Let's break down the **step-by-step solution** for optimal hedge ratio \( h^* \) and **verify** that it makes the portfolio risk-free, using the scenario described. --- ### **Step 1: Understanding the Problem** - **Spot price at \( t=0 \):** $3 - **At \( t=1 \):** Can go to $4 (up) or $2 (down) - **At \( t=2 \):** Stays at $4 (if up at t=1) or $2 (if down at t=1) - **Hedger's exposure:** Long 100 units spot (e.g., owns 100 units) - **Hedge:** Short \( 100h^* \) futures We want to **choose \( h^* \) so that the value of the portfolio is the same at \( t=2 \) regardless of whether the price rose or fell.** --- ### **Step 2: Portfolio Value at \( t=2 \) in Each Scenario** #### **Case 1: Price Rises (\$3 → \$4 → \$4)** - **Stock value at t=2:** \( 4 \times 100 = 400 \) - **Futures position:** Futures price at t=1 is \$4, at t=0 was \$3. - Loss at t=1 from futures: \( (3-4) \times 100h^* = -100h^* \) - This is a cash outflow at t=1. - This loss, if invested at 10% till t=2, grows to \( 1.1 \times (-100h^*) = -110h^* \) - **Net stock profit at t=1:** \( (4-3) \times 100 = 100 \) (but this is just the mark-to-market, actual value is \$400 at t=2) - **Total portfolio at t=2:** \[ \text{Stock value} + \text{Futures profit/loss} = 400 + (-110h^*) \] #### **Case 2: Price Falls (\$3 → \$2 → \$2)** - (Already shown in the image) - **Stock value at t=2:** \( 2 \times 100 = 200 \) - **Futures position:** Profit at t=1: \( (3-2) \times 100h^* = 100h^* \) - Invested to t=2: \( 1.1 \times 100h^* = 110h^* \) - **Total portfolio at t=2:** \[ 200 + 110h^* \] --- ### **Step 3: Setting the Portfolio Values Equal** To be **risk-free**, portfolio values must be equal **regardless of price movement at t=2**: \[ 400 - 110h^* = 200 + 110h^* \] --- ### **Step 4: Solve for \( h^* \)** \[ 400 - 110h^* = 200 + 110h^* \] \[ 400 - 200 = 110h^* + 110h^* \] \[ 200 = 220h^* \] \[ h^* = \frac{200}{220} = \frac{10}{11} \approx 0.909 \] --- ### **Step 5: Verification** - **If price goes up:** Portfolio = \( 400 - 110\times0.909 = 400 - 100 = 300 \) - **If price goes down:** Portfolio = \( 200 + 110\times0.909 = 200 + 100 = 300 \) **In both cases, final portfolio value is \$300.** --- ## **Final Answers** ### c) **Optimal hedge ratio, \( h^* \):** \[ \boxed{h^* = \frac{10}{11} \approx 0.909} \] ### d) **Verification:** With this \( h^* \), the portfolio value at \( t=2 \) is \$300 regardless of whether the price rises to \$4 or falls to \$2. Thus, the hedge is **perfect/risk-free**. --- **Summary Table:** | Scenario | Stock Value | Futures Value | Portfolio Value | |----------|-------------|--------------|----------------| | Up (\$4) | 400 | -100 | 300 | | Down (\$2)| 200 | +100 | 300 | --- If you need further breakdown or explanation of any step, let me know!