Problem 23 - A First Course in Probability

Question

You have \(\$ 1000,\) and a certain commodity presently sells for \(\$ 2\) per ounce. Suppose that after one week the commodity will sell for either \(\$ 1\) or \(\$ 4\) an ounce, with these two possibilities being equally likely. (a) If your objective is to maximize the expected amount of money that you possess at the end of the week, what strategy should you employ? (b) If your objective is to maximize the expected amount of the commodity that you possess at the end of the week, what strategy should you employ?

Step-by-Step Solution

Answer

For both objectives (a) and (b), the optimal strategy is to invest all available money ($1000) in the commodity, resulting in the purchase of 500 ounces. This maximizes the expected amount of money and commodity possessed at the end of the week.

Step 1: Objective (a): Maximize the expected amount of money at the end of the week

Let X be the amount of the commodity we buy in ounces. Then the amount of money invested in the commodity is 2X and the remaining money is 1000 - 2X. After one week, the expected value of the invested money is: \(E_{1} = \frac{1}{2}(1 \cdot 2X ) + \frac{1}{2}(4 \cdot 2X) = X + 4X = 5X\) Now, we need to maximize the total expected amount of money. This is the sum of the expected value E1 and the remaining money (1000 - 2X). Expected Total Money = E1 + (1000 - 2X) To find the maximum, we need to find the derivative of this function and set it to zero. \( \frac{d}{dX}(5X + (1000 - 2X)) = 0 \)

Step 2: Find the maximum point

Now, we'll differentiate the expression and find the optimal X value: \(\frac{d}{dX}(3X + 1000) = 3\) Since the derivative is a constant (unaffected by X), it means that the expected amount of money is constantly increasing with respect to X. Therefore, to maximize the expected amount of money, we should invest all 1000 dollars in the commodity: \(X = \frac{1000}{2} = 500\) ounces

Step 3: Objective (b): Maximize the expected amount of the commodity

After one week, we will have either X ounces (if the price drops to 1 dollar) or X/2 ounces (if the price goes up to 4 dollars). The expected value of the commodity is: \( E_{2} =\frac{1}{2}(X) + \frac{1}{2}(\frac{X}{2}) = \frac{3}{4}X \) Now we need to find X that maximizes E2. As E2 increases as X increases, we should also invest all 1000 dollars in the commodity for this objective: \(X = \frac{1000}{2} = 500\) ounces Thus, the best strategies for both (a) and (b) are the same: invest all available money ($1000) in the commodity, which results in buying 500 ounces.