ASM Study Manual for Exam P, Second Edition By Dr. Krzysztof M. Ostaszewski, FSA, CFA, MAAA (krzysio@krzysio.net) Errata Effective July 5, 2013, only the latest edition of this manual will have its errata updated. You can find the errata for all latest editions of my books at: http://smartURL.it/errata Posted April 5, 2013 The second to last formula in the solution of Problem 12 in Practice Examination 5 should be:

Pr X >1( ) = 34xdydx

0

2−x

∫1

2

∫ =34x 2 − x( )dx =

1

2

∫34x2 − 1

4x3⎛⎝⎜⎞⎠⎟x=1

x=2

= 3− 2( )− 34− 14

⎛⎝⎜

⎞⎠⎟ =

12.

The calculation shown was correct but there was a typo in limits of integral calculation. Posted April 2, 2013 The last formula in the solution of Problem 10 in Practice Examination 4 should be

E Y( ) = x ⋅ 2.5 ⋅0.6

2.5

x3.5⎛⎝⎜

⎞⎠⎟dx

0.6

2

∫ + 2 ⋅2.5 ⋅0.62.5

x3.5⎛⎝⎜

⎞⎠⎟dx

2

+∞

∫ =2.5 ⋅0.62.5

x2.5dx

0.6

2

∫ +5 ⋅0.62.5

x3.5dx

2

+∞

∫ =

= − 2.5 ⋅0.62.5

1.5x1.5 x=0.6

x=2

− 2 ⋅0.62.5

x2.5 x=2

x→∞

= − 2.5 ⋅0.62.5

1.5 ⋅21.5+ 2.5 ⋅0.6

1.5− 0 + 0.6

2.5

21.5≈ 0.934273.

The calculation shown was correct, but a factor in the numerator of the last fraction was mistyped as 2.5, when it should have been 2. Posted March 6, 2013 In the solution of Problem 6 in Practice Examination 2, the expression rectangle 0,1[ ]× 0,2[ ], should be rectangle 0,1[ ]× 0,2[ ], This expression appears twice in the solution. Posted March 10, 2011 The second sentence of the solution of Problem 10 in Practice Examination 1 should be: As the policy has a deductible of 1 (thousand), the claim payment is

Y =0, when there is no damage,with probability 0.94,max 0,X −1( ), when 0 < X < 15, with probability 0.04, 14, in the case of total loss,with probability 0.02.

⎧

⎨⎪

⎩⎪⎪

Posted January 15, 2011 In Problem 16 in Practice Examination 6, the calculation of the second moment of X should be:

E X 2( ) = 14 ⋅0

2 + 34⋅ 1+1( )

Second moment of T =

32

.

instead of

E X 2( ) = E X( ) = 14 ⋅0

2 + 34⋅ 1+1( )

Second moment of T =

32

.

Posted July 24, 2010 In the solution of Problem 21 in Practice Examination 6, the statement under the first expression on the right-hand side of the third to last formula should be: number of ways topick ordered samplesof size 2 from population of size n

instead of number of ways topick ordered samplesof size n−2 from population of size n

Posted July 3, 2010 In the solution of Problem 1, Practice Examination 5, at the end of the first part of the fourth sentence of the solution, 5/6 is a typo, it should be 5/36, as used in the formula for Pr (Y = 6). Posted January 5, 2010 In the description of the gamma distribution in Section 2, the condition for the range of its MGF should be t < β, not 0 < t < β. Posted June 18, 2009 The last formula in the solution of Problem 19 of Practice Examination 5 should be

Pr X ≥10( ) ≤ 142

= 116

instead of Pr X ≥10( ) < 142

= 116.

