abraham weishaus errata

Errata for ASM Exam MLC Study Manual (Tenth Edition Third Printing) Sorted by

Errata and Updates for ASM Exam MLC (Tenth Edition ThirdPrinting) Sorted by Page

[12/17/2011]On page 6, on the first line of Section 1.3, Pr(B 6= 0) should be Pr(B ) 6= 0.

[9/27/2011] On page 12, in the solution to exercise 1.3, on the first displayed line, change E[X +Y ]3 to E[(X +Y )3].

[5/29/2011] On page 14, in the solution to exercise 1.9, 7 lines from the end, change 32(7500) to 2

3(7500).

[5/29/2011] On page 27, in the solution to exercise 2.8, at the end, delete (C), since the question is not multiplechoice.

[12/20/2011]On page 41, in the solution to exercise 3.6, the second sentence is cut off and should read

It’s the usual CAS type of humor—it’s not that the answer cannot be determined from the giveninformation (almost never the right answer choice), rather there is too much information pro-vided.

[10/3/2011] On page 43, in the solution to exercise 3.15, on the second line, in the last two symbols, q shouldnot be in the subscript. Change them to 1|q 0 and 2|q 0.

[6/3/2011] On page 65, in the solution to exercise 4.37, on the last two displayed lines, add a right parenthesisbefore the last equals sign on each line.

[12/19/2011]On page 83, 4 lines above the graph, change �ex :n to �e u : n .

[1/7/2012] On page 143, in exercise 8.28, on the last line, put a period after 97.5.

[1/22/2012] On page 177, in the solution to exercise 9.18, on the first displayed line, put 1− before900

1000.

[7/26/2012] On page 181, one line above Quiz 13-1, the parentheses in the formula for R 2 are misplaced. Thecorrect formula is

R 2 =

�∑

xi yi

�2

�∑

x 2i

� �∑

y 2i

�

[6/7/2011] On page 183, on the fourth line, delete “actuarial”.

[4/11/2012] On page 188, on the second line of the answer to Example 10C, change “morality” to “mortality”.

[10/10/2012]On page 189, in Example 9F, on the second-to-last line, delete “actuarial”.

[1/23/2012] On page 207, in the solution to Quiz 10-1, on the second displayed line, replace A30:10 with A30:101 .

[1/30/2012] On page 216, in exercise 11.4, answer choice (B) should be 375.

[3/1/2012] On page 224, in the solution to exercise 11.12, on the fifth displayed line, change4

15to

4

14.

[6/3/2011] On page 249, in the solution to exercise 12.17, on the second line, remove the bar from 2Ax .

[6/7/2011] On page 264, on the fourth line, delete “actuarial”.

[6/3/2011] On page 271, in the solution to exercise 13.1, on the second line, change the second v10 to vt .

[12/27/2011]On page 276, in the solution to question 13.22, on the third displayed line, remove the negativesign at the start of the line.

[7/30/2012] On page 285, on the second line of the solution to Example 14E, change the subscript x : n tox : 10 .

Updated 11/3/2012

2 Errata for ASM Exam MLC Study Manual (Tenth Edition Third Printing) Sorted by Page

[8/12/2012] On page 296, exercise 14.22 may be worked out mechanically, but the given values are impossiblesince they imply (I A)x1:10 is negative.

[7/28/2011] On page 299, in the list of additional released exam questions, delete the “,35” in “M-S05:15,35”.

[3/24/2012] On page 304, in exercise 14.21, 2 lines from the end, delete one of the equal signs.

[8/13/2012] On page 313, exercise 15.4 can be solved mechanically, but it is impossible for the values to occursince the pure endowment would be worth more than v 30.

[8/13/2012] On page 316, in the solution to exercise 15.5, on the first line, Ax1

:20 should be Ax1

:20 .

[7/5/2012] On page 316, in the solution to exercise 15.7, on the second line, add “1000” before “20E 40 A60“.On the third line, add “1000” before “20E 40 A60“ and change the second equals sign to a minus.

[8/15/2011] On page 325, in Quiz 16-1, the values given are impossible, since the resulting 10-year pure en-dowment of 0.6 is greater than e −10δ. Therefore, change A60 to 0.42.

[6/6/2011] On page 325, on the third line of the paragraph beginning “Whole life”, replace the expression

e −(δ+µx (t )) with e −�

δt+∫ t

0µx (u )du

�

.

[6/6/2011] On page 334, in the solution to exercise 16.5, on the third line, replace µwith k +µx+t .

