerChild May 1 '12 at 3:00

The statement that each request is for a single salmon just means that no customer ever asks for more than one salmon at a time. Since $f(0)+f(1)+f(2)=1$, the demand on any given day is always $0,1$, or $2$; there is no chance that he will sell more than two salmon, so he certainly should not order more than two. This is a small enough problem that we can easily tabulate the possible outcomes:

$$\begin{array}{c} \qquad\qquad\qquad\qquad\qquad\text{Orders:}\\ \begin{array}{rc|ccc} &&0&1&2\\ \hline &0&0&-1&-2\\ \text{Gets Requests For:}&1&0&2&1\\ &2&0&2&4 \end{array} \end{array}$$

Now we calculate the expected profit for each of his three possible orders. (Well, okay: technically an order of $10$ salmon is possible, but we know that he loses money on it, so we rule it out.)

Clearly he is certain to have a profit of $\$0$if he orders$0$salmon. If he orders$1$salmon, his expected profit is$\frac14(-1)+\frac12\cdot2+\frac14\cdot2=\$2.25$.

And if he orders $2$ salmon, his expected profit is $\frac14(-2)+\frac12\cdot1+\frac14\cdot4=\$1\$.

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Awesome, thanks!!!! –  EulerChild May 1 '12 at 3:09