A couple takes out a medical insurance policy that reimburses them for days of work missed due to illness. Let [math]X[/math] and [math]Y[/math] denote the number of days missed during a given month by the wife and husband, respectively. The policy pays a monthly benefit of 50 times the maximum of [math]X[/math] and [math]Y[/math], subject to a benefit limit of 100. [math]X[/math] and [math]Y[/math] are independent, each with a discrete uniform distribution on the set {0,1,2,3,4}.

Calculate the expected monthly benefit for missed days of work that is paid to the couple.

• 70
• 90
• 92
• 95
• 140

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

Step 1: Understand the Random Variables and Benefit Structure

The number of days missed by the wife ([math]X[/math]) and husband ([math]Y[/math]) are independent random variables. Both [math]X[/math] and [math]Y[/math] follow a discrete uniform distribution on the set {0, 1, 2, 3, 4}. This means each of the 5 possible values for [math]X[/math] (or [math]Y[/math]) has a probability of [math]\frac{1}{5}[/math]. Since [math]X[/math] and [math]Y[/math] are independent, the probability of any specific pair [math](x,y)[/math] occurring is [math]P(X=x, Y=y) = P(X=x)P(Y=y) = \frac{1}{5} \times \frac{1}{5} = \frac{1}{25}[/math]. There are a total of [math]5 \times 5 = 25[/math] possible [math](x,y)[/math] pairs. The monthly benefit, denoted by [math]B[/math], is defined as 50 times the maximum of [math]X[/math] and [math]Y[/math], subject to a benefit limit of 100. Mathematically, this can be expressed as:

[[math]]B = \min(100, 50 \times \max(X, Y))[[/math]]

Step 2: Identify Possible Benefit Amounts

Let [math]Z = \max(X, Y)[/math]. The possible values for [math]Z[/math] are also {0, 1, 2, 3, 4}. We can analyze the benefit [math]B[/math] based on the values of [math]Z[/math]:

Possible Values for [math]Z[/math] and Corresponding Benefit

[math]Z = \max(X, Y)[/math]	Calculated Benefit ([math]50Z[/math])	Actual Benefit ([math]B = \min(100, 50Z)[/math])
0	[math]50 \times 0 = 0[/math]	0
1	[math]50 \times 1 = 50[/math]	50
2	[math]50 \times 2 = 100[/math]	100
3	[math]50 \times 3 = 150[/math]	100 (due to limit)
4	[math]50 \times 4 = 200[/math]	100 (due to limit)

From this table, we observe that there are only three possible benefit amounts: 0, 50, and 100.

Step 3: Calculate Probabilities for Each Benefit Amount

We need to find the probability of each of the three possible benefit amounts. Since each of the 25 possible [math](x,y)[/math] pairs has a probability of [math]\frac{1}{25}[/math], we can count the number of pairs that lead to each benefit amount.

• Case 1: Benefit = 0

The benefit is 0 when [math]\max(X, Y) = 0[/math]. This occurs only when both [math]X=0[/math] and [math]Y=0[/math].

• • Corresponding pair: (0,0)
  • Number of pairs: 1
  • Probability: [math]P(B=0) = \frac{1}{25}[/math]
• Case 2: Benefit = 50

The benefit is 50 when [math]\max(X, Y) = 1[/math]. This occurs when one of the variables is 1 and the other is 0 or 1.

• • Corresponding pairs: (0,1), (1,0), (1,1)
  • Number of pairs: 3
  • Probability: [math]P(B=50) = \frac{3}{25}[/math]
• Case 3: Benefit = 100

The benefit is 100 when [math]\max(X, Y) \geq 2[/math] (due to the cap). This covers all remaining [math](x,y)[/math] pairs.

• • Number of pairs for [math]\max(X,Y) \geq 2[/math]: Total pairs (25) - pairs for [math]\max(X,Y) = 0[/math] (1) - pairs for [math]\max(X,Y) = 1[/math] (3)

[[math]]25 - 1 - 3 = 21[[/math]]

• • Number of pairs: 21
  • Probability: [math]P(B=100) = \frac{21}{25}[/math]

We can verify that the probabilities sum to 1: [math]\frac{1}{25} + \frac{3}{25} + \frac{21}{25} = \frac{25}{25} = 1[/math].

Step 4: Calculate the Expected Monthly Benefit

The expected monthly benefit [math]E[B][/math] is calculated as the sum of each possible benefit amount multiplied by its probability:

[[math]]E[B] = \sum_{b} b \cdot P(B=b)[[/math]]

Using the probabilities calculated in Step 3:

[[math]]E[B] = (0 \times P(B=0)) + (50 \times P(B=50)) + (100 \times P(B=100))[[/math]]

[[math]]E[B] = \left(0 \times \frac{1}{25}\right) + \left(50 \times \frac{3}{25}\right) + \left(100 \times \frac{21}{25}\right)[[/math]]

[[math]]E[B] = 0 + \frac{150}{25} + \frac{2100}{25}[[/math]]

[[math]]E[B] = \frac{150 + 2100}{25}[[/math]]

[[math]]E[B] = \frac{2250}{25}[[/math]]

[[math]]E[B] = 90[[/math]]

The expected monthly benefit for missed days of work is 90.

Key Insights

• When dealing with independent discrete random variables [math]X[/math] and [math]Y[/math], the probability of a specific pair [math](x,y)[/math] is [math]P(X=x, Y=y) = P(X=x)P(Y=y)[/math].
• To calculate probabilities involving [math]\max(X,Y)[/math] for discrete uniform distributions, enumerate all possible outcomes [math](x,y)[/math] and count how many satisfy the condition.
• Benefit limits ([math]\min(A, B)[/math] function) mean that the payout amount can plateau, simplifying the number of distinct benefit values to consider.
• The expected value of a discrete random variable is the sum of each possible outcome multiplied by its respective probability. Ensure that all possible outcomes and their corresponding probabilities are accounted for.

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

Solution: B

Each (x,y) pair has probability 1/25. There are only three possible benefit amounts:

0: Occurs only for the pair (0.0) and so the probability is 1/25.

50: Occurs for the three pairs (0,1), (1,0), and (1,1) and so the probability is 3/25.

100: Occurs in all remaining cases and so the probability is 21/25.

The expected value is 0(1/25) + 50(3/25) + 100(21/25) = 2250/25 = 90.

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