The Lognormal Probability Distribution, Markov Chains, and Assorted Exam-Style Questions

Calculations performed with the Lognormal Distribution are highly similar to those performed with the Normal Distribution, discussed in Dr. Marcel Finan's free study guide for Exam 1/P.

Let X be a random variable that follows a Lognormal Distribution with parameters µ and σ. Then, for some value of k,

Pr(X ≤ k) = N[(ln(k) - µ)/σ]

Pr(X > k) = 1 - N[(ln(k) - µ)/σ]

A note on the function N(x): N(x) is called the cumulative normal distribution function. It is the probability that a randomly chosen number in the standard normal distribution (where mean = 0 and variance = 1) is less than x. It is impossible to directly integrate the normal probability distribution function to find N(x). On the exam, you will be given a table of values for N(x), so such integration will not be necessary. Here, however, we will use the Microsoft Excel function NormSDist. Try entering "=NormSDist(1)" into a cell in MS Excel. The result should be 0.84134474. Using this method will give us greater accuracy than using a table would.

Markov chains are illustrated by matrices which show the probability of moving from one state to another. Consider a 2-by-2 matrix pertaining to a Markov chain:

...X...Y
X(a....b)
Y(c....d)

This matrix represents the probabilities of moving between states X and Y within a single time period. The starting states appear on the vertical axis, while the states at the end of the period appear on the horizontal axis. For instance, a is the probability of an entity in state X staying in state X. b is the probability of an entity in state X moving to state Y. Note that a + b must equal 1 if X and Y are the only possible states.

If the matrices for each time period are the same, we call the resulting Markov chain homogeneous. If the matrices for each time period are not the same, we call the Markov chain non-homogeneous.

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

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

Problem S3L49-1. Gunnar the Gambler is playing a game of cards where his winnings follow a lognormal distribution with µ = 5 and σ = 2. Find the probability that his winnings are less than or equal to $4000.

Solution S3L49-1. We use the formula Pr(X ≤ k) = N[(ln(k) - µ)/σ], where k = 4000, µ = 5, and σ = 2. Thus, Pr(X ≤ 5000) = N[(ln(4000) - 5)/2] = N[1.64702482].

We use the function "=NormSDist(1.64702482)" to find our desired probability, 0.950223528.

Problem S3L49-2. In a certain class, the number of spelling errors per paper received by the professor follows a lognormal distribution with µ = 3 and σ = 1.5. Find the probability that a given paper will have more than 140 spelling errors in it.

Solution S3L49-2. We use the formula Pr(X > k) = 1 - N[(ln(k) - µ)/σ] where k = 140, µ = 3, and σ = 1.5. Thus, Pr(X > 140) = 1 - N[(ln(140) - 3)/1.5] = 1 - N[1.294428282].

We use the function "=1 - NormSDist(1.294428282)" to find our desired probability, 0.097758763.

Problem S3L49-3. Similar to Question 20 from the Casualty Actuarial Society's Fall 2006 Exam 3.

Lucretia owns a mine, where the only possible accidents are either due to cave-ins or noxious fumes, and only one accident occurs per year. The financial loss to Lucretia from a cave-in is three times the loss to her from noxious fumes. Noxious fumes have a likelihood of occurring that is four times greater than the probability that a cave-in will occur. The annual loss from noxious fumes follows a lognormal distribution with µ = 7 and σ = 4. Find the probability that Lucretia's loss will be greater than $3000 this year.

Solution S3L49-3. Let X be the random variable denoting the loss from noxious fumes and let Y be the random variable denoting the loss from cave-ins. We know that Y = 3X and

Pr(X > k) = 1 - N[(ln(k) - µ)/σ].

One accident will occur this year, and since noxious fumes are 4 times as likely as cave-ins, it follows that noxious fumes have a 4/5 probability of occurring and cave-ins have a 1/5 probability of occurring.

We want to find (4/5)Pr(X > 3000) + (1/5)Pr(Y > 3000) =

(4/5)Pr(X > 3000) + (1/5)Pr(3X > 3000) =

(4/5)Pr(X > 3000) + (1/5)Pr(X > 1000) =

(4/5)(1 - N[(ln(3000) - 7)/4]) + (1/5) (1 - N[(ln(1000) - 7)/4]) =

0.8(1 - N[0.2515918919]) + 0.2(1 - N[-0.0230611803])

We use the function

"=0.8*(1 - NormSDist(0.2515918919)) + 0.2*(1 - NormSDist(-0.0230611803))" to find our desired probability, 0.422382463.

Problem S3L49-4. Similar to Question 21 from the Casualty Actuarial Society's Fall 2006 Exam 3. Jawaharlal is currently 65 years old and has a survival function s₁(x) = (1 - x/140)^1/2.

Andrew is currently 89 years old and has a survival function s₂(y) = 1 - y/120. The two lives are completely independent.

Find the conditional probability that, given that both Jawaharlal and Andrew survive during the next 15 years, Jawaharlal will survive to the beginning of the 16^th year while Andrew will not.

Solution S3L49-4. The probability that Jawaharlal will survive to age 81, given that he survives to age 80 is ₁p^(x)₈₀ = s₁(81)/s₁(80) = (1 - 81/140)^1/2/(1 - 80/140)^1/2 = 0.991631652.

The probability that Andrew will not survive to age 105, given that he survives to age 104 can be found considering that Andrew's future lifetime is uniformly distributed, with a maximum of 120. At age 104, therefore, Andrew will have a 1/(120-104) = 1/16 probability of senselessly perishing in each year. Our desired probability is the product of the two probabilities we found:

0.991631652*(1/16) = about 0.0619769783.

Problem S3L49-5. The following non-homogeneous Markov Chain illustrates the transitions among states X, Y, and Z during three time periods.

Period 1:

...X...Y....Z

X[1.0..0....0..]

Y[0.4..0.3..0.3]

Z[0....0....1..]

Period 2:

...X...Y....Z

X[1.0..0....0..]

Y[0.2..0.7..0.1]

Z[0....0....1..]

Period 3:

...X...Y....Z

X[1.0..0....0..]

Y[0.5..0...0.5]

Z[0....0....1..]

Find the probability that an entity in state Y at the onset of Period 1 will be in state X at the end of Period 3.

Solution S3L49-5. We note that an entity in state X will always stay in state X, and the same holds for an entity in state Z.

Our desired probability is then equal to the sum of the following probabilities:
Pr(Y at onset → X after Period 1) = 0.4

Pr(Y at onset → Y after Period 1→ X after Period 2) = 0.3*0.2 = 0.06

Pr(Y at onset → Y after Period 1→ Y after Period 2→ X after Period 3) = 0.3*0.7*0.5 = 0.105

We add up the above probabilities to get our answer: 0.4 + 0.06 + 0.105 = 0.565.

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