In any given year, a male automobile policyholder will make...

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.

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.

Key Concepts

Mixture Distributions

Mixture distributions occur when a population is composed of distinct subgroups, each with its own probability characteristics. In this question, the overall population is a mix of male and female policyholders, each with different claim probabilities. This concept highlights that the overall probability is a weighted average of the subgroup probabilities based on their proportions in the population.

Conditional Probability

Conditional probability is the probability of an event occurring given that another event has already occurred. Here, knowing that a claim was made in one year provides information about the likely subgroup of the policyholder. Because the subgroups have different probabilities of making a claim, the probability of making a claim in the following year is influenced by this prior occurrence.

Bayesian Updating

Bayesian updating is the process of revising probabilities based on new information. When a claim occurs, it increases the likelihood that the policyholder belongs to the subgroup with a higher claim probability. Hence, the chance of a subsequent claim is higher than the unconditional probability, as the observed event updates our belief about the individual's underlying risk group.

Transcript

00:01 And hello calculus student.

00:04 We're looking at chapter 12, section 4, problem number 28.

00:08 So it starts off by an actuary assumption that the probability of the n plus one event is equal to one -fifth, the probability of the previous event.

00:27 So piece of n.

00:28 So piece of n is going to be the probability for n claims.

00:33 So what we're asked to find is the probability of more than one claim is going to happen.

00:40 So we start with the idea that p -sub -0 is the probability of no claims, and p -sub -1 is the probability of one claim.

00:59 So p -sop -2 would be the probability of two claims, etc.

01:04 And so p -sad -n, like i said, would be the probability for n claims.

01:10 So what we're trying to find then is that the probability of two or more claims, so probability two plus probability of three plus probability of four claims plus dot dot dot, which would be equal to the probability of more than one claim.

01:43 Okay, so what we're going to start with, i'll switch colors here, is that the probability of two claims occurring is one -fifth the probability of one claim, right? due to our actuary assumption that probability of n plus one is one -fifth probability of it, which means the probability of one more is one -fifth the previous.

02:10 Okay, so we can expand on that and say that the probability of three claims is equal to one -fifth the probability of two claims.

02:25 And so that is equal to one -fifth of one -fifth the probability of one claim, right? so just replacing piece of two with the one -fifth piece of one.

02:41 And so we can rewrite that as one -fifth quantity squared times the probability of one claim.

02:53 And we could keep doing this.

02:56 So we know that probability always adds to one.

03:00 So i'm going to write out the statement that the probability of zero claims plus the probability of one claim plus the probability of two claims plus the probability of three claims plus dot dot dot.

03:17 So if we keep going, it's going to total one.

03:22 And so we can say that the probability of two claims plus the probability of three claims dot dot dot is equal to one.

03:35 And then i'm going to subtract the probability of zero claims and subtract the probability of one claim to get this equation.

03:45 And what i want to recognize is that i can then replace the probability of two claims with, i'm going to need a little more room, so i'm going to jump over, and replace the probability of two claims with one -fifth, the probability of one, plus probability of three claims with one -fifth squared times the probability of one, plus and then i can keep doing that and we know that's equal to 1 minus p 0 minus p sub 1.

04:32 So just rewrote this first couple terms there but by doing that i recognize here that i have an infinite geometric series with a first term so a is 1 fifth piece of 1 and our common ratio is 1 5th right we're going to keep multiplying by 1 5th to get from one term to the next.

04:58 So we'll say one fifth.

05:00 So now we can find a sum, which we know a sum for our infinite geometric series, a over 1 minus r.

05:13 So we can replace a with 1 fifth piece of 1 over 1 minus r, and our r is 1 fifth...

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