The Progress of Liberty

The Actuary’s Free Study Guide for Exam 3L - Section 47

 

This section of sample problems and solutions is a part of The Actuary’s Free Study Guide for Exam 3L, authored by Mr. Stolyarov. This is Section 47 of the Study Guide. See an index of all sections by following the link in this paragraph.

 

The Pareto distribution is characterized by the following properties.

 

Survival function for Pareto Distribution:

s(x) = θ^α/(x + θ)^α for some specified parameters α and θ with α ≥ 1 and θ ≥ 0. Note that different notation is often used for these parameters, and (x + θ) is often expressed as a variable in itself (denoted, for instance, as y, such that s(y) = θ^α/(y)^α, for y ≥ x).

 

Cumulative distribution function for Pareto Distribution:

F(x) = 1 - θ^α/(x + θ)^α

 

Probability density function for Pareto Distribution:

f(x) = αθ^α/(x + θ)^α+1

 

We introduced Weibull’s Law of Mortality in Section 25. The Weibull distribution can also be expressed via different notation, however. We will focus on one alternative expression of this distribution here.

 

Survival function for Weibull Distribution:

s(x) = exp[-(x/θ)^τ] for some specified parameters θ and τ, for θ > 0 and τ > 0.

 

Cumulative distribution function for Weibull Distribution:

F(x) = 1 - exp[-(x/θ)^τ]

 

Probability density function for Weibull Distribution:

f(x) = τx^τ-1exp[-(x/θ)^τ]/θ^τ

 

The earlier problems in this section are designed to acquaint students with this new notation. Then, it will be possible to undertake some exam-style questions.

 

Some of the problems in this section were designed to be similar to problems from past versions of the Casualty Actuarial Society’s Exam 3L and the Society of Actuaries’ Exam MLC. They use original exam questions as their inspiration - and the specific inspiration for each problem is cited so as to give students a chance to see the original. All of the original problems are publicly available, and students are encouraged to refer to them. But all of the values, names, conditions, and calculations in the problems here are the original work of Mr. Stolyarov.

 

Source: Broverman, Sam. Actuarial Exam Solutions – CAS Exam 3 – Fall 2006.

 

Original Problems and Solutions from The Actuary’s Free Study Guide

 

Problem S3L47-1. The operational lifetimes of squigglewidgets in years follow a Pareto distribution with θ = 0.36 and α = 3. You have 94000 operational squigglewidgets. After 6 years, how many operational squigglewidgets would you expect to have? Round down to the nearest whole squigglewidget.

 

Solution S3L47-1. We use the formula s(x) = θ^α/(x + θ)^α for x = 6, θ = 0.36, and α = 3.

Thus, s(6) = 0.36³/6.36³ = about 0.0001813577651. Since we have 94000 squigglewidgets to begin with, at time 6, we can expect to have 94000s(6) = 94000*0.0001813577651 = 17.04762992 or about 17 squigglewidgets.

 

Problem S3L47-2. The number of millennia, starting at present, for which Prometheus Tower can be expected to stand is modeled by a Weibull distribution θ = 4 and τ = 2. What is the probability that Prometheus Tower will still be standing 1500 years from now?

 

Solution S3L47-2. We use the formula s(x) = exp[(-x/θ)^τ] for x = 1.5, θ = 4, and τ = 2.

Thus, s(1.5) = exp[-(1.5/4)²] = exp[-0.140625] = s(1.5) = 0.8688150563.

 

Problem S3L47-3. The lifetimes of gray gremlins follow a Pareto distribution with

θ = 4. You know that a newborn gray gremlin has a probability of 0.1051 of surviving to age 15. Find α for this Pareto distribution.

 

Solution S3L47-3. We use the formula s(x) = θ^α/(x + θ)^α for x = 15, θ = 4, and s(15) = 0.1051. Thus, 0.1051 = (4/19)^α and α*ln(4/19) = ln(0.1051). Thus, α = ln(0.1051)/ln(4/19) =

α = about 1.445849747.

 

Problem S3L47-4. You are given the following facts about a probability density function corresponding to a particular Weibull distribution: θ = 0.022 and τ = 0.4. Find f(6) for this distribution. Hint: The answer will be very small!

 

Solution S3L47-4. We use the formula f(x) = τx^τ-1exp[-(x/θ)^τ]/θ^τ.

Thus, f(6) = 0.4*6^-0.6exp[-(6/0.022)^0.4]/0.022^0.4 = about 0.00005068503794.

 

Problem S3L47-5.

Similar to Question 18 from the Casualty Actuarial Society’s Fall 2006 Exam 3.

A reckless real estate speculator can incur losses in accordance with a spliced loss distribution consisting of a Weibull distribution for modeling losses less than $1,500,000 and a Pareto distribution modeling losses greater than $1,500,000. The Weibull distribution has parameters θ₁ = 500,000 and τ = 2. The Pareto distribution has parameters θ₂ = 2,000,000 and α = 3. The probability of the speculator’s losses being less than $1,500,000 is 0.7. Find the probability that his losses are less than $3,000,000.

 

Solution S3L47-5. First, we consider the probability of a loss greater than $1,500,000 being also greater than $3,000,000.  This is s(3000000)/s(1500000) and is modeled by the given Pareto distribution. We use the formula s(x) = θ^α/(x + θ)^α for θ = 2,000,000 and α = 3.

Thus, s(3000000)/s(1500000) =

(3000000/5000000)³/(1500000/3500000)³ = (3/5)³/(5/7)³ = (21/25)³ = 0.592704.

This is the conditional probability of a loss greater than $1,500,000 being also greater than $3,000,000.  To get the absolute probability of a loss being greater than $3,000,000, we multiply

0.592704 by 0.3, the probability of a loss being greater than $1,500,000. This probability is

0.1778112. Our desired probability is the complement of this result, 1 - 0.1778112 = 0.8221888.

 

See other sections of The Actuary’s Free Study Guide for Exam 3L.

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