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# Quant Interview Questions

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Nov 22, 2010

### Quant Research Intern at Susquehanna International Group (SIG) was asked...

Mar 17, 2013
 Suppose you have two covariance matrices A and B. Is AB also a covariance matrix? Suppose that, by plain dumb luck, we also have that AB=BA. Is AB a covariance matrix under this additional condition?12 AnswersI suppose it's clear from how I wrote the question that the answer to the first question is no (for you: why?). For the second question, this is a little bit harder if you aren't experienced in linear algebra. I actually have a PhD in algebra, and the interviewer also had a PhD in algebra, so on some level this question might have been specifically targeting my background. There is a standard result that applies here; see if you can figure it out.1 is correct, 2 is wrong. Try again. :)Show More ResponsesIf AB=BA then yes it is symmetric. So the question boils down to: Is it in fact positive semi-definite? Why or why not? Work through that and you have your answer. My hint: Take a look at some results related to diagonalizing matrices that commute with each other.Hi for your telephone interview with doug, what was the focus for technical interview parts, everything from math, finance, programming? or mainly probability type of question? thanksI remember it being mostly probability and finance. Nothing too terrible.Hi Mr.interview candidate, what were the types of finance questions Doug asked? Thanks!Hi, Mr. interview. Do you remember what kind of data analysis example was for on-site interview? If so, can you share a little bit? What was the level of difficulty? Thanks!1, no 2, yesIt doesn’t. It provides an alternative way of establishing that the eigenvalues must be nonnegative. In fact, if you read the proof, all that’s established is that AB is similar to a positive semidefinite matrix and therefore must be positive semidefinite. It says nothing about whether or not AB is symmetric (which is also required for AB to be a covariance matrix). One or more comments have been removed. Please see our Community Guidelines or Terms of Service for more information.

### Quant Research Intern at Susquehanna International Group (SIG) was asked...

Mar 17, 2013
 Suppose you have 100 GB of data that you want to sort, but you only have 1 GB of memory. How would you sort this data?8 AnswersHint: This isn't really a difficult question (just was an unexpected one for me). You don't really need to know the answer to figure this out. As it turns out, the obvious thing actually works here (and it is a known sorting algorithm).Can you expand on this? What is sorting algorithm?Sorting algorithm = a computer algorithm to sort a list of objects. Well pretend you just have 2 GB of data (for simplicity, assume they are integers) and 1 GB of memory, since the technique is the same. And pretend you want to sort these integers in increasing order. What would you do? Like, what's the first idea that comes to your mind?Show More ResponsesYou do an on disk merge sort, bring chunks in to memory and sort using quick sort, then had the sorted data in to buckets (files). When your done merge them using a merge sort.Yep, exactly.External sortbucket sort. Sort each bucket, then merge.Mark

