The number of hurricanes that will hit a certain house in the next ten years is Poisson distributed with mean 4. Each hurricane results in a loss that is exponentially distributed with mean 1000. Losses are mutually independent and independent of the number of hurricanes.

Calculate the variance of the total loss due to hurricanes hitting this house in the next ten years.

• 4,000,000
• 4,004,000
• 8,000,000
• 16,000,000
• 20,000,000

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

Step 1: Define Random Variables and Their Distributions

Let's define the random variables involved and their given properties.

• • Number of Hurricanes ([math]N[/math]):**

The number of hurricanes in the next ten years is given as Poisson distributed.

• Mean: [math]\operatorname{E}[N] = 4[/math]
• Variance: [math]\operatorname{Var}[N] = 4[/math] (For a Poisson distribution, the mean is equal to the variance.)
  • Loss per Hurricane ([math]X_i[/math]):**

The loss due to the [math]i^{th}[/math] hurricane is exponentially distributed.

• Mean: [math]\operatorname{E}[X_i] = 1,000[/math]
• Variance: [math]\operatorname{Var}[X_i] = (\operatorname{E}[X_i])^2 = 1,000^2 = 1,000,000[/math] (For an exponential distribution with mean [math]\theta[/math], the variance is [math]\theta^2[/math].)

Important: Losses are mutually independent and independent of the number of hurricanes.

Step 2: Define Total Loss and State the Law of Total Variance

Let [math]X[/math] represent the total loss due to hurricanes in the next ten years. This total loss is a sum of the individual hurricane losses, where the number of losses itself is a random variable:

[[math]]X = \sum_{i=1}^{N} X_i[[/math]]

To calculate the variance of the total loss [math]X[/math], we use the Law of Total Variance (also known as the Conditional Variance Formula), which is crucial for compound distributions:

[[math]]\operatorname{Var}(X) = \operatorname{E}[\operatorname{Var}(X | N)] + \operatorname{Var}[\operatorname{E}(X | N)][[/math]]

Step 3: Calculate Conditional Expectation and Variance

We need to first find the conditional expectation and conditional variance of the total loss [math]X[/math], given a fixed number of hurricanes [math]N[/math].

• • Conditional Expectation of Total Loss ([math]\operatorname{E}(X | N)[/math]):**

If there are exactly [math]N[/math] hurricanes, the expected total loss is the sum of the expected losses from each of those [math]N[/math] hurricanes. Since each [math]\operatorname{E}[X_i] = 1,000[/math]:

[[math]]\operatorname{E}(X | N) = \operatorname{E}\left(\sum_{i=1}^{N} X_i \Big| N\right) = \sum_{i=1}^{N} \operatorname{E}(X_i) = N \cdot \operatorname{E}(X_1) = 1,000N[[/math]]

• • Conditional Variance of Total Loss ([math]\operatorname{Var}(X | N)[/math]):**

If there are exactly [math]N[/math] hurricanes, the variance of the total loss is the sum of the variances of individual hurricane losses, because the individual losses [math]X_i[/math] are mutually independent. Since each [math]\operatorname{Var}[X_i] = 1,000,000[/math]:

[[math]]\operatorname{Var}(X | N) = \operatorname{Var}\left(\sum_{i=1}^{N} X_i \Big| N\right) = \sum_{i=1}^{N} \operatorname{Var}(X_i) = N \cdot \operatorname{Var}(X_1) = 1,000,000N[[/math]]

Step 4: Apply the Law of Total Variance to Compute [math]\operatorname{Var}(X)[/math]

Now, we substitute the expressions for [math]\operatorname{E}(X | N)[/math] and [math]\operatorname{Var}(X | N)[/math] into the Law of Total Variance formula:

[[math]]\operatorname{Var}(X) = \operatorname{E}[\operatorname{Var}(X | N)] + \operatorname{Var}[\operatorname{E}(X | N)][[/math]]

Substitute the expressions from Step 3:

[[math]]\operatorname{Var}(X) = \operatorname{E}[1,000,000N] + \operatorname{Var}[1,000N][[/math]]

Using the properties of expectation and variance ([math]\operatorname{E}[cX] = c\operatorname{E}[X][/math] and [math]\operatorname{Var}[cX] = c^2\operatorname{Var}[X][/math]):

[[math]]\operatorname{Var}(X) = 1,000,000 \cdot \operatorname{E}[N] + (1,000)^2 \cdot \operatorname{Var}[N][[/math]]

[[math]]\operatorname{Var}(X) = 1,000,000 \cdot \operatorname{E}[N] + 1,000,000 \cdot \operatorname{Var}[N][[/math]]

From Step 1, we know that [math]\operatorname{E}[N] = 4[/math] and [math]\operatorname{Var}[N] = 4[/math] (since [math]N[/math] is Poisson distributed with mean 4). Substitute these values:

[[math]]\operatorname{Var}(X) = 1,000,000 \cdot 4 + 1,000,000 \cdot 4[[/math]]

[[math]]\operatorname{Var}(X) = 4,000,000 + 4,000,000[[/math]]

[[math]]\operatorname{Var}(X) = 8,000,000[[/math]]

The variance of the total loss due to hurricanes is $8,000,000.

Key Insights

• The problem involves a compound distribution where the number of events (hurricanes) and the severity of each event (loss) are random variables.
• The Law of Total Variance (conditional variance formula) is essential for calculating the variance of a compound sum: [math]\operatorname{Var}(X) = \operatorname{E}[\operatorname{Var}(X | N)] + \operatorname{Var}[\operatorname{E}(X | N)][/math].
• For a Poisson distribution, the mean equals the variance ([math]\operatorname{E}[N] = \operatorname{Var}[N] = \lambda[/math]).
• For an exponential distribution with mean [math]\theta[/math], the variance is [math]\theta^2[/math].
• When individual losses are independent, [math]\operatorname{E}(\sum X_i | N) = N \cdot \operatorname{E}(X_1)[/math] and [math]\operatorname{Var}(\sum X_i | N) = N \cdot \operatorname{Var}(X_1)[/math].
• Properties of expectation and variance are crucial for simplifying terms: [math]\operatorname{E}[cY] = c\operatorname{E}[Y][/math] and [math]\operatorname{Var}[cY] = c^2\operatorname{Var}[Y][/math].

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Solution: C

Let [math]N[/math] denote the number of hurricanes, which is Poisson distributed with mean and variance 4.

Let [math]X_i[/math] denote the loss due to the [math]i^{th}[/math] hurricane, which is exponentially distributed with mean 1,000 and therefore variance 1,000² = 1,000,000.

Let [math]X[/math] denote the total loss due to the [math]N[/math] hurricanes.

This problem can be solved using the conditional variance formula. Note that independence is used to write the variance of a sum as the sum of the variances.

[[math]] \begin{align*} \operatorname{Var}(X) &= \operatorname{Var}[ \operatorname{E}( X | N )] + E[\operatorname{Var}( X | N )] \\ &= \operatorname{Var}[ \operatorname{E}( X_1 + \cdots + X_N )] + E[\operatorname{Var}( X_1 + \cdots + X_N )] \\ &= \operatorname{Var}(1, 000 N ) + \operatorname{E}(1, 000, 000 N ) \\ &= 1, 000 ^ 2\operatorname{Var}( N ) + 1, 000, 000 \operatorname{E}( N ) \\ &= 1, 000, 000(4) + 1, 000, 000(4) = 8, 000, 000. \end{align*} [[/math]]

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