You are given the following information about [math]N[/math], the annual number of claims for a randomly selected insured:

[[math]] \operatorname{P}[N = 0] = \frac{1}{2}, \, \operatorname{P}[N = 1] = \frac{1}{3}, \, \operatorname{P}[N \gt1] = \frac{1}{6} [[/math]]

Let [math]S[/math] denote the total annual claim amount for an insured. When [math]N = 1 [/math], [math]S[/math] is exponentially distributed with mean 5. When [math]N \gt 1 [/math], [math]S[/math] is exponentially distributed with mean 8.

Calculate [math]\operatorname{P}(4 \lt S \lt 8) [/math]

• 0.04
• 0.08
• 0.12
• 0.24
• 0.25

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

Step 1: Understand the Problem and Given Information

We are given information about [math]N[/math], the annual number of claims, and [math]S[/math], the total annual claim amount, conditional on [math]N[/math]. The probabilities for the number of claims [math]N[/math] are provided as:

Probabilities for the Number of Claims ([math]N[/math])

Number of Claims ([math]N[/math])	Probability ([math]\operatorname{P}[N][/math])
[math]N=0[/math]	[math]\frac{1}{2}[/math]
[math]N=1[/math]	[math]\frac{1}{3}[/math]
[math]N\gt1[/math]	[math]\frac{1}{6}[/math]

The distribution of the total annual claim amount [math]S[/math] depends on the value of [math]N[/math]:

Conditional Distribution of Total Claim Amount ([math]S[/math])

Condition on [math]N[/math]	Distribution of [math]S[/math]	Mean ([math]\theta[/math])
[math]N=1[/math]	Exponential	[math]5[/math]
[math]N\gt1[/math]	Exponential	[math]8[/math]

Our goal is to calculate [math]\operatorname{P}(4 \lt S \lt 8)[/math].

Step 2: Formulate the Probability using the Law of Total Probability

To find the unconditional probability [math]\operatorname{P}(4 \lt S \lt 8)[/math], we utilize the Law of Total Probability. This law allows us to combine conditional probabilities over different scenarios for [math]N[/math]. The general formula is:

[[math]]\operatorname{P}(A) = \sum_{i} \operatorname{P}(A | B_i) \operatorname{P}(B_i)[[/math]]

In this problem, [math]A = (4 \lt S \lt 8)[/math], and [math]B_i[/math] are the mutually exclusive and exhaustive events for [math]N[/math], which are [math]N=0[/math], [math]N=1[/math], and [math]N\gt1[/math]. Applying the law, we get:

[[math]]\operatorname{P}[ 4 \lt S \lt 8] = \operatorname{P}[ 4 \lt S \lt 8 | N = 0] \operatorname{P}[ N = 0] + \operatorname{P}[ 4 \lt S \lt 8 | N = 1] \operatorname{P}[ N = 1] + \operatorname{P}[ 4 \lt S \lt 8 | N \gt 1] \operatorname{P}[ N \gt 1][[/math]]

If [math]N=0[/math], there are no claims, meaning the total claim amount [math]S=0[/math]. Therefore, the probability of [math]S[/math] being between 4 and 8 given [math]N=0[/math] is zero: [math]\operatorname{P}[ 4 \lt S \lt 8 | N = 0] = 0[/math]. The equation simplifies to:

[[math]]\operatorname{P}[ 4 \lt S \lt 8] = \operatorname{P}[ 4 \lt S \lt 8 | N = 1] \operatorname{P}[ N = 1] + \operatorname{P}[ 4 \lt S \lt 8 | N \gt 1] \operatorname{P}[ N \gt 1][[/math]]

Step 3: Calculate Conditional Probabilities

Next, we calculate the two conditional probabilities needed: [math]\operatorname{P}[ 4 \lt S \lt 8 | N = 1][/math] and [math]\operatorname{P}[ 4 \lt S \lt 8 | N \gt 1][/math]. For an exponentially distributed random variable [math]X[/math] with mean [math]\theta[/math], the probability [math]\operatorname{P}(a \lt X \lt b)[/math] can be efficiently calculated using its cumulative distribution function (CDF), [math]F(x) = 1 - e^{-x/\theta}[/math], for [math]x \gt 0[/math]. Thus, [math]\operatorname{P}(a \lt X \lt b) = F(b) - F(a) = (1 - e^{-b/\theta}) - (1 - e^{-a/\theta}) = e^{-a/\theta} - e^{-b/\theta}[/math].

