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If \(Y\) is a geometric random variable, define \(Y^{*}=Y-1 .\) If \(Y\) is interpreted as the number of the trial on which the first success occurs, then \(Y^{*}\) can be interpreted as the number of failures before the first success. If \(Y^{*}=Y-1, P\left(Y^{*}=y\right)=P(Y-1=y)=P(Y=y+1)\) for \(y=0,1,2, \ldots .\) Show that $$P\left(Y^{*}=y\right)=q^{y} p, \quad y=0,1,2, \dots$$ The probability distribution of \(Y^{*}\) is sometimes used by actuaries as a model for the distribution of the number of insurance claims made in a specific time period.

Short Answer

Expert verified
The distribution is \(P(Y^* = y) = q^y p\).

Step by step solution

01

Understand the relationship between Y and Y*

The problem establishes that if Y is the number of the trial on which the first success occurs, then \(Y^{*}=Y-1\) is the number of failures before the first success. Hence, \(Y = Y^* + 1\).
02

Identify the probability distribution of Y

Since Y is a geometric random variable, the probability that the first success occurs on the \(k\)-th trial is given by \(P(Y = k) = (1-p)^{k-1}p\), where \(p\) denotes the probability of success on each trial, and \(1-p = q\) represents the probability of failure.
03

Express \(P(Y^* = y)\) in terms of \(Y\)

We know \(Y^* = Y - 1\), and thus \(P(Y^* = y) = P(Y - 1 = y) = P(Y = y + 1)\) for \(y = 0, 1, 2, \ldots\).
04

Find \(P(Y = y+1)\)

Substituting \(y+1\) for \(k\) in the geometric probability distribution, we get \(P(Y = y + 1) = (1-p)^y p = q^y p\).
05

Conclude the distribution for \(Y^*\)

Since \(P(Y^* = y) = P(Y = y+1) = q^y p\), the distribution for the number of failures \(Y^*\) before the first success is \(P(Y^* = y) = q^y p\) for \(y = 0, 1, 2, \ldots\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Random Variables
Random variables are crucial in probability theory and statistics, serving as variables whose possible values are numerical outcomes of a random phenomenon. Imagine conducting a random experiment, like tossing a coin or rolling a dice. The outcomes, such as heads or tails, can be represented by random variables.

Geometric random variables, like in this exercise, tell us about the number of trials needed to get the first success in a series of independent and identically distributed Bernoulli trials (like repeated coin tosses). Here, the random variable \(Y\) represents the trial on which the first success occurs. If a new variable \(Y^* = Y-1\) is defined, it represents the number of failures before the first success occurs. This transformation is key to understanding variations and flexibility in probability modeling.
Probability Distribution
A probability distribution assigns a probability to each measurable event in a sample space, allowing us to understand the behavior of a random variable. In our exercise, we deal with the geometric distribution, which is a discrete probability distribution. It describes the number of Bernoulli trial failures before the first success, which is captured by \(Y^*\).
  • The probability of the first success occurring on the \(k\)th trial in a geometric distribution is given by \(P(Y = k) = (1-p)^{k-1}p\), where \(p\) is the probability of success, and \(q = 1-p\) is the probability of failure.
  • When transformed to \(Y^*\), the same distribution tells us about the number of failures \(y\), with \(P(Y^* = y) = q^y p\). This formula highlights a key feature: as the number of failures increases, probabilities decrease exponentially due to the \(q^y\) term.
Understanding probability distribution helps us predict outcomes and interpret random processes with greater accuracy.
Actuarial Models
Actuarial models use mathematical and statistical methods to assess risk in the insurance and finance industries. These models help actuaries make informed predictions about uncertain future events. Our exercise shows an application of geometric distribution where actuaries might use the probability distribution of \(Y^*\) to model the number of claims made.
  • The model can determine the likelihood of a certain number of events (like insurance claims) occurring over a set time period. This is paramount in setting premium rates and ensuring sufficient reserves for insurers.
  • By using \(P(Y^* = y) = q^y p\), actuaries can predict the expected frequency of claims, optimizing decision-making and financial planning.
The geometric distribution thus becomes a valuable tool in insurance modeling, aiding in the practical application of probability theory to real-world financial problems.

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Most popular questions from this chapter

A city commissioner claims that \(80 \%\) of the people living in the city favor garbage collection by contract to a private company over collection by city employees. To test the commissioner's claim, 25 city residents are randomly selected, yielding 22 who prefer contracting to a private company. a. If the commissioner's claim is correct, what is the probability that the sample would contain at least 22 who prefer contracting to a private company? b. If the commissioner's claim is correct, what is the probability that exactly 22 would prefer contracting to a private company? c. Based on observing 22 in a sample of size 25 who prefer contracting to a private company, what do you conclude about the commissioner's claim that \(80 \%\) of city residents prefer contracting to a private company?

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