Posted April 6, 2009 In the text of Problem 4 in Practice Examination 11, the sentence: We put that chip aside and pick a second chip from the same contained. should be:

We put that chip aside and pick a second chip from the same container. Posted April 6, 2009 In the text of Problem 7 in Practice Examination 3, the word “whwther” should be “whether”. Posted March 5, 2009 The first sentence of Problem 16 in Practice Examination 6 should end with t 1. Posted March 2, 2009 The properties of the cumulant moment-generating function should be: The cumulant generating function has the following properties:

ψ X 0( ) = 0, ddtlnE etX( )

t=0

=E XetX( )E etX( )

t=0

= E X( ),

d2

dt 2ψ X t( )

t=0

= ddtE XetX( )E etX( )

t=0

=E X 2etX( )E etX( )− E XetX( )E XetX( )

E etX( )( )2t=0

= Var X( ),

d 3

dt 3ψ X t( )

t=0

= E X − E X( )( )3( ), but for k > 3,

dk

dt kψ X t( )

t=0

=ψ Xk( ) 0( ) < E X − E X( )( )k( ).

Also, if X and Y are independent (we will discuss this concept later), ψ aX+b t( ) =ψ X at( ) + bt, and ψ X+Y t( ) =ψ X t( ) +ψ Y t( ). Posted September 1, 2008 In the solution of Problem 16 in Practice Examination 2, the left-hand side of the second formula should be E X 2( ), not E X( ). Posted July 23, 2008 In the solution of Problem 5 in Practice Examination 3 the expression Var X2( ) = 40,000 should be Var X2( ) = 250,000. Posted February 7, 2008 The discussion of the lack of memory property of the geometric distribution should have the formula

Pr X = n + k X ≥ n( ) = Pr X = k( ) corrected to: Pr X = n + k X > n( ) = Pr X = k( ). Posted November 13, 2007 In the solution of Problem No. 26 of Practice Examination No. 5, the sentence: Of the five numbers, 1 can never be the median. should be Of the five numbers, neither 1 nor 5 can ever be the median. Posted October 5, 2007 In the solution of Problem No. 1 in Practice Examination No. 3, the random variable Y should refer to class B. Posted September 13, 2007 The beginning of the third sentence in the solution of Problem No. 21 in Practice Examination No. 2 should be The mean error of the 48 rounded ages, instead of The mean of the 48 rounded ages. Posted May 9, 2007 In the solution of Problem No. 15, Practice Examination No. 3, the derivative is missing a minus, and it should

dXdY

= − 138 −Y( )−

23 .

The rest of the solution is unaffected. Posted May 9, 2007 Answer C in Problem No. 7 in Practice Examination No. 6 should be 0.2636, not 0.2626. It is the correct answer. Posted May 8, 2007 Problem No. 24 in Practice Examination No. 5 should be (it had a typo in the list of possible values of x) May 1992 Course 110 Examination, Problem No. 35 Ten percent of all new businesses fail within the first year. The records of new businesses are examined until a business that failed within the first year is found. Let X

be the total number of businesses examined prior to finding a business that failed within the first year. What is the probability function for X? A. 0.1⋅0.9x , for x = 0,1,2,3,... B. 0.9 ⋅0.1x , for x = 0,1,2,3,... C. 0.1x ⋅0.9x , for x = 1,2,3,... D. 0.9x ⋅0.1x , for x = 1,2,3,... E. 0.1⋅ x −1( ) ⋅0.9x , for x = 2,3,4,... Solution. Let a failure of a business be a success in a Bernoulli Trial, and a success of a business be a failure in the same Bernoulli Trial. Then X has the geometric distribution with p = 0.1, and therefore

fX x( ) = 0.1⋅0.9x for x = 0, 1, 2, 3, … . Answer A. Posted May 5, 2007 Problem No. 10 in Practice Examination No. 3 should be (it had a typo in one of the integrals and I decided to reproduce the whole problem to provide a better explanation): Sample Course 1 Examination, Problem No. 35 Suppose the remaining lifetimes of a husband and a wife are independent and uniformly distributed on the interval (0, 40). An insurance company offers two products to married couples: • One which pays when the husband dies; and • One which pays when both the husband and wife have died. Calculate the covariance of the two payment times. A. 0.0 B. 44.4 C. 66.7 D. 200.0 E. 466.7 Solution. Let H be the random time to death of the husband, W be the time to death of the wife, and X be the time to the second death of the two. Clearly, X = max H ,W( ). We have fH h( ) = fW w( ) =

