Errata for ASM Exam MFE/3F Study Manual (Ninth Edition) Sorted by Page 1

Errata and updates for ASM Exam MFE/3F (Ninth Edition) sortedby page.

Note the corrections to Practice Exam 6:9 (page 613) and 10:18. None of the answer choices iscorrect for Practice Exam 10:15.

[7/22/2011] On page xi, on the first line of the second paragraph under “The normal distribution table”, add“use” after “you may”.

[7/22/2011] On page xii, replace the first line with

http://www.beanactuary.org/exams/preliminary/register/Regpdf/MFE3F_Table.pdf

[8/30/2011] On page 6, third line, change ’s to seller’s.

[10/5/2012] On page 7, on the third line of the paragraph beginning “Butterfly spreads”, change K1 to K3. Onthe fourth line of the paragraph, change K3 to K1.

[8/17/2011] On page 43, in the enumerated list at the bottom of the page, item 1, change S −K1 to S −K2.

[12/1/2011] On page 54, on the last line of the page, change “borrow” to “lend”.

[7/5/2011] On page 60, in exercise 3.5, delete the period at the end of the first line and add “having a periodof 6 months. Also on the second-to-last line, after “‘stock”, add “expiring in 6 months”.

[1/13/2012] On page 93, in exercise 4.27 on the second line, change the upper-case in futures to lower-case.

[4/15/2012] On pages 110–111, in the answer to Example 5A, the calculation ofγ at the upper node is incorrectsince the replicating portfolio cannot be used at a node for which early exercise is optimal. Hereis the corrected solution:

α= 0.15 and δ= 0.10, while h , the period of the binomial tree, is 1/4 of a year. The probability ofan up movement is

p =e (α−δ)h −d

u −d=

e 0.05(0.25)−0.8

1.3−0.8=

1.012578−0.8

0.5= 0.425157

We calculate∆ and B at the d node and at the initial node. This calculation is no different fromthe ones we did in previous lessons; the same formulas are used. We will use the stock and optionvalues we already calculated. The option values could be recalculated using Ct = max(∆t St +Bt ,St −53).

∆d = e −0.025

�

1.08

54.08−33.28

�

= 0.05064

Bd = e −0.0075

�

−0.8(1.08)0.5

�

=−1.71509

∆0 = e −0.025

�

14.6−0.39158

67.6−41.6

�

= 0.53299

B0 = e −0.0075

�

(1.3)(0.39158)− (0.8)(14.6)0.5

�

=−22.17495

Now we calculate the return on the option. At the d node,

e γ(0.25) =0.05064(41.6)

0.39158e 0.0375+

−1.71509

0.39158e 0.0075

Updated 3/3/2013

2 Errata for ASM Exam MFE/3F Study Manual (Ninth Edition) Sorted by Page

= 1.172604

γ= 4 ln 1.172604= 0.63691

At the initial node,

e γ(0.25) =0.53299(52)

5.54027e 0.0375+

−22.17495

5.54027e 0.0075

= 1.161024

γ= 4 ln 1.161024= 0.59721

We cannot use the replicating portfolio at the u node, since the value of the option, the earlyexercise value, is higher than indicated by the replicating portfolio. Therefore we must use equa-tion (5.2) with the early-exercise value of C to back out γ:

C = e −γh�

p Cu + (1−p )Cd

�

14.6= e −0.25γ�

0.425157(34.88) + (1−0.425157)(1.08)�

e −0.25γ =14.6

15.45031= 0.944965

γ=−4 ln 0.944965= 0.22643

The results are shown in Figure (5.1).

[8/23/2012] On page 111, in line with the corrected solution to Example 5A, change γ= 0.59136 in Figure 5.1to γ= 0.22643. Also, 2 lines above Quiz 5-1, an equals sign should be inserted before −0.149302.

[6/17/2012] On page 112, replace the fourth through eighth lines with

At the u node, the replicating portfolio cannot be used, so the calculation is the same as before,and γ= 0.59136.

[3/8/2012] On page 124, in the solution to exercise 5.13, 4 lines from the end, change “one year” to “threemonths”.

[11/3/2011] On page 135, in the solution to exercise 6.3, on the last line, change > to <.

