Question 1:
Suppose Kylie has the following reference-dependent value function:
V(x) = ln(x + 1) where 𝑥≥0 and -3ln(x - 1) where 𝑥<0.
𝑥 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)

To determine which gamble Kylie prefers, we need to compute the expected value of her value function $ V(x) $ 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:
\[
V(x) = 
\begin{cases} 
\ln(x + 1) & \text{if } x \geq 0 \\
-3 \ln(1 - x) & \text{if } x < 0 
\end{cases}
\]

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.

Explore how Kylie makes decisions between two gambles, A and B, based on her reference-dependent value function. Calculate the expected values for each gamble and understand Kylie's preference according to utility theory and probability weighting.

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