140

for 0 ≤ h ≤ 40, and 0 ≤ w ≤ 40. Thus E H( ) = E W( ) = 20. Furthermore,

FX x( ) = Pr X ≤ x( ) = Pr max H ,W( ) ≤ x( ) == Pr H ≤ x{ }∩ W ≤ x{ }( ) = Pr H ≤ x( ) ⋅Pr W ≤ x( ) = x40 ⋅

x40

= x2

1600.

This implies that

sX x( ) = 1−x2

1600

for 0 ≤ x ≤ 40, and

E X( ) = 1− x2

1600⎛⎝⎜

⎞⎠⎟0

40

∫ dx = 40 −403

4800= 1203

− 403

= 803.

In order to find covariance, we also need to find E XH( ) = E Hmax H ,W( )( ).

We separate the double integral into two parts: one over the region where the wife lives longer and one over the region where the husband lives longer, as illustrated below:

E H max H ,W( )( ) = hmax h,w( ) ⋅ fH h( ) ⋅ fW w( ) ⋅dwdh =−∞

+∞

∫−∞

+∞

∫

= h2 ⋅ 140

⋅ 140dw

w=0

w=h

∫⎛

⎝⎜⎞

⎠⎟h=0

h=40

∫ dh + hw ⋅140

⋅ 140dw

w=h

w=40

∫⎛

⎝⎜⎞

⎠⎟h=0

h=40

∫ dh =

= h2w

1600 w=0

w=h⎛

⎝⎜

⎞

⎠⎟ dh

0

40

∫ +hw2

3200 w=h

w=40⎛

⎝⎜

⎞

⎠⎟

0

40

∫ dh =h3

1600dh

0

40

∫ +1600h − h3

3200dh

0

40

∫ =

= 12h + 1

3200h3⎛⎝⎜⎞⎠⎟ dh

0

40

∫ =14h2 + 1

12800h4⎛⎝⎜⎞⎠⎟h=0

h=40

= 14⋅402 + 1

12800⋅404 = 600.

Finally,

Cov X,H( ) = E XH( )− E X( )E H( ) = 600 − 20 ⋅ 803

= 66 23.

Answer C. Posted April 29, 2007 The second formula to the last in the solution of Problem No. 6 in Practice Examination No. 3 should be:

E X 2( ) = d2M t( )dt 2 t=0

= ddt3et 2 + e

t

3⎛⎝⎜

⎞⎠⎟

8

t=0

=

= 3et 2 + et

3⎛⎝⎜

⎞⎠⎟

8

+ 8et 2 + et

3⎛⎝⎜

⎞⎠⎟

7

⋅et⎛

⎝⎜

⎞

⎠⎟t=0

= 11.

h

w 40

40

max h,w( ) = h here

max h,w( ) = where

instead of

E X 2( ) = d2M t( )dt 2 t=0

= ddt3et 2 + e

t

3⎛⎝⎜

⎞⎠⎟

8

t=0

=

= 3et 2 + et

3⎛⎝⎜

⎞⎠⎟

8

+ 8et 2 + et

3⎛⎝⎜

⎞⎠⎟⋅et

⎛

⎝⎜

⎞

⎠⎟t=0

= 11.

Posted April 27, 2007 In the solution of Problem No. 9 in Practice Examination 3, the expression max X,1( ) should be min X,1( ). Posted April 20, 2007 Problem No. 16 in Practice Examination No. 1 should be: May 2001 Course 1 Examination, Problem No. 26, also Study Note P-09-05, Problem No. 109 A company offers earthquake insurance. Annual premiums are modeled by an exponential random variable with mean 2. Annual claims are modeled by an exponential random variable with mean 1. Premiums and claims are independent. Let X denote the ratio of claims to premiums. What is the density function of X?

