A city has just added 100 new female recruits to its police force. The city will provide a pension to each new hire who remains with the force until retirement. In addition, if the new hire is married at the time of her retirement, a second pension will be provided for her husband. A consulting actuary makes the following assumptions:

1. Each new recruit has a 0.4 probability of remaining with the police force until retirement.
2. Given that a new recruit reaches retirement with the police force, the probability that she is not married at the time of retirement is 0.25.
3. The events of different new hires reaching retirement and the events of different new hires being married at retirement are all mutually independent events.

Calculate the probability that the city will provide at most 90 pensions to the 100 new hires and their husbands.

• 0.60
• 0.67
• 0.75
• 0.93
• 0.99

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

Step 1: Determine Pension Probabilities for a Single Recruit

The problem describes three possible scenarios for each new recruit regarding pensions:

• 0 Pensions: The recruit does not remain with the police force until retirement.
• 1 Pension: The recruit remains with the police force until retirement but is not married at the time of retirement.
• 2 Pensions: The recruit remains with the police force until retirement and is married at the time of retirement.

Let's calculate the probability for each scenario based on the given information:

• Probability of remaining with the force until retirement = 0.4.
• Probability of not remaining with the force until retirement = [math]1 - 0.4 = 0.6[/math].
• Given she reaches retirement, probability of not being married = 0.25.
• Given she reaches retirement, probability of being married = [math]1 - 0.25 = 0.75[/math].

We can summarize these probabilities in the following table:

Pension Probabilities for a Single Recruit

Number of Pensions	Scenario	Probability Calculation	Probability
0	Recruit does not remain until retirement.	[math]P(\text{not retire})[/math]	0.6
1	Recruit retires AND is not married.	[math]P(\text{retire}) \times P(\text{not married } | \text{ retire}) = 0.4 \times 0.25[/math]	0.1
2	Recruit retires AND is married.	[math]P(\text{retire}) \times P(\text{married } | \text{ retire}) = 0.4 \times 0.75[/math]	0.3

Let [math]X[/math] be the number of pensions for a single recruit. The sum of probabilities is [math]0.6 + 0.1 + 0.3 = 1.0[/math].

Step 2: Calculate Mean and Variance for a Single Recruit

Using the probabilities from Step 1, we can calculate the expected number of pensions (mean) and the variance for a single recruit. Expected Number of Pensions ([math]E[X][/math]): The expected value is calculated as the sum of (each outcome multiplied by its probability).

[[math]]E[X] = (0 \times 0.6) + (1 \times 0.1) + (2 \times 0.3)[[/math]]

[[math]]E[X] = 0 + 0.1 + 0.6 = 0.7[[/math]]

Second Moment ([math]E[X^2][/math]): The second moment is calculated as the sum of (each outcome squared multiplied by its probability).

[[math]]E[X^2] = (0^2 \times 0.6) + (1^2 \times 0.1) + (2^2 \times 0.3)[[/math]]

[[math]]E[X^2] = (0 \times 0.6) + (1 \times 0.1) + (4 \times 0.3)[[/math]]

[[math]]E[X^2] = 0 + 0.1 + 1.2 = 1.3[[/math]]

Variance ([math]Var(X)[/math]): The variance is calculated using the formula [math]Var(X) = E[X^2] - (E[X])^2[/math].

[[math]]Var(X) = 1.3 - (0.7)^2[[/math]]

[[math]]Var(X) = 1.3 - 0.49 = 0.81[[/math]]

Step 3: Calculate Mean and Variance for 100 Recruits

We have 100 new hires, and the events for different new hires are mutually independent. Let [math]X_i[/math] be the number of pensions for the [math]i[/math]-th recruit. The total number of pensions, [math]T[/math], is the sum of pensions from all 100 recruits:

[[math]]T = \sum_{i=1}^{100} X_i[[/math]]

Total Expected Number of Pensions ([math]E[T][/math]): The expected value of the sum is the sum of the expected values.

[[math]]E[T] = \sum_{i=1}^{100} E[X_i] = 100 \times E[X][[/math]]

[[math]]E[T] = 100 \times 0.7 = 70[[/math]]

Total Variance of Pensions ([math]Var(T)[/math]): Since the events for different hires are independent, the variance of the sum is the sum of the variances.

[[math]]Var(T) = \sum_{i=1}^{100} Var(X_i) = 100 \times Var(X)[[/math]]

[[math]]Var(T) = 100 \times 0.81 = 81[[/math]]

Total Standard Deviation of Pensions ([math]\sigma_T[/math]): The standard deviation is the square root of the variance.

[[math]]\sigma_T = \sqrt{Var(T)} = \sqrt{81} = 9[[/math]]

Step 4: Apply Normal Approximation with Continuity Correction

Since the total number of pensions [math]T[/math] is the sum of 100 independent and identically distributed random variables, we can use the Central Limit Theorem to approximate the distribution of [math]T[/math] with a normal distribution. We need to calculate the probability that the city will provide at most 90 pensions, which is [math]P(T \le 90)[/math]. When using a normal approximation for a discrete distribution, we apply a continuity correction. "At most 90" means we are interested in the probability up to 90.5 on the continuous scale. So, we want to find [math]P(T \le 90.5)[/math]. Now, we standardize this value to a Z-score using the mean ([math]\mu_T = 70[/math]) and standard deviation ([math]\sigma_T = 9[/math]) calculated in Step 3.

[[math]]Z = \frac{T - \mu_T}{\sigma_T}[[/math]]

[[math]]Z = \frac{90.5 - 70}{9}[[/math]]

[[math]]Z = \frac{20.5}{9} \approx 2.2778[[/math]]

Rounding to two decimal places for standard normal tables, [math]Z \approx 2.28[/math]. Consulting a standard normal (Z) table for [math]P(Z \le 2.28)[/math]:

[[math]]P(Z \le 2.28) \approx 0.9887[[/math]]

Thus, the probability that the city will provide at most 90 pensions is approximately 0.9887.

Key Insights

• Complex probability scenarios for multiple individuals can be broken down by first analyzing the probabilities for a single individual.
• The expected value and variance for a sum of independent random variables can be found by summing their individual expected values and variances, respectively.
• For a large number of independent and identically distributed random variables, the Central Limit Theorem allows approximation of their sum's distribution with a normal distribution.
• When approximating a discrete probability distribution with a continuous one (like the normal distribution), a continuity correction is essential for accuracy. For [math]P(X \le k)[/math], use [math]P(X \le k + 0.5)[/math].
• The expected number of pensions per recruit (0.7) is a key metric, directly scaling up to the total expected pensions for the group.

This article was generated by AI and may contain errors. If permitted, please edit the article to improve it.

Solution: E

For a single recruit, the probability of 0 pensions is 0.6, of 1 pension is 0.4(0.25) = 0.1, and of 2 pensions is 0.4(0.75) = 0.3. The expected number of pensions is 0(0.6) + 1(0.1) + 2(0.3) = 0.7. The second moment is 0(0.6) + 1(0.1) + 4(0.3) = 1.3. The variance is 1.3 – 0.49 = 0.81. For 100 recruits the mean is 70 and the variance is 81. The probability of providing at most 90 pensions is (with a continuity correction) the probability of being below 90.5. This is (90.5 – 70)/9 = 2.28 standard deviations above the mean. From the tables, this probability is 0.9887.

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