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# Quantitative analyst Interview Questions

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### Quantitative Analyst at Goldman Sachs was asked...

13 May 2013
 You have 10 mice and 1000 bottles of wine. You also have 24 hours before a party, and one of the bottles has been tainted with a slow acting poison, which takes 24 hours to kill a mouse. In the 24 hours you have remaining, how many bottles can you guarantee safe for human consumption (assume humans and mice react identically)? Assume the lethal dosage is insignificant relative to the size of the bottle.9 AnswersI'll say 500. Since the dosage is insignificant, I'll divide the number of bottles in half, take samples from each of the first 500 bottles, mix them up, divide by 10 and feed to each mouse. If no mouse dies after 24 hours, then the first batch is safe. else, the second batch would be served.999. It is like a binary problem. First mouse tests the first #1-500 (mixed). Second tests #1-250 and #501-750. Third one tests #1-125, #251-375, #501-625, #751-875, and so on. 10 mice with 2 status each (death/alive) could encode number of bottles up to 2^10=1024. So 10 mice is enough to find out the single bottle that tainted.http://advanceddiscrete.wikispaces.com/Mice+and+WineShow more responsesChapter 7.2, P.183 in Zhou's bookAnswer is 900 10 mices vs 1000 bottles 1 mice per 100 bottles 24 hiurs one dies 9 alive 100*9=900The answer is 900. You have 1000 bottles divided between 10 mice. as dosage doesn't matter, 100 bottles can contribute to a single dosage, in which case, one mouse will die meaning the tainted batch needs to be discarded and 900 are confirmed untainted.The answer is 1023. You need to think in bit-wise way. 1023 can be represented in binary as (1111111111). Your goal should be: representing each wine label (i-th number) to each binary representation. 1000th wine will be represented with 1111101000 meaning (1,2,3,4,5,7th) mices will be used to check the toxicity of this wine. In binary way, you can assign label to up to 1023 wines. So by analysing the rats that die after 24 hrs, you can actually identify which wine is toxic or not. hope this helps.Brian was quite close, but to represent 1024 wines, you actually need 11 mice. So the maximum # of wines that one can guarantee is up to 1023. One or more comments have been removed. Please see our Community Guidelines or Terms of Service for more information.

### Quantitative Analyst at Goldman Sachs was asked...

21 Mar 2012
 a, b, c are integers. Such that a^2 + 2bc = 1; b^2 + 2ac=2012, find all the possibles values of c^2 + 2ab. 7 Answers2011 is primeMore details in addition to the last post. 1) subtracting the 1st equation from the 2nd one gives: (b-a)*(b+a-2c)＝2011 2) 2011 is prime, so there are only two possibilities: b-a=1, b+a-2c=2011; or b-a=2011, b+a-2c=1 3) subtracting the 1st equation from the 3rd equation gives: (c-a)(c+a-2b)=x-1 4) plugging c-a=0.5*[(b-a)-(b+a-2c)], c+a-2b=(c-a)-2(b-a) into 3) gives you xIn 2) there should be four rather than two possibilities. Both terms could be negative.Show more responses2 cand: heavy one... I have no idea how did you figure out that 2011 is prime! Do you remember the table of prime numbers from 2 to 10000?? I easily found trivial solution b=0 a=+-1, and then got stucked. however, later I found that this is unique solution, and (3)=c^2= 1006^2. I've used your approach. You've got a 4 possibilities, +-1 +-2011 and +-2011 +-1, but 2 of them lead to complex roots. You see, sum (1)+(2)+(3) gives you (a+b+c)^2=1+2012+x, thus x>=-2013, so (b-a) may be only +-1, not +-2011. The answer is (3)=1012036In the equation a^2 + 2bc = 1, obviously, a is either 1 or -1 and either b or c is zero. Then if you derive b from the second equation, b = sqrt(2012 - 2ac), c can't be zero, because sqrt(2012) is not an integer. Therefore b = 0. So, in order for b to be 0, 2012 - 2ac should be zero or 2ac = 2012. => it's either a = -1 and c = -606 or a =1 and c = 606. In any case, because b is zero, 2bc term in the 3rd equation disappears and c^2 gives an answer 367236.To the comment above, 2ab term in the 3rd equation, of coursesimple eq: (a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca. from given eq, we know: (a+b+c)^2 = (2012 + 1 + UNKNOWN) = K^2 with K being anything larger than 45. 45*45 = 2025 46*46 = 2116 .. so you can figure out the missing term by: (K^2 - 2013)