[8/15/2011] On page 338, revise the solution to Quiz 16-1 in line with the revised value of A60 given above:

Since A50 = A501

:10 + 10E 50 A60, we have 10E 50 = (0.22− 0.01)/0.42= 0.5. Then A50:10 = 0.01+ 0.5=0.51. Also, since Ax = 1−δax , we have δ= (1−0.22)/13= 0.06. Therefore

a50:10 =1− A50:10

δ=

1−0.51

0.06= 8 1

6

[7/5/2012] On page 339, in the third bullet, change “indicate” to “indicates”.

[7/27/2011] On page 349, on the first line of the answer to Example 17H, change “second” to third. It is refer-ring to the equation one line above Example 17H.

[7/28/2011] On page 366, in the list of additional released exam questions, remove SOA M-F06:4.

[10/23/2012]On page 375, in the solution to exercise 17.41, replace the paragraph beginning “Note” with

Note that there is a slight inconsistency in the four annuity values given, due to rounding. A 4-year pure endowment can be backed out from the two level annuities:

a20:4 −a20:4 = 1− 4E 20

3.41−3.04= 1− 4E 20

4E 20 = 0.63

On the other hand, using the increasing annuities,

(I a )20:4 − (I a )20:4 = a20:4 −4 4E 20

8.05−7.17= 3.41−4 4E 20

0.88= 3.41−4 4E 20

4E 20 = 0.6325

Thus you may get a slightly different answer if you do the calculation differently.

[7/27/2011] On page 383, the first sentence of the fifth paragraph (starting with “If I = 1”) skips a step. Replaceit with these two sentences:

Updated 11/3/2012


If I = 1, Y | I is v n times a whole life annuity on (x +n ), and we know the expected value and vari-ance for this annuity; the expected value is ax+n and the variance is (2Ax+n− A2

x+n )/δ2. Therefore,

E[Y | I ] = v n ax+n and Var(Y | I ) = v 2n (2Ax+n − A2x+n )/δ

2.

[6/15/2011] On page 388, in Table 18.1, formula (18.12) should have d instead of d 2 in the denominator.

[7/11/2011] On page 391, in exercise 18.10(iii), change 1250 to 125.

[7/28/2011] On page 396, add SOA M-F06:4 to the list of additional released exam questions.

[7/11/2011] On page 400, in the solution to exercise 18.10, on the 5th and 6th lies of the page, change 1250 to125. On the 7th line, change 625 to 62.5.

[9/8/2011] On page 411, on the 4th line of the second paragraph, change 14/5 to 14/6.

[7/29/2011] On page 440, on the 6th line, change (1+ i )(1/3) +2(1+ i )2/3 to 2(1+ i )1/3+ (1+ i )2/3.

[7/14/2011] On page 460, in the solution to exercise 21.8, on the second displayed line, change 5E x to 10E x .

[3/20/2012] On page 486, in exercise 22.51(ii), change “such than” to “such that”.

[7/28/2011] On page 487, in the list of additional released exam questions, add “,35” after “M-S05:14”

[7/15/2011] On page 498, the solution to exercise 22.38 is incorrect. Replace the part starting with “Thepresent value of the refund of premium. . . ” to the end with the following, which also includesan easier method:

The present value of the refund of premium benefits, since the premiums are refunded with in-terest, is the present value of the premiums themselves. If death occurs in year k , the premiumsup to that time are a k -year certain annuity-due. So the present value of the refund of premiumbenefits is

PV Refund= 0.5π10∑

k=1

k−1|q 20 ak

= 0.5π10∑

k=1

0.01(0.99k−1)

�

1−1/1.05k

0.05/1.05

�

= 0.005(21)π

10∑

k=1

0.99k−1−1

1.05

10∑

k=1

�

0.99

1.05

�k−1!

= 0.105

�

1−0.9910

1−0.99−

1

1.05

�

1− (0.99/1.05)10

1−0.99/1.05

��

= 0.225609

Equating the premiums with the benefits,

7.78379π= 9716.212+0.225609π

π=9716.212

7.78379−0.225609= 1285.523

An easier way to solve this question is to equate the accumulated premiums at time 10 to thepresent value of the annuity, 17,500. The nonrefundable premiums are accumulated with mor-tality and interest, and the refundable premiums are accumulated with interest only, so

π=17,500

0.5(s20:10 + s10 )

Updated 11/3/2012


We computed a20:10 above, and accumulating with mortality and interest,

s20:10 = 7.78379

�

1.0510

0.9910

�

= 14.01949

while

s10 =1.0510−1

0.05/1.05= 13.20679

so the premium is 17,500��

0.5(14.01949+13.20679)�

= 1285.523 .

