Quantitative Researcher Interview Questions | Glassdoor

# Quantitative Researcher Interview Questions

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Apr 17, 2011

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

Apr 17, 2011
 2) A. 10 ropes, each one has one red end and one blue end. Each time, take out a red and a blue end, make them together. Repeat 10 times. The expectation of the number of loops. B. 10 ropes, no color. All the other remains the same.7 Answers1/10 + 1/9 +...+ 1 ? B is similar..1/19+1/17+etc in BE[n] = 1/n + (n-1)/n*E[n-1] = 1/n + E[n-1] For the case of n=10, you would sum up all of the numbers from 1 to 10: 1/10+1/9+ 1/8 + 1/7 ... + 1/2Show More Responsesadd an extra 1 to the previous answerFor part A), the answer is 1+1/2+1/3+...+1/10. For part B), the answer is 1+1/3+1/5+...+1/19. Explanations: For part A), ctofmtcristo has the right approach but with a typo in the equation for E[n]. To obtain the expected number of loops, we note that the first red has a 1/n chance of connecting with its opposite blue end (and forming a loop) and a (n-1)/n chance of connecting with a different rope's blue end (and not yet forming a loop), so E[n] = 1/n*(1+E[n-1]) + (n-1)/n*E[n-1] = 1/n + E[n-1], with base case E=1. Then, by induction, we get E[n] = 1+1/2+1/3+...+1/n. Part B) is similar. We note that the first end now has 2n-1 possible ends to connect to, of which 1 of them is its opposite end and 2n-2 of them belong to a different rope. Then, E[n] = 1/(2n-1)*(1+E[n-1]) + (2n-2)/(2n-1)*E[n-1] = 1/(2n-1) + E[n-1], with base case E(1)=1. By induction, E[n] = 1+1/3+1/5+...+1/(2n-1).Ed's anwser is not right. Just check for the case of 3 pairs. So total cases is 3!=6. 1 case with 3 loops, 2 cases with all wrongly attached, and 3 cases with 1 loop. so expected value is (3/6)*(1) + (1/6)*(3) = 6/6 = 1... and Eds anwser gives 1+1/2 +1/3 = 11/6, which is wrong clearly.Timi, you are missing the fact that if they are "all wrongly attached" then they form a loop. Similarly, the case you are thinking of "with 1 loop" actually has 2 loops. The correct answer is still 11/6.

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

Dec 21, 2011
 A tosses n+1 coins. B tosses n coins. B wins if he has at least as many heads as A. What is the probability that B wins?6 AnswersSee the other guy's solution1/2 or 0.5Question rephrased: what's the prob that A has more heads? First n throws, they have equal number on the average. So, A gets a chance to have more heads on the last throw, 50% chance. -> P(A wins) = 1 - P(B wins) = 50% -> P(B wins) = 50%Show More Responses^ is a great way of thinking about it.Use symmetry. If both A and B had N coins, we can say that prob(A>B) = a, prob(AI don't think 50% can be right, that's already the chance for when they have an exactly equal amount.

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

Apr 17, 2011
 1) Tow coins, P(head)=1/3, P(tail)=2/3, design a way to get the effect of fair coin5 AnswersI guess Play 2 games , TH or HT = outcome 1, TT = outcome 2 . Both of probability 4/9 disregard HHmanipulate payouts. P(Tails) = 2/3, so if it lands on tails I get \$1. P(Heads) = 1/3, so if it lands on heads you get \$2. 2/3 * 1 = 1/3 * 2We need unbiased decision out of a biased coin. Throw the coin twice. Classify it as "heads" if we get HT and "tails" if we get TH. Disregard the other two occurrences i.e. HH and TT.Show More Responsesit's like you need to give heads another 'chance' (to double it's probability to match tails) if you get a tail, stop if you get a head, roll again and take the second resultSwift and anon are both correct, but Swift's solution is twice as efficient because 8/9 of the time, Swift only requires 2 flips, while 4/9 of the time, anon requires only two flips. Indeed, for Swift, we can show that the expected number of flips is 2.25, while for anon, the expected number of flips is double that, 4.5. Let X be the expected number of flips. Then, for Swift, EX = 2 + 1/9*EX ==> EX = 18/8 = 2.25, while for anon, EX = 2 + 5/9*EX ==> EX = 18/4=4.5.

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

May 22, 2013
 If X, Y and Z are three random variables such that X and Y have a correlation of 0.9, and Y and Z have correlation of 0.8, what are the minimum and maximum correlation that X and Z can have?7 Answers0.9http://www.johndcook.com/blog/2010/06/17/covariance-and-law-of-cosines/0.98 & 0.46Show More Responseshttp://wolfr.am/1i1XT4PHow'd you get 0.98 and 0.46?NND correlation matrix --> det(\Sigma)>=0 --> 0.98 and 0.46minimum: 0.9*0.8+sqrt(1-0.9^2)*sqrt(1-0.8^2) = 0.9815 maximum: 0.9*0.8-sqrt(1-0.9^2)*sqrt(1-0.8^2) = 0.4585

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

Jan 4, 2011
 5 babies in a room, 2 boys and 3 girls. one baby with unknown sex is added. Randomly choose one baby and the result is a boy. What's the prob that the added baby is a boy?4 AnswersBayesianBayes' theorem: P(A|B) = P(B|A)*P(A)/P(B) P(A)=1/2 by assumption; if the newborn is a boy, there is P(B|A)=3/6 chance of selecting a boy. now P(B)=P(B|A)*P(A)+P(B|A')*P(A') as you said, where A' = newborn is a girl. similarly, P(A')=1/2 by assumption, and P(B|A')=2/6. hence the answer is (1/2)*(3/6) / [(1/2)*(3/6)+(1/2)*(2/6)] = 0.6I think there is this typo in the last line, and the answer is (1/2)*(3/6) / [(1/2)*(3/6)+(1/2)*(4/6)] = 3/7.Show More Responses3C1/(3C1 + 2C1) = 60%

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

Feb 18, 2012
 You play a game with someone, the rule is to take turns to put a quarter on a round table ( can be any size, but must be of symmetrical shape). You are trying to cover the table up with quarters, you lose when there's no place for you to put down the quarter. Should you be the first person to play ?3 AnswersNo. You should be the second to play. The symmetry of round table guarantees the space for a 2nd coin for every coin placed.be the first to play with the coin dead center Then the symmetry rules will apply for the the rest of the coinsbe the first to play with the coin dead center Then the symmetry rules will apply for the the rest of the coins

### Quantitative Researcher at TGS Management Company was asked...

Sep 15, 2012
 How to generate a gaussian distribution with rand()2 Answersrand(m,n) produces a m*n matrix, where each element is a random observation from N(0,1).Use either inverse transform or Box-muller, rand(m,n) only gives uniform r.v.

### Quantitative Research Analyst at Numeric Investors LLC was asked...

Aug 4, 2012
 Suppose there is a series of 100 light bulbs labeled 1 through 100 that all start in the off position. Person 1 walks in the room and turns on every light bulb. Person 2 walks in the room and flips the switch on every 2nd light bulb (i.e. he turns off all even numbered light bulbs). Person 3 walks in the room and flips the switch on every 3rd light bulb (turning some on and some off). This process continues until person 100 goes through and flips the switch on the 100th light bulb. Which light bulbs will be lit up at the end?1 AnswerFor a light bulb to be left on, it must be flipped an odd number of times and in order to be flipped an odd number of times, it must have an odd number of factors. The only numbers that have an odd number of factors are the perfect squares so light bulbs 1,4,9,16,25,36,49,64,81,100 will be on.

Apr 9, 2012