• Case 1: [math]N = 1[/math]

When [math]N=1[/math], [math]S[/math] is exponentially distributed with a mean of [math]\theta_1 = 5[/math]. Using the formula for [math]\operatorname{P}(a \lt X \lt b)[/math]:

[[math]]\operatorname{P}[ 4 \lt S \lt 8 | N = 1] = e^{-4/5} - e^{-8/5}[[/math]]

• Case 2: [math]N \gt 1[/math]

When [math]N\gt1[/math], [math]S[/math] is exponentially distributed with a mean of [math]\theta_2 = 8[/math].

[[math]]\operatorname{P}[ 4 \lt S \lt 8 | N \gt 1] = e^{-4/8} - e^{-8/8} = e^{-1/2} - e^{-1}[[/math]]

Step 4: Combine Probabilities and Final Calculation

Now, we substitute the calculated conditional probabilities and the given probabilities for [math]N[/math] into the simplified formula from Step 2:

[[math]]\operatorname{P}[ 4 \lt S \lt 8] = \operatorname{P}[ 4 \lt S \lt 8 | N = 1] \operatorname{P}[ N = 1] + \operatorname{P}[ 4 \lt S \lt 8 | N \gt 1] \operatorname{P}[ N \gt 1][[/math]]

We use the values:

• [math]\operatorname{P}[ N = 1] = \frac{1}{3}[/math]
• [math]\operatorname{P}[ N \gt 1] = \frac{1}{6}[/math]
• [math]\operatorname{P}[ 4 \lt S \lt 8 | N = 1] = e^{-4/5} - e^{-8/5}[/math]
• [math]\operatorname{P}[ 4 \lt S \lt 8 | N \gt 1] = e^{-1/2} - e^{-1}[/math]

Substituting these into the equation:

[[math]]\operatorname{P}[ 4 \lt S \lt 8] = \frac{1}{3}(e^{-4/5} - e^{-8/5}) + \frac{1}{6} \left( e^{-1/2} - e^{-1}\right)[[/math]]

Now, we calculate the numerical values:

• [math]e^{-4/5} = e^{-0.8} \approx 0.449329[/math]
• [math]e^{-8/5} = e^{-1.6} \approx 0.201897[/math]
• [math]e^{-1/2} = e^{-0.5} \approx 0.606531[/math]
• [math]e^{-1} \approx 0.367879[/math]

So, the differences are:

• [math]e^{-4/5} - e^{-8/5} \approx 0.449329 - 0.201897 = 0.247432[/math]
• [math]e^{-1/2} - e^{-1} \approx 0.606531 - 0.367879 = 0.238652[/math]

Substitute these results back into the main equation:

[[math]]\operatorname{P}[ 4 \lt S \lt 8] \approx \frac{1}{3}(0.247432) + \frac{1}{6}(0.238652)[[/math]]

[[math]]\operatorname{P}[ 4 \lt S \lt 8] \approx 0.082477 + 0.039775[[/math]]

[[math]]\operatorname{P}[ 4 \lt S \lt 8] \approx 0.122252[[/math]]

Rounding to two decimal places, we get approximately [math]0.12[/math].

Key Insights

• The Law of Total Probability is crucial for combining probabilities across different scenarios or states of nature. It allows for the decomposition of a complex probability into simpler, conditional probabilities.
• Conditional probabilities are essential when the distribution of one variable (e.g., total claim amount [math]S[/math]) depends on the outcome of another variable (e.g., number of claims [math]N[/math]).
• For an exponential distribution with mean [math]\theta[/math], the probability [math]\operatorname{P}(a \lt X \lt b)[/math] can be efficiently calculated as [math]e^{-a/\theta} - e^{-b/\theta}[/math], simplifying calculations compared to integrating the PDF.
• Careful identification of the correct parameters (e.g., the mean [math]\theta[/math] for the exponential distribution) for each conditional case is vital for accurate calculations.
• Understanding edge cases, such as when [math]N=0[/math] implying [math]S=0[/math], can significantly simplify the application of the Law of Total Probability by eliminating zero-probability terms.

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

Solution: C

Observe

[[math]] \begin{align*} \operatorname{P}[ 4 \lt S \lt 8] &= \operatorname{P}[ 4 \lt S \lt 8 | N = 1] \operatorname{P}[ N = 1] + \operatorname{P}[ 4 \lt S \lt 8 | N \gt 1] \operatorname{P}[ N \gt 1] \\ &= \frac{1}{3}(e^{-4/5} - e^{-8/5}) + \frac{1}{6} \left( e^{-1/2} - e^{-1}\right) \\ &= 0.122. \end{align*} [[/math]]

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