### Quant Internship at Jane Street was asked...

May 6, 2011
 You have 100 marbles, 50 are blue, 50 are red. You want to distribute them between two drawers, in such a way that none is left outside and no drawer is left empty. After distributing them you are gonna select a drawer randomly and from that drawer you are gonna remove one marble randomly. How do you distribute the marbles in such a way that the probability of getting a red marble is maximized?6 AnswersPut one red marble in one drawer and all the others in the other drawer.It doesn't matter how you distribute the marbles. The fact that they are split into two drawers doesn't affect the sample space. The probability of choosing either draw is 1/2 (i.e. the same).The first answer is correct. Say you put one red marble in drawer 1, and drawer 2 has 24 red and 25 blue. The chances of choosing drawer 1 and a red marble are .5 or 49/98 since all the marbles are red in drawer 1. The chances of choosing drawer 2 and a red marble are .5*(24/49) so 12/49 or 24/98. The total chances of choosing a red marble in this case are 73/98, much higher than 1/2.Show More ResponsesIn drawer #1: 1 RED 0 BLUE In drawer #2: 49 RED 50 BLUE 50% chance of selecting drawer #1 - where there is a 100% chance of selecting RED = .5 * 1.00 = .5 50% chance of selecting drawer #2 - where there is a 49/99 or .4949% chance of selecting RED = .5 * .4949 = .2474 .5 + .2474 = .7475Stephen is correct. The answer is indeed 1/2*(1/1) + 1/2*(49/99) = 1/2*(148/99) = 74/99 = 74.75%. "P" has the right approach but has assumed the wrong number of marbles, as there should be 100 marbles in total, not 50 :)Here's a really intuitive way to see derive the solution without any actual math. You can obviously achieve 50% odds by putting all the red marbles in one drawer and all the blue marbles in the other. If you were to take a blue marble from the blue drawer and put it into the red drawer, then you slightly decrease the odds overall. Conversely, if you were to take a red marble from the red drawer and put it into the blue drawer, then you slightly *increase* the odds overall, since if you pick the red drawer you still have a 100% chance (at that point) of getting a red marble, and if you pick the blue drawer you now have a nonzero chance of getting a red marble. Repeating this, we arrive at the optimal solution posted by others: one drawer should contain one marble while the other drawer should contain all other marbles.

### Quant at Jane Street was asked...

Apr 1, 2012
 Unfair coin with P(H) = 1/3 and P(T) = 2/3. a) How to make an event with 50% probability? b) Expected number of flips until a realization occurs? c) Can you create a strategy to reduce the number of flips necessary? d) Can you create a strategy to reduce the number of flips necessary for an unfair coin with any bias?7 Answersa) Event 1 = {T,H}, Event 2 = {H,T}, if any other outcome, then re-roll b) The probability of Event 1 or Event 2 occurring is 1/9+1/9=2/9. The expected number of 2-roll "tries" is 9/2. And each "try" consists of two rolls so 9 expected rolls for a realization. c) Event 1 = {T,T}, Event 2 = any other combination. Probability of either event is 4/9. d) Many solutions. Trick is to not discard any rolls. Use strategy from part (a), but if you roll {T,T}, then continue and put {T,T,H,H} in Event 1 and {H,H,T,T} in Event 2.the c) is wrong. probability of event 2 is 5/9 not 4/9mokhlos is right. Discard HH.Show More ResponsesInterview Candidate, P(H) = 1/3 and P(T) = 2/3 so a H-T combo will have a 2/9 prob. For a), the probability of Event 1 or Event 2 occurring is 2/9 + 2/9 = 4/9. Therefore, the expected number of 2-roll "tries" is 9/4. Each try consists of two rolls - so an expected 9/2 rolls until a realization.best answer: TT is counted as case 1 -> p = 4/9 TH or HT is case 2 -> p = 2/9 + 2/9 = 4/9 only discard HH Expected # of flips E = 8/9*2 + 1/9*(2 + E) -> E = 9/4 or 2.25 flipsWhat is the best (minimum number of expected tosses) possible for any strategy ?Using Information Theory, Expected number of coin-tosses required is at least (1/ H(1/3))=1.08

### Quant at Jane Street was asked...

Nov 22, 2010
 suppose you live in middle town, your mom lives in downtown while your girl friend lives in uptown. You take subway to have dinner with either your mom or your girl friend every night. You always the first train that arrives at the station. why do you end up with have dinner with your girl friend 90% of the times? assume the uptown and downtown train run at the same intervals.6 AnswersThis would happen if suppose the downtown train always arrives slightly after the uptown. Suppose they each arrive in 10 minute intervals. Suppose the uptown arrives at 6:00, 6:10, 6:20, etc, and the downtown arrives at 6:01, 6:11, 6:21, etc. Then supposing you arrive at a random time (uniformly distributed), it is highly you arrive before an uptown train rather than before a downtown.And there is one more reason because boys will always prefer to stay with their girl friends than their mum.The downtown train should arrive six minutes after the uptwon train to visit mom 10% of the time.Show More Responsesif the trains arrive once an houri would see the girlfriend 100% of the timesAssuming the interval is X minutes. If uptown train arrive at t, t+X, t+2X...., the downtown train should arrive at t+0.1X, t+X+0.1X,....

