Problem 46 - A First Course in Probability

Question

In any given year, a male automobile policyholder will make a claim with probability \(p_{m}\) and a female policyholder will make a claim with probability \(p_{f}\) where \(p_{f} \neq p_{m} .\) The fraction of the policyholders that are male is \(\alpha, 0<\alpha<1 .\) A policyholder is randomly chosen. If \(A_{i}\) denotes the event that this policyholder will make a claim in year \(i,\) show that $$ P\left(A_{2} | A_{1}\right)>P\left(A_{1}\right) $$ Give an intuitive explanation of why the preceding inequality is true.

Step-by-Step Solution

Answer

In summary, the inequality \(P(A_2 | A_1) > P(A_1)\) is true and it implies that a policyholder who has already made a claim in year 1 is more likely to make a claim in year 2 than a random policyholder. This could be because policyholders with a history of claims might be more prone to accidents or have a higher tendency to engage in risky behavior, increasing the likelihood of future claims.

Step 1: Identify the given probabilities

We are given the following probabilities: 1. Probability of a male policyholder making a claim in a year: \(p_m\) 2. Probability of a female policyholder making a claim in a year: \(p_f\), with \(p_f \neq p_m\) 3. Fraction of policyholders that are male: \(\alpha\), with \(0 < \alpha < 1\)

Step 2: Calculate the probability of a policyholder making a claim in year 1

To find the probability of a policyholder making a claim in year 1, we need to consider the gender of the policyholder and the probability of them making a claim. We can calculate \(P(A_1)\) by using the law of total probability: \(P(A_1) = P(A_1 | M)P(M) + P(A_1 | F)P(F)\) Since the fraction of male policyholders is given by \(\alpha\), the fraction of female policyholders is given by \(1 - \alpha\). Therefore, \(P(A_1) = p_m\alpha + p_f(1 - \alpha)\)

Step 3: Calculate the probability of a policyholder making a claim in year 2, given that they made a claim in year 1

To find this probability, we will use the conditional probability formula: \(P(A_2 | A_1) = \frac{P(A_2 \cap A_1)}{P(A_1)}\) First, we need to calculate the probability of the intersection of the two events, \(P(A_2 \cap A_1)\). Since making a claim in year 1 and making a claim in year 2 are independent events for both male and female policyholders, we can write: \(P(A_2 \cap A_1) = P((A_1 \cap A_2) \cap M)P(M) + P((A_1 \cap A_2) \cap F)P(F)\) \(\Rightarrow P(A_2 \cap A_1) = p_m^2\alpha + p_f^2(1 - \alpha)\) Now, we can substitute these findings into the conditional probability formula: \(P(A_2 | A_1) = \frac{p_m^2\alpha + p_f^2(1 - \alpha)}{p_m\alpha + p_f(1 - \alpha)}\)

Step 4: Show that \(P(A_2 | A_1) > P(A_1)\)

To prove the given inequality, we need to show that: \(\frac{p_m^2\alpha + p_f^2(1 - \alpha)}{p_m\alpha + p_f(1 - \alpha)} > p_m\alpha + p_f(1 - \alpha)\) For this inequality to be true, the following must hold: \(p_m^2\alpha + p_f^2(1 - \alpha) > (p_m\alpha + p_f(1 - \alpha))^2\) This inequality is true because \(p_m^2 > p_mp_f\) and \(p_f^2 > p_mp_f\), as \(p_m \neq p_f\). The inequality is also true because the given probabilities are non-negative and the sum of the numerators in both sides is equal to 1.

Step 5: Provide an intuitive explanation

Intuitively, the inequality \(P(A_2 | A_1) > P(A_1)\) suggests that a policyholder who has already made a claim in year 1 is more likely to make a claim in year 2 than a policyholder chosen at random. This could be due to the fact that policyholders who have made a claim in the past may be more prone to accidents or more likely to engage in risky behavior, making them more likely to make a claim in the future as well.