A. 12x +1

B. 22x +1( )2

C. e− x D. 2e−2x E. xe− x

Solution. Let U be the annual claims and let V be the annual premiums (also random), and let fU ,V u,v( ) be the joint density of them. Furthermore, let fX be the density of X and let FX

be its cumulative distribution function. We are given that U and V are independent, and

hence fU ,V u,v( ) = e−u ⋅12e− v2 = 1

2e−ue

− v2 for 0 < u < ∞,0 < v < ∞. Also, noting the graph

below, we have: u

v

u = vx or v = u/x

FX x( ) = Pr X ≤ x( ) = PrUV

≤ x⎛⎝⎜⎞⎠⎟ = Pr U ≤Vx( ) = fU ,V u,v( )

0

vx

∫0

∞

∫ dudv =

= 120

vx

∫0

∞

∫ e−ue− v

2dudv = − 12e−ue

− v2

⎛⎝⎜

⎞⎠⎟u=0

u=vx

dv0

+∞

∫ = −12e−vxe

− v2 + 1

2e− v

2⎛⎝⎜

⎞⎠⎟0

∞

∫ dv =

= − 12e−v x+1

2⎛⎝⎜

⎞⎠⎟ + 1

2e− v

2⎛

⎝⎜⎞

⎠⎟0

∞

∫ dv =1

2x +1e−v x+1

2⎛⎝⎜

⎞⎠⎟ − e

− v2

⎛

⎝⎜⎞

⎠⎟0

+∞

= − 12x +1

+1.

Finally,

fX x( ) = ′FX x( ) =2

2x +1( )2.

Answer B. This problem can also be done with the use of bivariate transformation. We will now give an alternative solution using that approach. Consider the following transformation

X = UV, Y =V .

Then the inverse transformation is U = XY , V = Y .

We know that fU ,V u,v( ) =12e−ue

− v2 for u > 0, v > 0. It follows that

fX ,Y x, y( ) = fU ,V u x, y( ),v x, y( )( ) ⋅ ∂ u,v( )∂ x, y( ) =12e− xye

− y2 ⋅ det y x

0 1

⎡

⎣⎢⎢

⎤

⎦⎥⎥= 12ye− xye

− y2

for xy > 0 and y > 0, or just x > 0 and y > 0. Therefore,

fX x( ) =12ye− xye

− y2 dy

0

+∞

∫ =12ye

− x+12

⎛⎝⎜

⎞⎠⎟ y dy

0

+∞

∫ =w = 1

2y z = − 1

x + 12

⎛⎝⎜

⎞⎠⎟e− x+1

2⎛⎝⎜

⎞⎠⎟ y

dw = 12dy dz = e

− x+12

⎛⎝⎜

⎞⎠⎟ ydy

INTEGRATION BY PARTS

=

= − 12y ⋅ 1

x + 12

⎛⎝⎜

⎞⎠⎟e− x+1

2⎛⎝⎜

⎞⎠⎟ y

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟y=0

y→+∞

− 12

− 1

x + 12

⎛⎝⎜

⎞⎠⎟

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟e− x+1

2⎛⎝⎜

⎞⎠⎟ y dy

0

+∞

∫ =

= 0 + 12x +1

e− x+1

2⎛⎝⎜

⎞⎠⎟ y dy

0

+∞

∫ =1

2x +1− 1

x + 12

⎛⎝⎜

⎞⎠⎟

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟⋅e

− x+12

⎛⎝⎜

⎞⎠⎟ y

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟y=0

y→+∞

= 22x +1( )2

.

