Question

Suppose that the stock prices in the following three scenarios are \(\begin{array}{lrrr}\text { Scenario } & S(0) & S(1) & S(2) \\ \omega_{1} & 100 & 110 & 120 \\ \omega_{2} & 100 & 105 & 100 \\ \omega_{3} & 100 & 90 & 100\end{array}\) with probabilities \(1 / 4,1 / 4,1 / 2\), respectively. Find the expected returns \(E(K(1)), E(K(2))\) and \(E(K(0,2)) .\) Compare \(1+E(K(0,2))\) with \((1+\) \(E(K(1)))(1+E(K(2)))\)

Short answer

1 + E(K(0,2)) = 1.05 and (1 + E(K(1)))(1 + E(K(2))) = 1.0791.

Step-by-step solution

Calculate Returns

First, calculate the returns for each scenario at each time point. The return is given by the formula \( K(t) = \frac{S(t)}{S(t-1)} - 1 \).

Calculate Returns at Time 1, \( K(1) \)

For each scenario, calculate the return at time 1, \( K(1) \): For \( \omega_1 \): \( K(1) = \frac{S(1)}{S(0)} - 1 = \frac{110}{100} - 1 = 0.10 \). For \( \omega_2 \): \( K(1) = \frac{105}{100} - 1 = 0.05 \). For \( \omega_3 \): \( K(1) = \frac{90}{100} - 1 = -0.10 \).

Calculate Returns at Time 2, \( K(2) \)

For each scenario, calculate the return at time 2, \( K(2) \): For \( \omega_1 \): \( K(2) = \frac{120}{110} - 1 = 0.0909 \). For \( \omega_2 \): \( K(2) = \frac{100}{105} - 1 = -0.0476 \). For \( \omega_3 \): \( K(2) = \frac{100}{90} - 1 = 0.1111 \).

Calculate Returns from Time 0 to Time 2, \( K(0,2) \)

For each scenario, calculate the return from time 0 to time 2, \( K(0,2) \): For \( \omega_1 \): \( K(0,2) = \frac{120}{100} - 1 = 0.20 \). For \( \omega_2 \): \( K(0,2) = \frac{100}{100} - 1 = 0 \). For \( \omega_3 \): \( K(0,2) = \frac{100}{100} - 1 = 0 \).

Calculate Expected Returns Using Probabilities

Calculate the expected returns using the given probabilities. \( E(K(1)) = \frac{1}{4} \times 0.10 + \frac{1}{4} \times 0.05 + \frac{1}{2} \times (-0.10) = 0.0125 \). \( E(K(2)) = \frac{1}{4} \times 0.0909 + \frac{1}{4} \times (-0.0476) + \frac{1}{2} \times 0.1111 = 0.0661 \). \( E(K(0,2)) = \frac{1}{4} \times 0.20 + \frac{1}{4} \times 0 + \frac{1}{2} \times 0 = 0.05 \).

Compare \(1 + E(K(0,2))\) with \((1 + E(K(1)))(1 + E(K(2)))\)

Calculate \(1 + E(K(0,2))\) and \((1 + E(K(1)))(1 + E(K(2)))\): \( 1 + E(K(0,2)) = 1 + 0.05 = 1.05 \). \( (1 + E(K(1)))(1 + E(K(2))) = (1 + 0.0125)(1 + 0.0661) = 1.0791 \).

Key concepts

Stock Price Scenario Analysis

To better understand expected returns, consider a stock that has different prices in various scenarios. Imagine we have three stock price scenarios, as given:
Scenario 1: Prices start at 100 and then increase to 110 and 120.
Scenario 2: Prices begin at 100, go up to 105, and then return to 100.
Scenario 3: Prices start at 100, drop to 90, and then return to 100.
These scenarios are assigned probabilities of occurring: 1/4 (or 25%) for Scenario 1, 1/4 (or 25%) for Scenario 2, and 1/2 (or 50%) for Scenario 3. This kind of stock price scenario analysis lets us see how the stock performs in different market conditions and helps to calculate expected returns.

Expected Return

The 'expected return' is the average return you can expect to earn from an investment, taking into account all possible scenarios and their probabilities. To find the expected returns, first calculate the returns for each scenario at each time point. The return formula is:
\[ K(t) = \frac{S(t)}{S(t-1)} - 1 \]
For time 1 \[ K(1) \] Scenario 1: (110/100) - 1 = 0.10
Scenario 2: (105/100) - 1 = 0.05
Scenario 3: (90/100) - 1 = -0.10
For time 2 \[ K(2) \] Scenario 1: (120/110) - 1 = 0.0909
Scenario 2: (100/105) - 1 = -0.0476
Scenario 3: (100/90) - 1 = 0.1111
Now calculate returns from time 0 to time 2 \[ K(0,2) \] Scenario 1: (120/100) - 1 = 0.20
Scenario 2: (100/100) - 1 = 0
Scenario 3: (100/100) - 1 = 0
By taking the probability-weighted returns, we find the expected returns:

Probability-Weighted Average

A probability-weighted average takes into account the different probabilities of various outcomes to give a more accurate expected value. To calculate expected returns using probabilities, multiply each return by its probability and sum them up.
For time 1:
\[ E(K(1)) = \frac{1}{4} \times 0.10 + \frac{1}{4} \times 0.05 + \frac{1}{2} \times (-0.10) = 0.0125 \]
For time 2:
\[ E(K(2)) = \frac{1}{4} \times 0.0909 + \frac{1}{4} \times (-0.0476) + \frac{1}{2} \times 0.1111 = 0.0661 \]
From time 0 to 2:
\[ E(K(0,2)) = \frac{1}{4} \times 0.20 + \frac{1}{4} \times 0 + \frac{1}{2} \times 0 = 0.05 \] Compare \[ 1 + E(K(0,2)) \] and \[ (1 + E(K(1)))(1 + E(K(2))) \]:
\[ 1 + E(K(0,2)) = 1 + 0.05 = 1.05 \]
\[ (1 + E(K(1)))(1 + E(K(2))) = (1 + 0.0125)(1 + 0.0661) = 1.0791 \]
This method provides a comprehensive view of possible returns by accounting for both the varying returns and their likelihood of occurring.