[8/5/2011] On page 135, the solution to exercise 6.4 is incorrect. The correct solution is

The gain in exercising is the gain in dividends on the stock, or S (1− e −0.06) = 0.058235S . The lossis the loss in interest on the strike price, 50(1− e −0.04) = 1.96053, plus the loss of the implicit put.Calculate the gain and loss starting at 64 and working down:

Stock price Dividend gain Interest and put loss

64 64(0.058235) = 3.72707 1.96053+1.6= 3.5605363 63(0.058235) = 3.66883 1.96053+1.7= 3.6605362 62(0.058235) = 3.61060 1.96053+1.8= 3.76053

When the stock price is 62, it is not optimal to exercise, since the loss is greater than the gain.We don’t know what the put’s price is when the stock price is between 62 and 63, but it is worthat least 1.7. So the gain is definitely larger than the loss when 0.058235S > 1.96053+ 1.7, or S =3.66053/0.058235 = 62.858. The lowest price for which it may be optimal to exercise the option

is 62.858 .

[8/23/2012] On page 138, in the solution to exercise 6.14, the heading of the third column of the table shouldbe ln(St /St−1).

Updated 3/3/2013

Errata for ASM Exam MFE/3F Study Manual (Ninth Edition) Sorted by Page 3

[6/17/2012] On page 144, on the second line of the paragraph beginning with “1.”, change 125/100 to 125/100−1.

[10/2/2011] On page 145, two lines after the answer to Example 7B, change µ = 0.14 to µ = 0.15. On thesecond line of the paragraph beginning “You will notice”, change e 0.14 to e 0.15 and change 46.01to 46.47. Change 46.01 to 46.47 on the next line as well. On that line, change 40e 0.14−0.32

= 42.05 to40e 0.15−0.32

= 42.47. On the fourth line, change 40�

e 0.14+0.5(0.32)�

= 48.13 to 40�

e 0.15+0.5(0.32)�

= 48.61.On page 146, in the caption for Figure 7.1, change µ= 0.14 to µ= 0.15.

[7/14/2012] On page 150, on the second line from the end of the page, “is the stock” should be “if the stock”.

[11/2/2011] On page 156, in the solution to exercise 7.2, on the first line, change “lognormal” to “normal” intwo places.

[9/27/2011] On page 168, on the first line of Table 8.2, change xi to Si .

[9/27/2011] On page 169, change the first sentence of exercise 8.3 to

A stock’s price follows a lognormal model.

[8/14/2012] On page 178, on the fourth displayed line of the page, changep

t top

T .

[2/26/2012] On page 200, on the first displayed line, the signs should be reversed:

dP

dt=

dC

dt−δS e −δ(T−t )+ r K e −r (T−t )

[8/23/2012] On page 202, on the fourth line of Subsection 10.1.6, change 0.01 to 100.

[8/13/2011] On page 214, in exercise 10.6(ii), change the dividend yield from 0.05 to 0.06.

[9/27/2011] On page 217, in exercise 10.19, replace (vii) with

The risk premium is positive.

[11/3/2011] On page 220, in the solution to exercise 10.6, the final answer is missing a negative sign and shouldbe −0.52757.

[10/29/2012]On page 223, in the solution to exercise 10.18, on the third line, change “volatility of the put” to“elasticity of the put”. On the eighth line, delete −1 from the right hand side.

[9/27/2011] On page 223, in the solution to exercise 10.19, on the last line, replace “where we used . . . sincethe risk-free rate is nonnegative” with “We used−0.62 since using 0.62 results in an r greater than0.19, making the risk premium negative.”

[1/4/2013] On page 243, in the solution to exercise 11.7, on the last line, change+0.00536e 0.03/4 to−0.00536e 0.03/4

[9/7/2011] On page 261, in exercise 12.6(vi), add “%.” at the end of the line.

[2/21/2012] On page 256, in Example 12G(i), add “%.” after 26.

[2/21/2012] On page 258, on the fourth line, add “%.” after 10.

[8/20/2011] On page 268, in the box before question 12.39, (iii) and (iv) are cut off. (iii) should finish “5% perannum” and (iv) should finish “20% per annum”.