### Quantitative Analyst at Morgan Stanley was asked...

4 Nov 2011
 You have a deck of 52 cards, and you keep taking pairs of cards out of the deck. if a pair of cards are both red, then you win that pair; if a pair of cards are both black, then I win that pair; if a pair of cards has one red and one black, then it's discarded. If, after going through the whole deck, you have more pairs than I do, then you win \$1, and if I have more pairs than you do, I win \$1. What is the value of this game in the long run?9 Answers00 lol. nice oneby symmetry, probability of losing = probability of winning therefore, -1p(lose)+1p(win)=0. done, ko, finished, fiendin'Show more responsesNo one bothered to ask if you're paying the loser -- e.g. when you win, you get \$1, but when you lose, are you losing \$1? It's not clear from the question above, though the interviewer may have been more specific.good point J.D It can either be 0 or 0.5 depending on how you interpret the questionI think you will always have the same number of pairsWhy is it 0 for us dumb people plsNo value because same probability to win and to lose One or more comments have been removed. Please see our Community Guidelines or Terms of Service for more information.

### Quantitative Analyst at Credit Suisse was asked...

3 Nov 2013
 A = Sqrt(E(Variance)) B = E(Sqrt(Variance)) Is A larger or B larger? Please explain. 5 AnswersA is larger than B. Maybe use Jensen's inequality and sqrt() is a concave functionA = B = sqrt(Variance). Variance is not a random variable, it is a constant. Then we get E[Variance] = Variance and E[ sqrt(Variance) ] = sqrt(Variance) .Francesco, have you ever heard of stochastic volatility models ? Do you still think that "Variance is not a random variable, it is a constant." ? I agree with Jolene using Jensen inequality on a concave function implies that sqrt(E(...)) > E(sqrt(...))Show more responsesIt depends on the variance structure. If it is constant, price of an option is a convex function of volatility and we know from Jensen's B>A. However, if the process is stochastic, we need to check volga, derivatives of vega to vola, which is vega * d1*d2 / vola. Since vega and vola are positive for an option holder, d1*d2 sets the direction. Theoretically, we can find conditions, where the option is close to ATM, that make vega * d1*d2 / vola < 0 . Thus, the function is not always convex.Let's call sqrt(Variance) -> X: And then A = sqrt(E(X^2)) B= E(X) V(X) >= 0 V(X) = A^2 - B^2 (definition) Which implies A >= B.

6 May 2017
 Round 2: If you have balls weighting from 1, 2, ... 40g, and you have a fair balance, how many of their weights you have to know in advance such that you can measure all the rest?5 AnswersIf you can weigh 1,...,n-1 for n > 2, then you can weigh n by using 1 and n-1 (this does not hold for weighing 2). So you just need 1 and 2.Not sure I fully follow the question, can't you just get everything in terms of the lightest, so just weigh the lightest to get your unit? e.g. number the weights w1, w2, ... ,w40 in increasing order (we can determine this without external knowledge, right?). Then we have w1 + w2 = w3 w1 + w3 = w4 w1 + w4 = w5 and also stuff like w2 + w3 = w5 substitute come stuff in and you find that w2 = 2w1, hence wn = nw1 So just 1?Show more responsesNot sure I fully follow the question, can't you just get everything in terms of the lightest, so just weigh the lightest to get your unit? e.g. number the weights w1, w2, ... ,w40 in increasing order (we can determine this without external knowledge, right?). Then we have w1 + w2 = w3 w1 + w3 = w4 w1 + w4 = w5 and also stuff like w2 + w3 = w5 substitute come stuff in and you find that w2 = 2w1, hence wn = nw1 So just 1?I think no prior knowledge of the weights needed. it is kind of like a sorting algorithm, where you use a fair balance to compare any two balls so that you can rank all the 40 balls by their weights. Once you rank them, you will know the weight for each ball since you already know there are 1g, 2g ... 40g. One or more comments have been removed. Please see our Community Guidelines or Terms of Service for more information.

