24th Edition (September 2023) ISBN: ISBN 978-1-9911554-6-7

• Answer 4.7, Footnote 3. Skewness of Binomial distribution: Numerator should have parentheses. (q-p)/sqrt(Npq).

23rd Edition (August 2022) ISBN: ISBN 978-1-9911554-4-3

• Myron and Fischer escalator question: It says "Myron's absolute speed up the incline holding the escalator is e + v_M." It should say "Myron's absolute speed up the incline holding the escalator is v_e + v_M." That is, "e" should be "v_e".
• Q1.40 problem statement. "a women deduces" should be "a woman deduces"
• Solution to Q1.40 (matriarchal town). Second to last paragraph. Wrong word: "It follows my mathematical induction" should be "It follows by mathematical induction".
• Solution to Question 4.58 has a typo. It says var(X_i)=\sigma, but it should say var(X_i)=\sigma^2.

20th Edition (October 2019) ISBN: 978-0-9951173-8-9

• Spelling mistakes. Last page of the introduction. The words absoluely and prevously appears where they should be absolutely and previously.

19th Edition (July 2018; Corrected September 2018) ISBN: 978-0-9941386-9-9.

• Spelling mistakes. Last page of the introduction. The words absoluely and prevously appears where they should be absolutely and previously.
• Posted April 21, 2019. There is a typographical error on p195. In the middle of the page, it says "E^∗[S^α(T)|S(T) ≥ K] = S^α(t)e^[α(r+[(α−1)/2]σ^2)(T−t)]N(d_{2−α})", but the LHS of the equation is not stated correctly. The LHS should be E^∗[PAYOFF(S(T))], where PAYOFF(S(T))=S^α(T); if S(T) ≥ K, and 0 otherwise. If you take the expectation as stated in the book, however, the RHS would be the stated answer divided by N(d_2), which is not what we are looking for. The key to understanding this error is that the payoff being considered is not PAYOFF(S(T))=S^α(T)|S(T) ≥ K, but PAYOFF(S(T))=S^α(T) if S(T) ≥ K and 0 otherwise, and that these two payoffs are different. All the formulae around this error are correct, and, with the correction mentioned here, this formula can be used to correctly derive the powered option payoffs, as indicated. If you have my book Basic Black-Scholes, you can see a discussion of the distinction between these sorts of conditional payoffs in Section 8.3.6 Interpretation VI (of Black-Scholes).

• Posted August 8, 2018. The book was revised/corrected late September 2018 to correct for the following error. The corrected version has a footnote on p296 saying that it is a revised and corrected answer. If you purchased the book in July, August, or September of 2018, however, then the answer to Question 4.57 on page 296-297 (i.e., choose p to maximize my payoff given that my opponent will choose p'(p)) is not correct. The first nine lines are correct, down to Equation D.14. After that the answer is simply incorrect and should be ignored.

  In fact, your opponent looks at your expected payoff -20pp'+9p+9p'-4 and sees that it may be written as p'(9-20p)+9p-4. For fixed p, your opponent sees a linear function b.p'+a, where the coefficient is b=(9-20p) and the intercept is a=9p-4.

  If b>0, your opponent wants to minimize b.p'+a and chooses the lowest possible p' value (p'=0). In this case, the expected payoff is a=9p-4. Note, however, that b>0 implies 9-20p>0 implies p<9/20. So, the expected payoff is a=9p-4<81/20-4=$1/20 (So, I get an expected payoff strictly less than $1/20).

  If b<0, your opponent wants to minimize b.p'+a and chooses the highest possible p' value (p'=1). In this case, your expected payoff is 5-11p. Note, however, that b<0 implies 9-20p<0 implies p>9/20. So, the expected payoff is 5-11p<5-99/20=$1/20 (So, you get an expected payoff strictly less than $1/20).

  If, however, b=0 (i.e., p=9/20), then your expected payoff is 9p-4=$1/20 for every p' value, which is superior to all of the above. So, you want to choose p=9/20.

18th Edition (July 2017) ISBN: 978-0-9941386-4-4.

• "Descartes' Rule of Signs" is spelled incorrectly as "Descarte's Rule of Signs" in three places (pp. 128, 130, 302).
• There is a typo on page 229. The equation on the last line should read "E(Nt) = pt · 1 + (1 − pt) · [1+E(Nt)]." The answer is otherwise correct.

17th Edition (November 2016) ISBN: 978-0-9941386-3-7.

• None yet.

16th Edition (September 2015) ISBN: 978-0-9941182-5-7.

• Posted October 7, 2015. Middle of page 97 "k=2,...n" should be "k=1,...,n". Page 98, "tirangle" should be "triangle."
• Posted October 16, 2015. Question 1.69 on p22. An "even number of heads" is what the interviewer wrote, but that is ambiguous (it is not about odd versus even). What he meant was "the same number of heads"
• Posted December 4, 2016. Answer 4.1 p199. First two equations should use x instead of v.
• Posted December 4, 2016. Last line p.226 "to rare" should be "too rare"
• Posted December 4, 2016. p.255 Peijnenburg and Atkinson 2013 is accidentally indented in the references. It is there but easy to miss.
• Posted April 6, 2016. p.20 Q1.59 "three men and one women" should be "three men and one woman"

15th Edition (August 2014) ISBN: 978-0-9941038-6-4.

• Posted January 23, 2015. In the third solution to the 90-coin problem (Q1.18 of the 15th edition, but appearing in earlier editions also), I mistakenly overlooked a case. In the case where A and B do not balance, and I swap the {27}s, it is possible that the scales tilt the opposite way. In this case, {27}B contains the coin known to be heavy or known to be light. Similarly if you go to the next step and swap {9}s.

13th Edition (July 2012) ISBN: 0-9700552-8-5.

• Posted July 24, 2012: Answer 1.6 on page 59 should say 15 plusminus 10 = 5, or 25. These correspond to S=10, or S=50. The case S=10 has a physical interpretation where the 5x10 box touches the circle on the other side of the square, but that is not consistent with the picture.

12th Edition (July 2009) ISBN: 0-9700552-7-7.

• Posted Sept 3, 2009: Question 1.16 about the ants is mis-stated. In the original question the ants can only walk parallel to the edge of the ruler. As such, they can only ever meet head on. With this amendment, the given answer is correct. Thank you to Craig Smith for finding this.
• Posted May 20, 2010: Answer 4.5 on page 191 says that 8/3 is the same as 2 1/3. Obviously 8/3 = 2 2/3.
• Posted November 24, 2010: Question 4.10 on page 37 is not worded quite clearly enough. There are multiple interpretations. My interpretation is that the game could be played repeatedly where the guest reveals a door to be empty.

Last Updated: August 4, 2024