[8/20/2011] On page 278, in the solution to question 12.39, if 5-place rounding of the normal distributionfunction is used, then N (d1) = 0.53206 and the final answer is 46,794 shares. The solution toquestion 12.40 with 5-place rounding is as follows:

In this question, they didn’t want you to include the changes of prices of the stock in the cost;they ask for the cost of the hedge, not the cost of the instrument that is being hedged.

Updated 3/3/2013

4 Errata for ASM Exam MFE/3F Study Manual (Ninth Edition) Sorted by Page

The values of d1 that are provided are rounded to 2 decimal places. We will use them, but we willcalculate the normal distribution to 5 decimal places.

N (−0.10) = 0.46017, and N (−0.05) = 0.48006. Initially we buy 46,794 shares. After 1 week, weneed 46,017 shares so we sell 46, 017− 46, 794= 777 shares. After 2 weeks we need 48,006 sharesso we buy 48,006−46,017= 1989 shares. Each week’s purchase or sale is at the week’s stock price.These purchases are then accumulated at interest for 2 weeks, 1 week, or 0 weeks. The followingtable summarizes the calculation of the cost of the hedge:

Total Shares Stock Interest AccumulatedShares Purchased Price Factor Cost

46,794 46,794 49.16 1.001925 2,304,82146,017 −777 49.33 1.000962 −38,36648,006 1,989 49.09 1 97,640

2,364,095

The interest factors in this table were computed as e 0.05(2)/52 = 1.001925 and e 0.05/52 = 1.000962.The final column is the product of shares purchased, stock price, and interest factor. The accu-mulated cost of the hedge is 2,364,095 .

[2/21/2012] On page 261, in the introductory box for exercises 12.8 and 12.9, in (i), add “%.” after 4.

[11/29/2011]On page 271, in the solution to exercise 12.11, on the fifth line, delete the first “is”. Put a negativesign before N (−d1). On the sixth line, delete one of the negative signs in the exponent of e −r .

[9/27/2011] On page 273, in the solution to exercise 23, replace the first two expressions in the displayed linewith

Γd =∆d u −∆d d

Sd u −Sd d

[2/21/2012] On page 264, in exercise 12.22(iv), add “%.” after 5.

[8/17/2011] On page 294, on the second line of the page, change 45.5 to 45 and change 4.5923 to 4.5968. Onthe eighth line, change 4.5923 to 4.5968.

[8/30/2012] On page 294, in the answer to Example 13I, on the second displayed line, change −0.87816 to−0.77124. On the third displayed line. change P (40, to P (40.5,.

[10/2/2012] On page 321, on the first line of part 2 of the answer to Example 14D, change−0.26406 to−0.26372.

[10/2/2012] On page 347, in the solution to exercise 14.31, on the last line, change 65.50+ to 65.50= .

[11/5/2011] On page 363, in the exercise 15.1(v), the word after “risk-free” should be “interest”.

[8/4/2012] On page 379, in the third paragraph from the bottom in point 1, delete the first “Given that X (t ) =k ,” and capitalize the “t” immediately after that comma.

[8/27/2011] On page 387, the solution to exercise 16.7 is incorrect. The correct solution is

Let X (t ) be the exchange rate, and Y (t ) = ln X (t ). Logging the expression we’re given,

Y (t ) = Y (0) +0.005t +0.1Z (t )

so Y (5) is a normal random variable with mean ln X (0) + 0.005(5) = ln 0.9+ 0.025 and variance(0.12)(5) = 0.05. The probability that X (t ) is less than 1, or Y (t ) less than 0, is

N

�

0− (ln 0.9+0.025)p

0.05

�

=N (0.35938) = 0.64035

Updated 3/3/2013

Errata for ASM Exam MFE/3F Study Manual (Ninth Edition) Sorted by Page 5

[8/23/2011] On page 388, the solution to Quiz 16-3 is incorrect. The correct solution is

ln�

X (t )/X (0)�

= 0.05t + 0.3Z (t ) is an arithmetic Brownian motion. By memoryless properties of

Brownian motion, the distribution of ln�

X (2.5)/30�

is normal with mean 0.05(0.5) = 0.025 andvariance (0.32)(0.5) = 0.045. The probability of 30≤ X (t )≤ 33 is

Pr�

30≤ X (t )≤ 33�

=N

�

ln 33/30−0.025p

0.045

�

−N

�

ln 30/30−0.025p

0.045

�

=N (0.33145)−N (−0.11785)

= 0.62985−0.45309= 0.17676

[10/5/2011] On page 390, on the fifth line, the paragraph should start “The time-t value of a geometric . . . ”Add a hyphen after “time”, delete “nbd”, and move t after the hyphen.