[9/5/2011] On page 507, 3 lines below formula (23.2), change the expression before “is wrong” to (b +P /δ)2.

[2/24/2012] On page 516, in the solution to exercise 23.9, on the third line, change 0.1875 to −0.1875.

[6/21/2011] On page 536, in the caption for Figure 25.2, change fuction to function.

[3/21/2012] On page 558, in the solution to exercise 26.12, on the last line, change 1.06 to 1.08.

[8/15/2011] On page 581, on the third displayed line below Table 28.2, change the denominator to h−k P x+k1

:n−kand change h < k to k < h .

[3/12/2012] On page 582, on the last line of the answer to Example 28B, change Ax+5 to Ax+5.

[3/12/2012] On page 608, in question 29.8, on the last line of the first paragraph, add “payment” after “lump-sum”. Also, on the second line from the end of the question, change i from 0.025 to 0.03.

[10/5/2011] On page 632, in exercise 30.18(i), “full” should be “fully”. Also, the first column of the table repre-sents the number of survivors on July 1, 2009.

[10/5/2011] On page 663, in the solution to exercise 31.23, on the third line, change π45 to π19.

[3/12/2012] On page 667, in Example 32C, on the first line, change (50) to (x ).

[8/14/2011] On page 672, on the four displayed lines after “The fractional premium for the endowment in-surance”, there are two errors. Before the first line, add:

A45:20 = A451

:20 +A45:201 = 0.09+0.25= 0.34

On the line for a (4)45:20

, replace 0.3475558 with 0.342001. Replace the last line with

P (4)45:20

=0.34

11.3743= 0.029892

On the last line of the answer, change 0.030324 to 0.029892.

[11/13/2011]On page 687, on the last line of the solution to exercise 32.21, the left side should be 2V (A40:10 ).

[3/12/2012] On page 691, in the solution to Quiz 32-2, on the second line, the reference should be to equa-tion (17.11) instead of (12.4).

[10/22/2011]On page 733, in the solution to exercise 35.17, on the first line, replace qx qy with t q x t q y .

[4/10/2012] On page 762, in the solution to exercise 37.4, on the second displayed line, Ax+n :y+n at the end ofthe line should be Ax+n :y+n .

[9/27/2011] On page 783, in exercise 38.23, replace the first sentence with

For two independent lives, (30) and (50), a continuous life annuity pays 100 per year to (50) afterthe death of (30), but only until (50) reaches age 80.

Updated 11/3/2012


[11/3/2012] On page 800, in Table 39.1, in formula (39.9), the first line to the right of the brace should ben1− n 2

2a band the second line should be 1− a

2b. The formula is correctly stated on page 797.

[7/29/2011] On page 843, delete SOA M-F06:24 from the list of additional released exam questions.

[9/13/2012] On page 863, in the solution to exercise 42.14, on the second line, change 0.0015 to 0.0005.

[4/11/2012] On page 867, in the solution to exercise 42.29, on the second displayed line, in the denominator,0.02 should be 0.02t .

[7/5/2011] On page 885, in the title of Section 44.1 and in the next two paragraphs, “double decrement”should be replaced by “multiple decrement” wherever it appears. It appears four times.

[7/5/2011] On page 887, in Quiz 44-1(iii), replace “if” with “is”.

[7/29/2011] On page 901, add SOA M-F06:24 to the list of additional released exam questions.

[7/5/2011] On page 908, in the solution to Quiz 44-1, on the left side of the first line, the superscript (1) shouldbe (2).

[4/11/2012] On page 924, in the solution to exercise 45.6, in the first line of the page, change “that” to “than”.

[4/17/2012] On page 925, in the solution to exercise 45.15, on the second line, replace the 20 in e −(0.06+0.02+0.04)(20)

with t .

[3/25/2012] On page 1003, in the solution to exercise 49.6, on the first line, change “by” to “be”.

[4/24/2012] On page 1101, in question 4, in the table, change “All years” to “Renewal years”.

[7/19/2011] On page 1167, in question 21(iv), delete “actuarial”.

[4/3/2012] On page 1179, in the solution to question 21, on the fifth displayed line, put a bar over Ax+5.

