An investor invests 100 dollars in a stock. Each month, the investment has probability 0.5 of increasing by 1.10 dollars and probability 0.5 of decreasing by 0.90 dollars. The changes in price in different months are mutually independent.

Calculate the probability that the investment has a value greater than 91 dollars at the end of month 100.

• 0.63
• 0.75
• 0.82
• 0.94
• 0.97

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Step 1: Define the Monthly Change and its Moments

Let [math]X_k[/math] be the random change in the investment's value during month [math]k[/math]. We are given that for each month, the investment has a 0.5 probability of increasing by $1.10 and a 0.5 probability of decreasing by $0.90. The expected value of [math]X_k[/math] is calculated as the sum of each outcome multiplied by its probability:

[[math]]\operatorname{E}(X_k) = (0.5)(1.10) + (0.5)(-0.90) = 0.55 - 0.45 = 0.10[[/math]]

The variance of [math]X_k[/math] is calculated using the formula [math]\operatorname{Var}(X) = \operatorname{E}(X^2) - (\operatorname{E}(X))^2[/math]. First, we calculate [math]\operatorname{E}(X_k^2)[/math]:

[[math]]\operatorname{E}(X_k^2) = (0.5)(1.10)^2 + (0.5)(-0.90)^2 = (0.5)(1.21) + (0.5)(0.81) = 0.605 + 0.405 = 1.01[[/math]]

Now, we can find the variance:

[[math]]\operatorname{Var}(X_k) = 1.01 - (0.10)^2 = 1.01 - 0.01 = 1.00[[/math]]

Step 2: Calculate the Expected Value and Variance of the Total Change

Let [math]S[/math] be the total change in the investment's value over 100 months. The changes in price in different months are mutually independent. This means that the expected value of the sum of these changes is the sum of their individual expected values, and the variance of the sum is the sum of their individual variances. The total change [math]S[/math] is given by:

[[math]]S = \sum_{k=1}^{100} X_k[[/math]]

The expected value of the total change [math]S[/math] is:

[[math]]\operatorname{E}(S) = \sum_{k=1}^{100} \operatorname{E}(X_k) = 100 \times 0.10 = 10[[/math]]

The variance of the total change [math]S[/math] is:

[[math]]\operatorname{Var}(S) = \sum_{k=1}^{100} \operatorname{Var}(X_k) = 100 \times 1.00 = 100[[/math]]

Step 3: Formulate the Probability and Standardize the Variable

The initial investment is $100. We need to calculate the probability that the investment's value is greater than $91 at the end of month 100. This can be expressed as:

[[math]]\operatorname{P}(\text{Initial Investment} + S \gt 91)[[/math]]

Substituting the initial investment value:

[[math]]\operatorname{P}(100 + S \gt 91)[[/math]]

Rearranging the inequality to isolate [math]S[/math]:

[[math]]\operatorname{P}(S \gt 91 - 100) = \operatorname{P}(S \gt -9)[[/math]]

Since [math]S[/math] is the sum of a large number (100) of independent and identically distributed random variables, we can apply the Central Limit Theorem (CLT). The CLT states that [math]S[/math] will be approximately normally distributed with mean [math]\operatorname{E}(S)[/math] and variance [math]\operatorname{Var}(S)[/math]. To find this probability using a standard normal distribution, we standardize [math]S[/math] by converting it to a Z-score:

[[math]]Z = \frac{S - \operatorname{E}(S)}{\sqrt{\operatorname{Var}(S)}}[[/math]]

Using the calculated values for [math]\operatorname{E}(S)[/math] and [math]\operatorname{Var}(S)[/math], and the value [math]S = -9[/math], the corresponding Z-score is:

[[math]]Z = \frac{-9 - 10}{\sqrt{100}} = \frac{-19}{10} = -1.9[[/math]]

Step 4: Calculate the Final Probability

We need to find the probability [math]\operatorname{P}(S \gt -9)[/math], which is equivalent to finding [math]\operatorname{P}(Z \gt -1.9)[/math] for a standard normal random variable [math]Z[/math]. Using the properties of the standard normal distribution:

[[math]]\operatorname{P}(Z \gt -1.9) = 1 - \operatorname{P}(Z \le -1.9)[[/math]]

Due to the symmetry of the standard normal distribution, [math]\operatorname{P}(Z \le -1.9) = \operatorname{P}(Z \ge 1.9)[/math]. Also, [math]\operatorname{P}(Z \ge 1.9) = 1 - \operatorname{P}(Z \lt 1.9)[/math]. Therefore, [math]\operatorname{P}(Z \gt -1.9) = 1 - (1 - \operatorname{P}(Z \lt 1.9)) = \operatorname{P}(Z \lt 1.9)[/math]. From standard normal distribution tables, the cumulative probability for [math]Z = 1.9[/math] is approximately 0.9713.

[[math]]\operatorname{P}(Z \gt -1.9) \approx 0.9713[[/math]]

Thus, the probability that the investment has a value greater than $91 at the end of month 100 is approximately 0.9713. This value is closest to 0.97 among the given options.

Key Insights

• For a discrete random variable, the expected value is [math]\operatorname{E}(X) = \sum x_i P(X=x_i)[/math] and the variance is [math]\operatorname{Var}(X) = \operatorname{E}(X^2) - (\operatorname{E}(X))^2[/math].
• For a sum of independent random variables [math]S = \sum X_k[/math], the expected value is the sum of individual expected values [math]\operatorname{E}(S) = \sum \operatorname{E}(X_k)[/math].
• For a sum of independent random variables [math]S = \sum X_k[/math], the variance is the sum of individual variances [math]\operatorname{Var}(S) = \sum \operatorname{Var}(X_k)[/math]. This property relies on the independence of the variables.
• The Central Limit Theorem (CLT) is applicable when summing a large number of independent and identically distributed (i.i.d.) random variables, allowing the approximation of their sum with a normal distribution.
• To calculate probabilities for an approximately normal variable, it is standardized to a Z-score using the formula [math]Z = \frac{\text{Value} - \text{Mean}}{\text{Standard Deviation}}[/math].
• The symmetry of the standard normal distribution can be used to simplify probability calculations, e.g., [math]\operatorname{P}(Z \gt -a) = \operatorname{P}(Z \lt a)[/math].

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