Key Concepts

Law of Total Probability

The Law of Total Probability is a fundamental rule that serves as the backbone of several probability calculations. It allows us to calculate the probability of an event by considering all possible ways that event can occur. In the context of our exercise, it helps us find the probability of a policyholder making a claim in Year 1, regardless of whether they are male or female.

To apply this law, we recognize that the total probability of a claim in Year 1 can be dissected into two parts:

• The probability of a male making a claim, multiplied by the proportion of male policyholders.
• The probability of a female making a claim, multiplied by the proportion of female policyholders.

Thus, the formula becomes:\[ P(A_1) = P(A_1 | M)P(M) + P(A_1 | F)P(F) \]where:

• \( P(A_1 | M) \) is the probability that a male policyholder makes a claim.
• \( P(A_1 | F) \) is the probability that a female policyholder makes a claim.
• \( P(M) \) and \( P(F) \) are the fractions of male and female policyholders, respectively.

This approach ensures that all possible avenues for the event are captured, thereby giving us a complete picture of the event's likelihood.

Independent Events

The concept of independent events is critical when understanding the correlations between events over time. In probability, two events are said to be independent if the occurrence of one event has no effect on the occurrence of the other event. In our exercise, making a claim in Year 1 and making a claim in Year 2 are considered independent events.

This assumption implies that the probability of making a claim in Year 2 doesn't change based on what happened in Year 1. Therefore, the formula \( P(A_2 \cap A_1) = P(A_1)P(A_2) \) holds for both male and female policyholders.

• For a male policyholder, \( P(A_2 \cap A_1 | M) = (p_m)^2 \alpha \)
• For a female policyholder, \( P(A_2 \cap A_1 | F) = (p_f)^2 (1 - \alpha) \)

Thus, to find the overall probability of making a claim in both years (in our example), we take an aggregated view:\[ P(A_2 \cap A_1) = p_m^2 \alpha + p_f^2 (1 - \alpha) \]Understanding the nature of independent events ensures a correct application of probability formulas and avoids assumptions where past events affect future outcomes.

Probability Inequality

Probability inequality is a way to compare the likelihoods of different events. In some situations, as we have in this exercise, it shows us that the occurrence of a specific event can lead to higher probability of another event occurring. Our inequality suggests that if a policyholder has made a claim in Year 1, the likelihood of them making a claim again in Year 2 is higher compared to making a claim in Year 1 without any conditions applied.

This is mathematically shown by:\[ P(A_2 | A_1) > P(A_1) \]To reason this, consider the factors:

• The intersection probability, \( P(A_2 \cap A_1) \), adds a layer of likelihood when a claim has already been made.
• The simplification of the inequality \( p_m^2 \alpha + p_f^2 (1 - \alpha) > (p_m \alpha + p_f (1 - \alpha))^2 \) backs up this intuitive understanding.
• This is grounded in behavioral assumptions; for example, policyholders who previously made claims might be more at risk, engaging more often in actions that lead to claims.

Such inequalities enhance our grasp of how interconnected events can shift overall probabilities. They allow us to set a precedent for evaluating probability in sequential events.

Conditional Probability Formula

Conditional probability provides a framework for understanding how the probability of an event alters when additional information is known. It quantifies how likely an event is, given that another event has occurred. In our problem, it ties the probability of making a claim in Year 2 with the knowledge that a claim was already made in Year 1.

The conditional probability formula is:\[ P(A_2 | A_1) = \frac{P(A_2 \cap A_1)}{P(A_1)} \]Breaking this down, we calculate:

• \( P(A_2 \cap A_1) \), which measures the combined occurrence of both events.
• \( P(A_1) \), the baseline probability of a claim in Year 1, sets a context for comparison.

Utilizing the intersection probability accounts for both gender-based claim behavior:

• \[ P(A_2 \cap A_1) = p_m^2 \alpha + p_f^2 (1 - \alpha) \]

This formula helps provide clarity and depth to our probability assessments, explicitly highlighting the influence of past events on future outcomes.