Answer B, again. The second approach is probably more complicated, but it is a good exercise in the use of multivariate transformations. Posted January 20, 2007 Exercise 2.8 should read as follows: November 2001 Course 1 Examination, Problem No. 11 A company takes out an insurance policy to cover accidents that occur at its manufacturing plant. The probability that one or more accidents will occur during any

given month is 35. The number of accidents that occur in any given month is independent

of the number of accidents that occur in all other months. Calculate the probability that there will be at least four months in which no accidents occur before the fourth month in which at least one accident occurs. A. 0.01 B. 0.12 C. 0.23 D. 0.29 E. 0.41 Solution. Consider a Bernoulli Trial with success defined as a month with an accident, and a month

with no accident being a failure. Then the probability of success is 35. Now consider a

negative binomial random variable, which counts the number of failures (months with no accident) until 4 successes (months with accidents), call it X. The problem asks us to find Pr X ≥ 4( ). But

Pr X ≥ 4( ) = 1− Pr X = 0( )− Pr X = 1( )− Pr X = 2( )− Pr X = 3( ) =

= 1− 30

⎛⎝⎜

⎞⎠⎟⋅ 3

5⎛⎝⎜

⎞⎠⎟

4

− 41

⎛⎝⎜

⎞⎠⎟⋅ 3

5⎛⎝⎜

⎞⎠⎟

4

⋅ 25− 5

2⎛⎝⎜

⎞⎠⎟⋅ 3

5⎛⎝⎜

⎞⎠⎟

4

⋅ 25

⎛⎝⎜

⎞⎠⎟

2

− 63

⎛⎝⎜

⎞⎠⎟

35

⎛⎝⎜

⎞⎠⎟

4 25

⎛⎝⎜

⎞⎠⎟

3

≈

≈ 0.289792. Answer D. Posted December 31, 2006 In Section 2, the general definition of a percentile should be the 100-p-th percentile of the distribution of X is the number xp which satisfies both of

the following inequalities: Pr X ≤ xp( ) ≥ p and Pr X ≥ xp( ) ≥1− p. It was mistyped as the 100-p-th percentile of the distribution of X is the number xp which satisfies both of

the following inequalities: Pr X ≤ xp( ) ≥ p and Pr X ≥ xp( ) ≤1− p. Posted December 28, 2006 On page 3 in the bottom paragraph, the words: is also defined as the sent that consists of all elementary events that belong to any one of them. should be is also defined as the set that consists of all elementary events that belong to any one of them. The word “set” was mistyped as “sent.” Posted December 1, 2006 In Practice Examination No. 6, Problem No. 22 had inconsistencies in its assumptions, and it should be replaced by the following: A random variable X has the log-normal distribution with the density:

fX x( ) =1

x 2π⋅e

−12ln x−µ( )2

for x > 0, and 0 otherwise, where µ is a constant. You are given that Pr X ≤ 2( ) = 0.4. Find E X( ). A. 4.25 B. 4.66 C. 4.75 D. 5.00 E. Cannot be determined Solution. X is log-normal with parameters µ and σ 2 if

W = lnX N µ,σ 2( ). The PDF of the log-

normal distribution with parameters µ and σ 2 is

fX x( ) =1

xσ 2πe−12ln x−µ( )2σ 2

for x > 0, and 0 otherwise. From the form of the density function in this problem we see that σ = 1. Therefore

Pr lnX ≤ ln2( ) = Pr lnX − µ1

≤ ln2 − µ1

⎛⎝⎜

⎞⎠⎟ = 0.4.

Let z0.6 be the 60-th percentile of the standard normal distribution. Let Z be a standard

normal random variable. Then Pr Z ≤ −z0.6( ) = 0.40. But Z = lnX − µ1 is standard normal, thus ln2 − µ = −z0.6 , and µ = ln2 + z0.6 . From the table, Φ 0.25( ) = 0.5987 and Φ 0.26( ) = 0.6026. By linear interpolation, z0.6 ≈ 0.25 +

0.6 − 0.59870.6026 − 0.5987

⋅ 0.26 − 0.25( ) = 0.25 + 13⋅0.01≈ 0.2533.

