Research Interview Questions in New York City, NY | Glassdoor

# Research Interview Questions in New York City, NY

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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...

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

Dec 21, 2011

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

Feb 7, 2011
 V~U[0,1]. Make a bet B, for V, and will be awarded 1,5 times V if B>V, otherwise your bet is worthless - you do not lose anything as well. What is your bet?5 Answers00.75I think you mean to say "if BV, bet 1 and get 1.5. If payout = 1.5B when BShow More Responsesprofit=1.5*V-B (if VB). prob (VB)=1-B Expected value of V=0.5 E(profit)=(1.5*E(V)-B)*B+0*(1-B)=0.75*B-B*B To search for max(E(profit)): d(E(profit))/dB=0 => 2B=0.75 => final result is B=0.375, and E(profit)=0.14my previous post was wrong, since E(V)=b/2. so, actually, after you calculate it, B=0.

### 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 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?6 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%TEstUsing Bayes' Theorem with: Scenario A1 - added baby is a boy --> P(A1) = 1/2 Scenario A2 - added baby is a girl --> P(A2) = 1/2 A1, A2 for a complete set of scenarios P(B|A1) = 3/6 P(B|A2) = 2/6 P(A1|B) = [P(B|A1)*P(A1)] / [P(B|A1)*P(A1) + P(B|A2)*P(A2)] = [3/6 * 1/2] / [3/6*1/2 + 2/6*1/2]= 3/5 = 0.6

### 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

### Research Analyst at Acuris Global was asked...

Sep 18, 2012
 If i give you \$1,000, and I ask you to invest it for me, what are two questions you should ask me?3 Answers1) Timeframe, in how long do you want your return? 2) How risk averse are you?Exactly the answer that I expected. Thanks mate1) What's your expected rate of return? 2) How long do you want to invest?

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

Mar 20, 2012
 Write a function to append two lists in OCaml.2 Answerslet rec append_list l1 l2 = match l1 with [] -> l2 | h :: t -> h :: (append_list t l2) Usage: # append_list [1;2;3] [4;5] - : int list = [1;2;3;4;5]L1 @ L2
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