Quantitative researcher Interview Questions | Glassdoor.co.uk

# Quantitative researcher Interview Questions

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### 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 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 Researcher at Maven Securities was asked...

16 Jul 2020
 Why did you leave the job before your last job.2 AnswersDo you want me to get your CV across to the CEO? I know him personally and we are on good terms.Lol, salty are we? They do not need to test you for 'good' fit considering the review you wrote. Doubt most people would work with you. The level of arrogance because you have a MSc LOL. Most PhDs fail prop interviews.

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

16 Oct 2014
 Assume that you have an NxN Matrix, filled with decreasing numbers if you go from left to right and from top to bottom. Can you design an efficient search algorithm on this matrix. Hint: the most efficient one should have a Log(N) complexity...2 AnswersI would do something similar to quicksort; in this case matrix seems to be sorted. I would pick N/2,N/2 as my pivot point and compare the target with element at N/2,N/2. On the left and top of this matrix (ie interval 0 to N/2 -1 for both row and column indices) are numbers greater than element on current location. I would keep narrowing down my data set to 1/4 (by halving both row and column sets) until I get to the target. This has O(log n).You start at the middle, then if it's greater you check (n, n/2) and (n/2, n), if less check (1, n/2) and (n/2, 1), which will tell you which quadrant it is in. Since n is cut down by 1/2 each time in a constant number of steps the algorithm is O(log n).

17 Jul 2013

### Associate - Quantitative Research at J.P. Morgan was asked...

5 Oct 2015
 given a series of stock prices, describe an algorithm to find the best buy and sell point. ( only one buy and one sell, buy before sell)2 Answersstraight forward O(n) through memorization.This is similar to the maximum sub-array algorithm questions. Can look it up.

### Quantitative Researcher at GSA Capital was asked...

27 Apr 2018
 I met a coule of quants with average MSc degrees from average schools, sub-par work experiences but with egos of Nobel Prize winners. I was very surprised to see the mediocrity of these two quants because I had read that GSA had only recruited 25 profiles out of 10000 in the last few years so I was expecting people working there to be Quants from the top 10 schools with Quantitative PhDs (which is my case). The recruiter had mentioned to me prior the interview that GSA looked as much at personality fit than skills so I was really on my best behaviour and within the Quant community I am usually known to be pleasant and easy to work with. However, both of these interviewers had severe personal issues. The first one wouldn't tell me about his background because "he was the one asking the questions" and the second thought that a multi linear regression (MLR) was the best Machine Learning Techniques and would scowl at me when I would confront him with the fact that the assumptions behind an MLR around returns being i.i.d were violated by the observed data and that as a result taking a Bayesian approach had more potential in my opinion. As a result of giving my honest opinion and seing through his facial expression the kind of tantrum anger you see on pre-adolescent children, he then went on a rampage trying to make me fail in his next few questions which I answered correctly but instead of moving on to the next ones quickly, he tried to make me fail with stupid details instead of help me show my best through asking additional conceptual questions and bringing the conversation to an interesting level of abstraction in which interesting trading ideas could emerge. I lost interest in GSA as a result of these interviews (GSA was amongst my top 10 preferred places to work for before that). It's a shame. I don't understand why management decided to put these two clowns as the face of GSA. They really give a poor image of the company. Currently, my best offers are with Goldman Sachs and the Man Group. It would have been good to be able to compare an offer with GSA.2 AnswersSounds like your the one with ego issues my friend.I can't believe people like you exist.

### Associate - Quantitative Research at J.P. Morgan was asked...

5 Oct 2015
 throwing a dice until the face 6 comes up two times, what is the expected stopping time3 Answersnegative binomial with success probability p = 1/6, expected value is 12With a=2, p = 1/6 (and q = 5/6); the expected value is 10. One or more comments have been removed. Please see our Community Guidelines or Terms of Service for more information.

### Quantitative Researcher at G-Research was asked...

2 Aug 2016
 If I break a stick of unit length into three random pieces, what’s the expected length of the largest piece?3 Answersdepends how you break: break a random piece first with mean 0.5 and then another piece from what remains (again with mean=0.5) or pick two randoms both with mean 0.5 and break simultaneously. first gives ~0.66, second ~0.61Previously I answered the question by writing a small MatLab script; clearly, not an option for interview. After revising geometrical probabilities (and closed-form anti-derivatives) I finally managed to get a proper solution for both options: (a) 11/18 for the 2nd case (2 simultaneous independent break points); this is a classical problem, which was most likely the intended one (solution can be found on e.g. math.stackexchange). (b) 1st case (break the rod in a first random point, then use a random point on the remaining piece to break it) can be solved in a similar manner but involves heavy integrals (x ln x, x^2/(1-x) etc). After lots of algebra I got a solution of 2ln2 - ln3 + 3/8. Concept is the same, just heavy calculations, probably not a good option for an interview. After doing that, my advice would be to definitely practice solving problems that involve 2 random variables (x,y) on a square (0,1)x(0,1). If you know what to do, it becomes purely algebraic. A much more simple example of such problems - 2 people agree to meet each other between noon and 1pm. each comes at a random moment and waits for 20 minutes before leaving. What is the probability of them meeting each other? One or more comments have been removed. Please see our Community Guidelines or Terms of Service for more information.

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

11 Jan 2021
 If I throw two regular dice and tell you one of them is a six, what is the probability that I got two sixes?3 AnswersOne six is already fixed so the question is equivalent to 'what is the probability of getting one six?'. 1/6There are 11 cases where at least one dice is six, but there's only 1 case where both dice are six. Therefore, the answer is 1/11. One or more comments have been removed. Please see our Community Guidelines or Terms of Service for more information.
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