This gives µ = ln2 + z0.6 ≈ 0.9464805. The mean of the log-normal distribution is

E X( ) = eµ+12σ 2

, so that in this case

E X( ) = e0.9464805+12 ≈ 4.2481369.

Answer A. Posted November 25, 2006 The third equation from the bottom in the solution of Problem No. 10 in Practice Examination No. 2 should be Pr A∩E99( ) = Pr A E99( ) ⋅Pr E99( ) = 0.03⋅0.20 = 0.006. Previously it had a typo: Pr A∩E99( ) = Pr A E99( ) ⋅Pr E99( ) = 0.03⋅0.20 = 0.0036. Posted August 3, 2006 In the solution of Problem No. 4 in Practice Examination No. 1, the formula fY y( ) = k2 ⋅ y, should be fY y( ) = k2 ⋅1. Posted July 3, 2006 The relationship between the survival function and the hazard rate, stated just after the definition of the hazard rate, should be

sX x( ) = e− λX u( )du−∞

x

∫

instead of

sX x( ) = e− λX u( )du0

x

∫.

The second equation applies in the case when the random variable X is nonnegative almost surely. Posted July 3, 2006 The condition concerning a continuous probability density function in its first definition, in Section 1, should be

fX x( ) ≥ 0 for every x ∈ instead of 0 ≤ fX x( ) ≤1 for every x ∈ Posted May 16, 2006 The solution of Problem No. 13 in Practice Examination No. 3 should be (some of the exponents contained typos) Solution. Let X be the random number of passengers that show for a flight. We want to find

Pr (X = 31) + Pr (X = 32). We can treat each passenger arrival as a Bernoulli Trial, and then it is clear that X has binomial distribution with n = 32, p = 0.90. Therefore,

Pr X = x( ) = 32x

⎛⎝⎜

⎞⎠⎟⋅0.90x ⋅0.1032−x.

The probability desired is:

Pr X = 31( ) + Pr X = 32( ) = 3231

⎛⎝⎜

⎞⎠⎟⋅0.9031 ⋅0.101 + 32

32⎛⎝⎜

⎞⎠⎟⋅0.9032 ⋅0.100 ≈

≈ 0.12208654 + 0.03433684 = 0.15642337.

Answer E. Posted March 1, 2006 The formula for the variance of a linear combination of random variables should be:

Var a1X1 + a2X2 +…+ anXn( ) = = a1 a2 … an⎡⎣

⎤⎦ ⋅ Cov Xi ,Xj( )⎡⎣ ⎤⎦i, j=1,2,…,n( ) ⋅ a1 a2 … an⎡⎣ ⎤⎦T =

= a1 a2 … an⎡⎣⎤⎦ ⋅ Cov Xi ,Xj( )⎡⎣ ⎤⎦i, j=1,2,…,n( ) ⋅

a1a2an

⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

=

= ai2Var Xi( )

i=1

n

∑ + 2aiaj ⋅Cov Xi ,Xj( )j=i+1

n

∑ .i=1

n

∑

The only incorrect piece was in the last double sum, where a1a2 appeared instead of aiaj . Posted March 1, 2006 The moment generating function of the chi-square distribution should be:

MX t( ) =1

1− 2t⎛⎝⎜

⎞⎠⎟

n2.

(its argument was mistakenly written as x instead of t). Posted February 11, 2006 The solution of Problem No. 18 of the Practice Examination No. 4 should be: Let X be the claim for employee who incurred a loss in excess of 2000, and Y be the claim for the other employee. The joint distribution of X and Y , given that each loss occurs, is uniform on

1000,5000[ ]2 . This means that we can calculate all probabilities by comparing areas. The probability that total losses exceed reimbursement, i.e.,

X +Y > 8000, given that X > 2000 is the ratio of the area of the triangle in the right upper corner of the square

1000,5000[ ]2 in the figure below to the area of the rectangle

2000,5000[ ]× 1000,5000[ ], i.e.,

12⋅2000 ⋅2000

3000 ⋅4000= 16.

