Math Problem Statement
Question 1: Suppose Kylie has the following reference-dependent value function: V(x) = ln(x + 1) where š„ā„0 and -3ln(x - 1) where š„<0. š„ is the change in Kylie's position relative to her reference point. Kylie weights probabilities linearly.(c) Kylie considers a choice between:
- gamble A: (0.5, $200; 0.5, -$100). That is, gamble A gives a 50% chance of winning $200 and a 50% chance of losing $100.
- gamble B: (0.5, $100; 0.5, -$50). That is, gamble B gives a 50% chance of winning $100 and a 50% chance of losing $50. Will Kylie prefer gamble A or gamble B? (2 marks)
Solution
To determine which gamble Kylie prefers, we need to compute the expected value of her value function for each gamble. This involves calculating the value Kylie assigns to each outcome using her reference-dependent value function and then taking the weighted average based on the probabilities.
Step 1: Define Kylie's Value Function
Kylie's value function is given by:
\begin{cases} \ln(x + 1) & \text{if } x \geq 0 \\ -3 \ln(1 - x) & \text{if } x < 0 \end{cases}$$ where $$x$$ is the change in her position relative to the reference point. ### Step 2: Calculate the Expected Value of Each Gamble **For Gamble A:** - **Outcome 1:** $$x = 200$$ with probability 0.5 $$V(200) = \ln(200 + 1) = \ln(201)$$ - **Outcome 2:** $$x = -100$$ with probability 0.5 $$V(-100) = -3 \ln(1 - (-100)) = -3 \ln(101)$$ The expected value $$E[V(x)]$$ for Gamble A is: $$E[V(x)]_A = 0.5 \cdot V(200) + 0.5 \cdot V(-100)$$ $$E[V(x)]_A = 0.5 \cdot \ln(201) + 0.5 \cdot (-3 \ln(101))$$ **For Gamble B:** - **Outcome 1:** $$x = 100$$ with probability 0.5 $$V(100) = \ln(100 + 1) = \ln(101)$$ - **Outcome 2:** $$x = -50$$ with probability 0.5 $$V(-50) = -3 \ln(1 - (-50)) = -3 \ln(51)$$ The expected value $$E[V(x)]$$ for Gamble B is: $$E[V(x)]_B = 0.5 \cdot V(100) + 0.5 \cdot V(-50)$$ $$E[V(x)]_B = 0.5 \cdot \ln(101) + 0.5 \cdot (-3 \ln(51))$$ ### Step 3: Compare Expected Values of the Two Gambles Now, let's compute these expected values numerically to compare them. #### Numerical Calculation Let's calculate the natural logarithm values using a calculator: - $$\ln(201) \approx 5.303$$ - $$\ln(101) \approx 4.615$$ - $$\ln(51) \approx 3.932$$ Plug these values into the expressions for the expected values: $$E[V(x)]_A = 0.5 \cdot 5.303 + 0.5 \cdot (-3 \cdot 4.615)$$ $$E[V(x)]_A = 0.5 \cdot 5.303 - 1.5 \cdot 4.615$$ $$E[V(x)]_A = 2.6515 - 6.9225 = -4.271$$ $$E[V(x)]_B = 0.5 \cdot 4.615 + 0.5 \cdot (-3 \cdot 3.932)$$ $$E[V(x)]_B = 0.5 \cdot 4.615 - 1.5 \cdot 3.932$$ $$E[V(x)]_B = 2.3075 - 5.898 = -3.5905$$ ### Step 4: Conclusion Since $$E[V(x)]_B = -3.5905 > E[V(x)]_A = -4.271$$, Kylie will prefer **Gamble B** over Gamble A, as it has a higher expected value according to her value function. Would you like more details on this explanation, or have any questions about how the calculations were done? ### Related Questions: 1. How would Kylie's preference change if the probabilities of the gambles were different? 2. What if Kylie had a different reference point for her value function? 3. How does the coefficient -3 in the loss function affect Kylie's decision-making? 4. Could Kylie be risk-averse or risk-seeking based on this value function? Why? 5. How would Kylie's preferences change if we introduce a third gamble with different outcomes? **Tip:** When analyzing decision-making under uncertainty, consider both the probabilities of outcomes and how the decision-maker values gains and losses.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Expected Value
Utility Theory
Probability
Formulas
V(x) = ln(x + 1) for x ā„ 0
V(x) = -3ln(x - 1) for x < 0
Theorems
-
Suitable Grade Level
College
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