19 Oct 2011

### Quantitative Analyst at Morgan Stanley was asked...

16 Nov 2013
 What is the price of a contract that at time T=2 pays out S_2 / S_1. Interest rate is continuous and equal to r?3 Answersexp(-r-q)A typical change of numaraire. Change measure into that of S2 and find the new drift of S1 under it, then get the expectation of a GeoBrownian. A more common method is to decompose the two brownian motions W1, W2 of stocks S1, S2 to W1 and W2 = pW1 + sqrt(1-p^2)W, and the distribution of S1/S2 can be represented by a single brownian motion, then the coefficients of s1/s2 can be found, then similar integral bounds yield the answer in terms of CDF of Standard Normal and some coefficient.In fact you don't need any CDF here because you are not asked to price any form of option, just a forward and therefore you are just interested in computing an expectation. There are 2 ways to do this: - If you assume a single brownian motion for both stocks then using S2 as a numeraire is straightforward and will give you a pretty simple result since S1/S2 is a martingale under S2 and you know the radon-Nikodym density for dP(S2)/dQ as well as how to write a brownian motion under S2 as a brownian motion under Q - If you assume 2 different brownian motions (correlation with rho) then you can not directly use the usual numeraire change result but you can still derive an ito calculus to write the risk-neutral dynamics of 1/S1, S2 and then S1/S2. When it is done will will notice that you get a lognormal distribution for S1/S2 and you can write the expectation as the exponential of something with sigma1^2, sigma2^2 and roh*sigma1*sigma2

### Quantitative Researcher at Jane Street was asked...

6 Oct 2017
 First round: basic probability, combinatorics. A bear wants to catch 3 fish from a river. when he has caught 3 fish, he'll leave. when a fish comes, there is a 1/2 chance he'll catch it. what's the probability that the 5th fish will not be caught?4 AnswersP: The 5th fish has been caught, P=P1+P2+P2, where, P1: The 5th fish been caught as the first one P2: ... second one... P3: ....third one (or the last one). P1 = 0.5^5 P2 = 4*0.5^5 P3 = 6*0.5^5 The fifth fish will not be caught Pc = 1-P = 1-P1-P2-P3I got a different answer :/ There's 2 ways the fish can survive: either the bear leaves before the 5th fish, or the bear fails to catch it. The probability the bear leaves is the probability the bear gets 3 fish in 4 attempts, or 4C3/16 = 1/4 The probability the bear tries * fails to catch the fish is 3/4 (the probability of trying) * 1/2 (the probability of failing) = 3/8. The total probability is thus 5/8? Not sure what I did differentlyShow more responsesLet P(C) be the probability that the fish is caught, and P(F) be probability the bear is fishing. We know that P(!C|F) is 1/2 since the bear is fishing, and P(!C|!F) = 1 since the bear is not fishing. The law of total probability lets us include this information: P(!C) = P(!C|F)*P(F)+P(!C|!F)*P(!F) = 1-P(F)/2 The bear fishes until 3 are caught so P(F) is the bear has caught 0, 1, or 2 of the first four fish. The binomial distribution gives P(F) = (1/2)^4*(4 choose 0 + 4 choose 1 + 4 choose 2) = 11/16. Now we can calculate P(!C) = 1-(11/16)/2 = 21/32. One or more comments have been removed. Please see our Community Guidelines or Terms of Service for more information.

### Quantitative Analyst at Willis Towers Watson was asked...

30 Jun 2016
 If a car takes one lap with 20 k/h, how fast should it go the second lap to get the average of the first and second laps equal to 25 k/h?3 AnswersI could not answer it.Idk, maybe I'm also wrong but it looks extremely simple. If the first lap is 20 km/h, then to get an average of 25 for two laps the second one should be 30 km/h. (20+30)/2 = 25.There is another option. By definition, average speed is the total length of the route taken divided by the total time it took to cover that distance. Hence in this case it'll be V_avg = (V1*t1 + V2*t2)/(t1 + t2), where t1 and t2 are the times taken to cover the laps and V1*t1 and V2*t2 are the distances. Obviously, as it's the same lap the distances are equal: V1*t1 = V2*t2 = S, and t1 = S/V1, T2 = S/V2 Thus, equation for V_avg becomes V_avg = 2S/(S/V1 + S/V2). S is canceled and V_avg = 2/(1/V1 + 1/V2) = 2V1*V2/(V1 + V2). Solve for V2: V2 = V_avg*V1/(2V1 - Vavg) = 33.3 km/h