### Software Engineer/Quant at TGS Management Company was asked...

Aug 20, 2014
 Write numbers 2^1000 and 5^1000 in a decimal notation, next to each other. How many digits does the newly obtained number have? 8 AnswersI solved this using logarithms, and obviously needed a calculator to get a correct answer. Also, you can get a very good idea what the solution is by playing with smaller exponents, but I still don't know how to strictly prove the claim using a simple high school math.Using logs. 10^(x-1) = 2^1000 or 5^1000. Solve for x, then round down. In general: x=(exponent value)*[ln(base number)/ln(10)] + 1 Therefore 2^1000 will have 303 digits. 5^1000 will have 699 digits.1001 integer between 1000*lg2+1000*lg5 and 1000*lg2+1000*lg5 + 2 so the answer is 1000+1Show More Responsesint(1000*lg2)+1 + int(1000*lg5)+1 = 1001we don’t need a calculator here because lg2 + lg5 = 1The answer is 1001, but to prove it you need to note that 1000log2 is not an integer One or more comments have been removed. Please see our Community Guidelines or Terms of Service for more information.

### Quant at Jane Street was asked...

Dec 21, 2010
 find a subset of {1,2,3,...,30} so that numbers in that subset is coprime to each other. 6 Answers2,3,5,7,11,13,17,19,23,29MMXMV, you forgot about 1: 1 is coprime to all intergers Should be 1,2,3,5,7,11,13,17,19,23,29I believe this question was supposed to be "find a subset such that the sum is the greatest" so the answer would be 1,2,7,11,13,17,19,23,25,27,29Show More Responsessorry, delete 2 add 16woops, take out 16 and 7 and add 28! since 16+7 < 28 One or more comments have been removed. Please see our Community Guidelines or Terms of Service for more information.

### Quant at Jane Street was asked...

Nov 22, 2010
 suppose you, player A, player B are in a tennis tournament. Player A has a better chance of beating you than player B does. If you would like to maximize your chance of winning two games in a row, which sequence would you prefer, ABA or BAB?4 AnswersABA.If you want to win 2 games, you must win the middle game. So you should play the weaker player in the middle: ABA.P(Beating A)=x, P(Beating B)=y. Winning two in a row (including winning all three games)= 2xy-x^2y for ABA Winning two in a row (including winning all three games)= 2xy-y^2x for BABShow More ResponsesBAB win 1，2; win 2,3; win 1,2,3 calculate the probability

### Quant at Renaissance Technologies LLC was asked...

Jan 10, 2012
 how fast can you find the max in a sorted list of numbers4 AnswersO(1). The answer is simply max(List(0), List(-1)) in python notation. Because it is a sorted list, the maximum has to be either the first or last element.O(1) is correct if we assume a random access model where we can access each address in memory in constant time.The answer isn't as simple as others are saying. Notice that the question refers to a list of numbers, not an array. Finding the max of a list depends on whether the it is sorted in increasing or decreasing order. In the case it is decreasing it takes O(1) time; just get the first element. Otherwise you know the max is the last element in the list. Getting to the last element requires iterating over each element. O(n) time.Show More ResponsesI think there are a couple things that you don't know here. First, even if the list is ordered, you don't know the metric. It could go as the square, in which case you would be O(n). Since this is unlikely, the other thing that could be amiss is where the list really starts. Just because order has been forced upon a set of numbers, that doesn't mean you know where the minimum is. 1, 2, 3, 4, 5 and 3, 4, 5, 1, 2 are also ordered if you don't mind looping around. In that case I think you're at O(log(n)). Finally, if you know that the list starts with the max or min, then you get O(1).
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