[3/19/2012] On page 1188, in the solution to question 20, on the fourth line, in the last integrand, t is missingfrom the exponent: e −0.075t 0.005 dt .

[3/15/2012] On page 1190, in the solution to question 26, on the last two lines of the page, put a bar overa40 0.05 and a40 0.10.

[4/25/2012] On page 1198, in the solution to question 19, change 0.02768 to 0.02766 on the first two displayedlines (once apiece).

[8/7/2011] On page 1199, in the solution to question 22, on the last line, a right parenthesis is missing fromthe numerator, after 0.104251.

[7/19/2011] On page 1203, in the solution to question 6, on the second line, change the integrand to µx+s ds .

[3/29/2012] On page 1226, in the solution to question 12, on the first displayed line, change the denominatorfrom α−1 to α+1.

[7/19/2011] On page 1243, in the solution to question 24, 4 lines from the bottom, change the first numeratorfrom a10 0.1 to a10 0.1.

[7/19/2011] On page 1252, in the solution to question 20, on the second line at the end, change 10p x = 0.2 to10p x −0.2.

[7/19/2011] On page 1253, in the solution to question 24, on the last line, change 10L to 30L .

[4/27/2012] On page 1256, in the solution to question 3, on the last line of the page, add dt at the end of theline.

[7/19/2011] On page 1258, in the solution to question 11, on the last line of the page, change 0.000890 to0.001780.

Updated 11/3/2012


[7/19/2011] On page 1261, the solution to question 22 is correct, but here is a more straightforward solution,using the prospective formula for reserves.

The expected present value of the benefits at time 19 is the sum of the EPV of the death benefitand the survivorship benefit:

Ax+19:1 = 1000

∫ 1

0

e −0.07t 0.02 dt +1000e −0.07 = 1000

�

2e −0.07

7+ e −0.07

�

= 951.71

The expected present value of the premiums at time 19 is, using ax :n = (1− Ax :n )/δ,

P ax+19:1 = 30

�

1−0.95171

0.05

�

= 28.97

So the reserve is 951.71−28.97= 922.74 . (D)

[11/5/2011] On page 1267, in the solution to question 6, on the first displayed line, remove the bar from Ax+10.On the third displayed line, change + to −.

[7/19/2011] On page 1269, in the solution to question 10, on the last line, change t q (2)0 to 5q (2)0 .

[3/31/2012] On page 1275, in the solution to question 24, on the last line, change T (50)2 to T (50).

[7/28/2011] On pages 1285 and 1286, in the solution to question 15, on the last line of page 1285, remove theline from the presubscript of 9|p 40 at the end of the line, so that it becomes 9p 40. Make the samecorrection on page 1286, four lines from the end of the solution.

[3/12/2012] On page 1301, in the solution to question 36, on the third line, the integrand should be e −(0.08+0.05)t dt .

[7/29/2011] On page 1313, in the solution to question 1, on the second displayed line, q45 should be q (s )45 .

[8/26/2012] On page 1323, in the solution to question 27, on the second line below the table, l65−1000 shouldbe l65 = 1000.

[8/26/2012] On pages 1344, in the solution to question 22, the first displayed line should read

A40 =a70

70=

1−1/(1.07170)70 ln 1.071

= 0.206557

On the second displayed line, replace the two Ax ’s with A40’s.

[10/24/2012]On page 1345, in the solution to question 4, on the last line, change the first4

3to

1

3.

[8/26/2012] On page 1349, in the solution to question 1, on the second line from the end, change l61 − l58 tol58− l61.

[8/26/2012] On page 1350, in the solution to question 8, on the first line, replace d (1)x = l (τ)x with d (1)x = l (τ)x (1−e −0.06).

[10/28/2012]On page 1361, in the solution to question 5, at the end of the first line, change 5|q x y to 5|q x y .

[4/1/2012] On page 1362, in the solution to question 12, on the second line, change v q to v p .

[7/28/2011] On page 1366, the lesson number for SOA Spring 2005 question 35 should be 22 instead of 14.The lesson number for SOA Fall 2006 question 4 should be 18 instead of 17. The lesson numberfor Fall 2006 question 24 should be 44 instead of 41. The lesson number for SOA Spring 2007question 29 should be 16 instead of 33.

[3/16/2011] On page 399, in the solution to exercise 18.10, on the second displayed line, remove the δ2 fromthe denominator.

Updated 11/3/2012

abraham weishaus errata

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