[9/27/2011] On page 391, on the second displayed line of the answer to Example 17A, the first expressionshould be

Pr

�

lnS (t )S (0)

> ln 1.05

�

[1/15/2012] On page 397, in the solution to exercise 17.9, on the fifth line, change “then Z2” to “then Z3”. Amore accurate version of the sentence is

. . . then Z3 is a Brownian motion with variance equal to the sum of the variances of 0.24Z1‘ and0.1Z2 . . .

[1/17/2012] On page 403, on the second line of the answer to Example 18F, changed2Z (t )

dt 2to

d2X (t )dZ (t )2

.

[9/7/2011] On page 405, in exercise 18.9, on the third line from the end, change T − t to 1− t .

[9/7/2011] On page 408, the solution to exercise 18.9 has several errors. The correct solution is

The Itô process for X (t ) is

dX (t )X (t )

= (r − r f )dt +σdZ (t ) = (0.03−0.08)dt +0.15 dZ (t )

Since Y (t ) = F Pt ,1

�

X (t )�

= X (t )e −0.08(1−t ), by Itô’s lemma, we have

dY = e −0.08(1−t )dX (t ) +0.08Y dt

= e −0.08(1−t )�

−0.05X (t )dt +0.15 dZ (t )�

+0.08Y dt

=−0.05Y dt +0.15Y dZ (t ) +0.08Y dt = 0.03Y dt +0.15 dZ (t )

so µ�

t , Y (t )�

= 0.03 .

[10/17/2011]On page 412, four lines below equation (19.1), change (r −δ)C to (r −δC )C .

[7/9/2012] On page 414, on the first line, change d1 to d2. On the second line, change d2 to d1.

[10/9/2011] On page 414, on the third line in the S a paragraph, add an a before r −δ:

0.5a (a −1)σ2S a +a (r −δ)S a = r S a

[2/24/2012] On page 417, in exercise 19.6, change (ii) to “ Var�

ln�

S (t )/S (0)�

�

= 0.01t .”

Updated 3/3/2013

6 Errata for ASM Exam MFE/3F Study Manual (Ninth Edition) Sorted by Page

[4/3/2012] On page 418, in exercise 19.9, add the following line after (ii):

(iii) The price of the stock follows the Black-Scholes framework.

[1/25/2012] On page 420, in the solution to exercise 19.7, the expression for Vt on the second line should haveS (t )0.5 instead of S (t ), so

Vt = 0.055S (t )0.5e 0.055t

Similarly, on the fourth line, change the term before the equals sign to 0.055S (t )0.5e 0.055t .

[7/24/2012] On page 435, in the solution to exercise 20.1, on the fourth line, change (ii) to (iii). On the seventhline, change (iii) to (iv).

[8/19/2011] On page 442, on the first displayed line in the answer to Example 21C, change 0.3 dZ (t ) to 0.2 dZ (t )in two places.

[8/19/2011] On page 443, in Example 21D(ii), change S (0) = 10 to S (0) = 40.

[4/1/2012] On page 443, in Example 21D(i), on the displayed line, change the dS (t ) on the right to dt . In theanswer, on both displayed lines, change E to E∗, since these are risk-neutral expected values.

[7/15/2012] On page 449, in exercise 21.11(iii), change S (0) to W (0).

[10/9/2011] On page 454, 5 lines above Quiz 22-1, add dt at the end of the line.

[8/26/2011] On page 454, 2 lines above Quiz 22-1, replace Y at the end with dt and put a dt after the secondsummand, so that the line reads:

dY

Y=

aS

S

�

(α−δ)dt +σdZ (t )�

+0.5a (a −1)σ2dt −�

a (r −δ) +0.5a (a −1)σ2�

dt

[3/21/2012] On page 462, in the solution to exercise 22.8, on the first displayed line, in the exponent of thenumerator of the second fraction, change 0.5(4)(3) to 0.5(3)(2).

[10/9/2012] On page 463, in the solution to exercise 22.10, on the second line, change a to c .