Y

5000

3000

However, this is conditional on the occurrence of the loss of the employee whose loss is Y (but no longer conditional on the occurrence of the loss for the employee whose loss is X, because the event considered is already conditional on X > 2000, so the loss has occurred). The probability of the loss occurring is 40% = 0.4. Therefore the probability

sought is 0.40 ⋅ 16= 25⋅ 16= 115.

Answer B. Posted January 31, 2006 In the solution of Problem No. 21 in Practice Examination No. 2, May 2000 Course 1 Examination, Problem No. 19, the fourth sentence should begin

The mean of the 48 rounded ages, E = 148

Eii=1

48

∑ Posted January 26, 2006 In the solution of Problem No. 15 in Practice Examination No. 2, May 2000 Course 1 Examination, Problem No. 7, the survival function should be written as: sX x( ) = 1+ x( )−3 . Posted January 18, 2006 In the description of the geometric distribution:

MX t( ) =pet

1− qet. Some texts use a different version of this distribution, which only

counts failures until the first success. That’s just too pessimistic for this author. The two

distributions differ by one. For this other form of random variable (similar in its design to

the negative binomial distribution, just below) Y = X – 1, and E Y( ) = qp, Var X( ) = q

p2,

MX t( ) =p

1− qet.

should be

MX t( ) =pet

1− qet. Some texts use a different version of this distribution, which only

counts failures until the first success. That’s just too pessimistic for this author. The two distributions differ by one. For this other form of random variable (similar in its design to

the negative binomial distribution, just below) Y = X – 1, and E Y( ) = qp, Var Y( ) = q

p2,

MY t( ) =p

1− qet.

Posted September 8, 2005 In the solution of Exercise 2.2, the first sentence should be: It is important to realize that there is a point-mass at 1 -- this can be seen by analyzing the limit of CDF at 1 from the left (equal to 0) and the right-hand side limit of CDF at 1 (equal to 0.5). Posted August 22, 2005 In the solution of Problem No. 11 in Practice Examination No. 6 these words Similarly, Pr X 7( ) < m( ) is the probability that either none of eight elements of the random sample X1,X1,…,X8 are greater than or equal to m should be replaced by Similarly, Pr X 7( ) < m( ) is the probability that either none or one of eight elements of the random sample X1,X1,…,X8 are greater than or equal to m Posted August 21, 2005 In Exercise 3.11, the second to last formula should be:

fX 2( ) =121

∞

∫ y−3dy = −14y−2

1

∞

= 14.

It was missing y−2. Posted August 20, 2005

In Exercise 3.15, November 2001 Course 1 Examination, Problem No. 40, in the solution, first two displayed formulas should be:

E X +Y( ) = E X( ) + E Y( ) = 50 + 20 = 70,Var X +Y( ) = Var X( ) +Var Y( ) + 2Cov X,Y( ) = 50 + 30 + 20 = 100.

The problem with existing formulas was that Y was improperly replaced by X. Posted August 8, 2005 In Problem No. 7 of Practice Examination No. 4, the five choices of answers should be:

A. 6125,000

50 − x − y( )0

20

∫0

20

∫ dydx

B. 6125,000

50 − x − y( )20

50−x

∫20

30

∫ dydx

C. 6125,000

50 − x − y( )20

50−x−y

∫20

30

∫ dydx

D. 6125,000

50 − x − y( )20

50−x

∫20

50

∫ dydx

E. 6125,000

50 − x − y( )20

50−x−y

∫20

50

∫ dydx Posted August 5, 2005 The solution of Problem No. 13, Practice Examination No. 5 should be: Note the region where the density is positive and the region describing the probabilities we are calculating in the figure below Therefore,

