For a certain insurance company, 10% of its policies are Type A, 50% are Type B, and 40% are Type C. The annual number of claims for an individual Type A, Type B, and Type C policy follow Poisson distributions with respective means 1, 2, and 10.

Let [math]X[/math] represent the annual number of claims of a randomly selected policy. Calculate the variance of [math]X[/math].

• 5.10
• 16.09
• 21.19
• 42.10
• 47.20

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

Step 1: Understand the Problem and Define Parameters

The problem describes an insurance company with three types of policies: Type A, Type B, and Type C. The annual number of claims for an individual policy of each type follows a Poisson distribution, but with different mean claim rates. We are asked to calculate the variance of the annual number of claims ([math]X[/math]) for a randomly selected policy. Here are the given parameters for each policy type:

Policy Details

Policy Type	Probability ([math]P(\text{Type})[/math])	Poisson Mean ([math]\lambda[/math])
Type A	10% (0.1)	1
Type B	50% (0.5)	2
Type C	40% (0.4)	10

For a random variable [math]N[/math] following a Poisson distribution with mean [math]\lambda[/math], the following properties are essential for this problem:

• The expected value is [math]\operatorname{E}[N] = \lambda[/math].
• The variance is [math]\operatorname{Var}(N) = \lambda[/math].
• The second moment is [math]\operatorname{E}[N^2] = \operatorname{Var}(N) + (\operatorname{E}[N])^2 = \lambda + \lambda^2[/math].

We will use the formula [math]\operatorname{Var}(X) = \operatorname{E}[X^2] - (\operatorname{E}[X])^2[/math] to calculate the variance of [math]X[/math]. This requires us to first calculate [math]\operatorname{E}[X][/math] and [math]\operatorname{E}[X^2][/math] using the Law of Total Expectation.

Step 2: Calculate the Expected Value of X ([math]\operatorname{E}[X][/math])

To find the overall expected value of [math]X[/math], the annual number of claims for a randomly selected policy, we use the Law of Total Expectation. This law states that we can find the total expected value by weighting the expected value of claims for each policy type by its respective probability:

[[math]]\operatorname{E}[X] = \sum_{\text{all policy types } i} P(\text{Type } i) \cdot \operatorname{E}[X | \text{Type } i][[/math]]

Given that the expected value for a Poisson distribution is equal to its mean [math]\lambda[/math], we have:

• Expected claims for Type A: [math]\operatorname{E}[X | \text{Type A}] = \lambda_A = 1[/math]
• Expected claims for Type B: [math]\operatorname{E}[X | \text{Type B}] = \lambda_B = 2[/math]
• Expected claims for Type C: [math]\operatorname{E}[X | \text{Type C}] = \lambda_C = 10[/math]

Now, we substitute these values along with the probabilities into the Law of Total Expectation:

[[math]]\operatorname{E}[X] = (0.1 \times 1) + (0.5 \times 2) + (0.4 \times 10)[[/math]]

[[math]]\operatorname{E}[X] = 0.1 + 1.0 + 4.0[[/math]]

[[math]]\operatorname{E}[X] = 5.1[[/math]]

Step 3: Calculate the Second Moment of X ([math]\operatorname{E}[X^2][/math])

Next, we calculate the second moment of [math]X[/math], [math]\operatorname{E}[X^2][/math], also using the Law of Total Expectation. Similar to the expected value, we weight the second moment of claims for each policy type by its probability:

[[math]]\operatorname{E}[X^2] = \sum_{\text{all policy types } i} P(\text{Type } i) \cdot \operatorname{E}[X^2 | \text{Type } i][[/math]]

For a Poisson distribution with mean [math]\lambda[/math], the second moment is given by [math]\operatorname{E}[N^2] = \lambda + \lambda^2[/math]. Applying this formula for each policy type:

• Second moment for Type A: [math]\operatorname{E}[X^2 | \text{Type A}] = \lambda_A + \lambda_A^2 = 1 + 1^2 = 1 + 1 = 2[/math]
• Second moment for Type B: [math]\operatorname{E}[X^2 | \text{Type B}] = \lambda_B + \lambda_B^2 = 2 + 2^2 = 2 + 4 = 6[/math]
• Second moment for Type C: [math]\operatorname{E}[X^2 | \text{Type C}] = \lambda_C + \lambda_C^2 = 10 + 10^2 = 10 + 100 = 110[/math]

Now, we substitute these values along with the probabilities into the Law of Total Expectation for [math]X^2[/math]:

[[math]]\operatorname{E}[X^2] = (0.1 \times 2) + (0.5 \times 6) + (0.4 \times 110)[[/math]]

[[math]]\operatorname{E}[X^2] = 0.2 + 3.0 + 44.0[[/math]]

[[math]]\operatorname{E}[X^2] = 47.2[[/math]]

Step 4: Calculate the Variance of X ([math]\operatorname{Var}(X)[/math])

With the expected value [math]\operatorname{E}[X][/math] and the second moment [math]\operatorname{E}[X^2][/math] calculated, we can now find the variance of [math]X[/math] using the fundamental formula:

[[math]]\operatorname{Var}(X) = \operatorname{E}[X^2] - (\operatorname{E}[X])^2[[/math]]

Substitute the values we found in the previous steps:

• [math]\operatorname{E}[X^2] = 47.2[/math]
• [math]\operatorname{E}[X] = 5.1[/math]

So, the calculation proceeds as follows:

[[math]]\operatorname{Var}(X) = 47.2 - (5.1)^2[[/math]]

[[math]]\operatorname{Var}(X) = 47.2 - 26.01[[/math]]

[[math]]\operatorname{Var}(X) = 21.19[[/math]]

The variance of the annual number of claims for a randomly selected policy is 21.19.

Key Insights

• When dealing with mixed distributions (where a primary random variable determines the distribution of a secondary random variable), the Law of Total Expectation is fundamental for calculating overall moments (like [math]\operatorname{E}[X][/math] and [math]\operatorname{E}[X^2][/math]).
• The variance of any random variable [math]X[/math] can be efficiently calculated using the formula [math]\operatorname{Var}(X) = \operatorname{E}[X^2] - (\operatorname{E}[X])^2[/math]. This is often preferred over the Law of Total Variance when [math]\operatorname{E}[X^2][/math] is straightforward to compute conditionally.
• For a Poisson distribution with mean [math]\lambda[/math], it's crucial to remember that [math]\operatorname{E}[N] = \lambda[/math], [math]\operatorname{Var}(N) = \lambda[/math], and consequently, [math]\operatorname{E}[N^2] = \lambda + \lambda^2[/math]. These properties simplify calculations significantly in actuarial and statistical problems involving Poisson claims.
• Understanding how to combine probabilities and conditional expectations (or moments) across different segments of a population is a core skill in actuarial science for analyzing aggregate risk.

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Solution: C

For the Poisson distribution the variance is equal to the mean and hence the second moment is the mean plus the square of the mean. Then,

[[math]] \operatorname{E}[ X ] =0.1(1) + 0.5(2) + 0.4(10) =5.1 [[/math]]

[[math]] \operatorname{E}[ X^2 ] = 0.1(1 + 12 ) + 0.5(2 + 22 ) + 0.4(10 + 102 )= 47.2 [[/math]]

[[math]] \operatorname{Var}( X ) = 47.2 − 5.12 = 21.19. [[/math]]

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