[2/29/2012] On page 464, in the solution to exercise 22.14, on the third displayed line, remove the negativesign in −0.154375.

[10/30/2011]On page 464, in the solution to exercise 22.15, 2 lines from the bottom of the page, change the rin the exponent to α.

[11/9/2011] On page 472, on the 7th displayed line of the page, there should be a dt at the end of the line, sothat the line looks like this:

dX (t ) =�

−λX (0)e −λt +αλe −λt�

dt + e −λt d

�

∫ t

0

σe λs dZ (s )

�

−

�

λe −λt

∫ t

0

σe λs dZ (s )

�

dt

[11/9/2011] On page 486, on the last line of the warning box, change “year n” to “year k ”.

[10/25/2011]On page 492, in exercise 24.20, in the graph, replace “Month 5” with “Month 6”.

[9/27/2011] On page 499, in the solution to exercise 24.20, on the second to last line, replace 0.965523 with0.983570. On the last line, replace 0.965523 with 0.983570 and replace the final answer 0.9681with 0.9758.

[10/22/2011]On page 510, in the solution to exercise 25.3, on the second to last line, put a negative sign in frontof 0.10433: N (−0.10433).

[9/27/2011] On page 515, on the third displayed line of Section 26.1, the one ending with (*), change P (t , t , T )to P (r, t , T ). Also, put a 0.5 factor in front of (T − t )2.

Updated 3/3/2013

Errata for ASM Exam MFE/3F Study Manual (Ninth Edition) Sorted by Page 7

[9/27/2011] On page 516, put a 0.5 factor before the expression under the brace with ¬.

[9/27/2011] On page 516, in the table of the answer to Example 26A, on the second line under the heading,change −2.608047 to −2.608074.

[3/19/2012] On page 522, on the third line from the bottom, change B (t ) to B (T ).

[9/27/2011] On page 530, on the third line of Subsection 26.3.2, change “rats” to “rates”.

[4/25/2012] On page 531, 2 lines above Example 26F, change∂ P

∂ tto∂ P /∂ t

P (r, t , T ).

[8/4/2012] On page 543, in exercise 26.22, on the last displayed line, add “dt” after the large right parenthesis.

[10/26/2011]On page 546, on the last line of exercise 26.38, delete the phrase “, using a negative sign to indicatea sale”, and delete the negative signs in the five choices.

[9/24/2012] On page 547, in the solution to exercise 26.2, on the displayed line, remove the minus sign before1000N e −0.05.

[2/26/2012] On page 549, in the solution to exercise 26.13, on the first line, change B (0, 3) to B (0, T ).

[10/26/2011]On page 555, in the solution to exercise 26.38, on the third displayed line, delete the negative signbefore 1.64065. On the fifth displayed line, put a negative sign before 0.74177. On the next line,delete the negative sign before 1.9247. On the last line, replace “sell” with “buy”.

[5/22/2011] On page 556, in the solution to exercise 26.39, on the second line, the exponent is based on theformula in McDonald, but McDonald’s formula is erroneous; there should be a negative sign be-fore 0.5σφ(52). This has no effect on the solution sinceφ = 0. Since McDonald has not correctedthis error, it is unlikely you would be expected to know the correct formula.

[11/4/2011] On page 613, in question 9(iv), the right side of the equation should be e 0.2952t .

[5/24/2011] On page 629, in question 28, replace the first sentence with

Let S (t ) be the time-t value of a stock index, and Q (t ) the time-t value of an annuity contract.The annuity’s contract value grows at the same rate as the value of the stock index, except that acontinuously compounded 1% management fee is assessed. More precisely,

Q (t +dt ) =Q (t )

�

S (t +dt )−0.01S (t )dt

S (t )

�

[10/26/2011]On page 656, question 18 is defective. In (iii), change 0.07 to 0.05. Divide all the answer choicesby 10, so that they are 0.01, 0.015, 0.02, 0.025, and 0.03.

[11/10/2011]On page 678, the solution to question 15 is incorrect. The correct solution is

The prepaid forward price of currency, as indicated in Table 1.2, is

xt e −r f (T−t )

Here, for dollars in terms of euros, xt = 1/1.50 and dollars are the foreign currency so r f = 0.04.