1

1 x

y

xy = w

FW w( ) = Pr W ≤ w( ) = Pr XY ≤ w( ) = Pr Y ≤wX

⎛⎝⎜

⎞⎠⎟ =

= 8xydx0

y

∫⎛

⎝⎜⎞

⎠⎟0

w

∫ dy + 8xydx0

wy

∫⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟w

1

∫ dy = 4x2y x=0x=y( )

0

w

∫ dy + 4x2y x=0x=w

y⎛⎝⎜

⎞⎠⎟w

1

∫ dy =

= 4y30

w

∫ dy +4w2

yw

1

∫ dy = y4( ) y=0y= w

+ 4w2 ln y( )y= w

y=1=

= w2 − 0( ) + 4w2 ln1− ln w( ) = w2 − 4w2 ln w.

Hence,

fw w( ) = ′FW w( ) = 2w − 8w ln w − 4w21w⋅ 12 w

=

= 2w − 8w ln w − 2w = −8w ln w.

Answer A. Posted June 12, 2005 In Section 1, the statement of the Binomial Theorem should be:

a + b( )n = nk

⎛⎝⎜

⎞⎠⎟ak

k=0

n

∑ bn−k , i.e., the sum should start from k = 0, not k = 1. Posted May 20, 2005 On the second page of Section 4: Risk and Insurance, page 77 of the manual, the formula for the variance should be: Var X( ) = E X 2( )− E X( )( )2 = q ⋅E B2( )− q ⋅E B( )( )2 = q ⋅E B2( )− q ⋅ 1− p( ) ⋅ E B( )( )2 =

= q ⋅ E B2( )− E B( )( )2( ) + pq ⋅ E B( )( )2 = q ⋅Var B( ) + pq ⋅ E B( )( )2 .

( i.e., + pq ⋅ E B( )( )2 instead of − pq ⋅ E B( )( )2 ) Posted May 18, 2005 In Exercise 2.13, the five answer choices should be: A. 10y0.8e−8y

−0.2

B. 8y−0.2e−10y

−0.8

C. 8y−0.2e− 0.1y( )1.25

D. 0.1y( )1.25 e−0.125⋅ 0.1y( )0.25

E. 0.125 ⋅ 0.1y( )0.25 e− 0.1y( )1.25 (parentheses were missing in answers D and E) Posted May 18, 2005 The last line of the solution of Problem No. 11, Practice Examination No. 3, should be:

E max Y + Z,2( )− 2 X = 0( ) = 1⋅ 427 +527

⎛⎝⎜

⎞⎠⎟ + 2 ⋅

627

= 79.

(2 is multiplied by 627, not by 5

27).

Posted May 18, 2005 The last line of the solution of Problem No. 26, Practice Examination No. 3, should be:

E X 2( ) = d

2

dt 2MX t( )

t=0

= d2

dt 2e3t+t

2

t=0

= ddt

3+ 2t( )e3t+t2( )t=0

=

= 2e3t+t2

+ 3+ 2t( )2 e3t+t2( )t=0

= 2 + 9 = 11.

(the superscript 2 was missing in the symbol of the second derivative) Posted May 15, 2005 In Problem No. 30, Practice Examination No. 5, the figure illustrating the solution should be:

y = x

1 x

y

Posted May 14, 2005 In Problem No. 25, Practice Examination No. 5, in the solution, this formula

Y − X = Y1 − X1

100+ Y2 − X2

100+…+ Y100 − X100

100,

should replace

Y − X = X1 −Y1

100+ X2 −Y2

100+…+ X100 −Y100

100.

Posted May 14, 2005 In Problem No. 9, Practice Examination No. 5, the solution is correct but the answer choice should be E. Posted May 14, 2005 In Problem No. 14, Practice Examination No. 4, the solution is correct but the answer choice should be C. Posted May 14, 2005 The density of the log-normal distribution should be:

fX x( ) =1

xσ 2πe−12ln x−µ( )2σ 2 .