Also, T − t = 2. We conclude that the prepaid forward price is 100e −0.04(2)/1.50= 61.54 .

[9/27/2011] On page 693, the answer key for question 3 should be (A) Correct the answer key on page 692 aswell.

[10/2/2011] On page 694, the answer key for question 5 should be (D). Correct the answer key on page 692 aswell.

Updated 3/3/2013

8 Errata for ASM Exam MFE/3F Study Manual (Ninth Edition) Sorted by Page

[11/1/2011] On page 704, in the solution to question 6, 3 lines from the bottom of the page, change the 0.09531in the denominator to 0.125.

[10/2/2011] On page 707, the answer key for question 19 should be (B). Correct the answer key on page 703as well.

[11/1/2012] On page 715, the solution to question 11 has multiple typos involving the 45’s and 50’s. The cor-rect solution is

We use put-call parity to calculate the value of a 1-year put.

P (50, 45, 1) =C (50, 45, 1) +45e −0.1−50e −0.1 = 3.00

The value of the chooser option, denoted by V , by formula (??), is (note that (r −δ)(T − t ) = 0)

V =C (50, 45, 4) + e −δ(T−t )P (50, 45, 1)

so we have 12 = C (50, 45, 4) + e −0.3(3.00) from which it follows that C (50, 45, 4) = 12 − 3e −0.3 =9.78 . (E)

As indicated in Subsection 14.4.1, you could derive equation (14.9) using put-call parity; the max-imum of P (S , 45, 3) and C (S , 45, 3) at time 1 is

C (S , 45, 3) +max�

0, P (S , 45, 3)−C (S , 45, 3)�

=C (S , 45, 3) +max�

0, 45e −0.3−S e −0.3�

=C (S , 45, 3) + e −0.3 max�

0, 45−S )

which becomes C (S , 45, 4) + e −0.3P (S , 45, 1)when discounted to time 0.

[4/1/2012] On page 716, in the solution to question 15, on the line below the first 3 displayed lines, change0.0159 to 0.016065. Two lines lower, change $pounds640,000 to £640,000.

[4/1/2012] On page 725, in the solution to question 9, on the first displayed line, change −0.5a (a −1)σ2 to+0.5a (a −1)σ2.

[11/4/2011] On page 727, in the solution to question 15, the column St /St−1 is incorrect, except that 1.0025 iscorrect. The five entries in that column should be 1.0025, 1.0948, 0.9180, 0.9429, 1.0579.

[11/5/2011] On pages 773–774, in the solution to question 15, the payoffs should be discounted at 0.02. Themultiplication by e −0.02 may be postponed to the end, however, since multiplying the five payoffsby e −0.02 will multiply the standard deviation by e −0.02. Therefore, replace the last phrase of thesolution, “then take the square root. . . ” with “then take the square root and multiply by e −0.02 to

get the standard deviation of the call option price, e −0.02p

0.000020272 = 0.0044 . None of thefive answer options are correct.

[10/26/2011]On page 774, correct the solution to question 18 (in accordance with the revised question) asfollows:

• On the fourth displayed line, change 0.07 to 0.05.

• The last two lines should read:

−0.05σ2−0.01+0.02= 0

σ2 = 0.02 (C)

[8/8/2012] On page 782, in the solution to question 5, on the first displayed line, delete “C(50,60,3)+” on theleft hand side.

Updated 3/3/2013

Errata for ASM Exam MFE/3F Study Manual (Ninth Edition) Sorted by Page 9

[11/15/2011]On page 789, on the first line of the solution to question 28, change “put” to “call”.

[8/4/2012] On page 825, in the solution to question 37, 4 lines from the end, change −0.01 to 0.01.

[4/21/2012] On pages 837–838, in the solution to question 69, 8 lines from the bottom of page 837, change76.9 to 76.8 and change 43.1 to 43.2 twice. 4 lines from the bottom of the page, change 43.1 to43.2 and 16.8401 to 16.8692. In Figure B.8 on page 838, change 43.1 (third column, lowest node)to 43.2.

[10/10/2011]On page 838, the answer key for question 69 should be (C) rather than (A).

[3/3/2013] On page 840, in the solution to question 74, on the last two lines, replace the expression 27.65724e −0.08 =25.53 with 27.65724/1.024 = 25.55 .

Updated 3/3/2013