Quantitative research analyst Interview Questions | Glassdoor

# Quantitative research analyst Interview Questions

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

Nov 11, 2011
 What is the probability of a Brownian motion hit 1 before hitting -2. The Brownian motion starts at 0.7 Answers2/3Why ?Let p be the required probability If the first move is to the right(with prob 0.5), it hits 1 If the first move is to the left, then we have to find the prob of hitting 1 before hitting 2, starting from -1, which is equal to (1-p). So p = 0.5 + 0.5*(1-p) p=2/3Show More Responseswhat is browian motion??A martingale approach is easier and more intuitive. Standard Brownian motion is martingale, and with stopping time it is still a martingale. Given a stopping time tao, when it first hits 1 or -2, E[Btao] = B0 = 0 = p*1 + (1-p)*(-2). Therefore, 2/3.let desired probability be p if the first step is 1, then p(-2 reached after 1) = 1 if the first step is -1, then if second stop is -1 , we get to -2 without getting to 1 if the first step is -1, then if second step is 1, we get back to origin hence: p = 0.5*1 + 0.5*0.5*p + 0.5*0.5*0 p = 2/3a stopped martingale is still a martingale. Ans: 2/3

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

Oct 24, 2014
 Interesting question: From a deck of 52 cards pick 26 at random. From this set of 26 you pick two cards. You win if the both of these cards are of the same color. Is this a game you would prefer over one in which you win by picking two (first two picks) of the same color at random from a deck of 26 with equal number of black and red cards8 Answersrandom is betterRandom is not better, both give equal win rates according to simulation.No difference, you can think of the first 26 cards in the shuffled deck as the randomly selected 26 cards, and then you pick the first two. So the winning probability will be exactly the same.Show More ResponsesFirst option is slightly better. One way to argue if the 26 random cards are even, then its the same as the 2nd situation, but if its uneven (12R 14B), then probability is 12/26 * 11/25 + 14/26 * 13/25 > the probability of the 2nd option. And the probability just gets better as the draw is more skewed. Another way to argue is the 1st option is picking from a deck of 52 even cards, and 2nd option is picking from 26. First option probability is 25/51, second option is 12/25. As u choose from more and more cards, the probability increases and tends towards 1/2.Intuitive Solution. 26 is an arbitrary selection. For a two card case: Case 2: deck of two cards. 1 black and 1 white card gives P = 0. Case 1: pick two cards from 52. P > 0.Random is better. Here is the solution: For latter game(fixed 26 cards with equal red and black), the probability to win is: p2=1-13*13/C(2,26)=0.48. (1-probability of picking two different color cards) For the random game, although the expectations of number of red card and black card are equal, but they may not be the exactly same. Assume R is the number of red cards, and B is the number of black cards. with constraints: R+B=26 then the probability to win for this game becomes: p1=1-R*B/C(2,26) With the constraints R+B=26, then R*B = p2 always.fix is better, since in random case more colors are mixed in. The prob of hitting same color pair got lowered. Keep in mind a deck of cards is composed of 4 colors and each of 13 cards. So that 12+14 is not happening.oops I got this wrong. random is better. 12+14 is happening. We are talking about colors...

Apr 30, 2012

Dec 21, 2011

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

Feb 22, 2011
 From (0,0,0) to (3,3,3) in 3D space how many paths are there if we move only right, forward or up?7 Answers1449! / (3! 3! 3!)Can you explain that please?Show More ResponsesA dynamic programming solution is applicable here. Considering each decision hop (a point where a decision is made whether to move right/up/forward) is placed unit length apart, the number of decision hops one might encounter in moving from (0,0) to (3,3) is 4 , making it a length=5 path.Also, there are 3 directions of movement corresponding to the 3 dimensions. A 2-Dimensional *dynamic programming based Forward Table* of length 5 along the X-axis (5 steps from (0,0) to (3,3)) and length 3 along the Y-axis (3-directions of movement), can be used to solve the problem.here is a MUCH easier way to think of this, no dynamic programming required: Consider a much simpler problem - you have 2 white books and 2 black books and want to arrange them on a shelf. How many ways can you do this? Basic permutation theory tells us that when you have m distinct groups of n items, the number of ways they can be arranged is given by: n! / (i_1)!(i_2)!...(i_m)! where i_x is the number of items in group x. So, the answer to this problem is 4! / 2!2! = 6. Now, let's return to the original problem: you can ONLY move forward, right, or up. This means you MUST get closer to (3,3,3) each move. It doesn't take a rocket scientist to figure out that this will ALWAYS require 3+3+3 = 9 moves. Now, with these 9 moves, you MUST do 3 moves of each type. Now, using the formula given above you get 9! / 3!3!3!If this was 2 dimensional, ie a square instead of a cube, the answer would be 2. The coordinate of the far corner can be at (1,1), (3,3) or (1000, 1000), there are only two distinct paths to get from (0,0) to the far corner, if we can only move up and right. Same deal for the cube. There are a total of 8 distinct paths.We know there has to be 3 up, 3 forward and 3 right to arrive at the final destination, the only difference is how many combinations are there for up, forward and right. So we choose 3 up from 9 steps, that is C{_9}{^3}, then we choose 3 forward from the following 6 steps, which is C{_6}{^3}, lastly the three steps have to be right, which is C{_3}{^3}. So the total is 1680 paths.

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

Aug 28, 2013
 You call the home of a family w/ two children and a kid "billy" answers the phone. What is the probability that both children are boys? What is a virtual constructor in C++?7 AnswersBasic Bayesian question Pr(#boys = 2) = (1 / 3) No such thing.We are asked what is the Probability that 2 children are Boys, given that the one of them is a Boy? So given that 1 is a Boy, the probablity of the 2nd one being a Boy is 50/50. So the answer is the Prob (both of kids being Boys| given that one is a Boy) = .5The question is slightly ambiguous and perhaps was phrased in this manner to induce further discussion. See http://en.wikipedia.org/wiki/Boy_or_Girl_paradox. I believe further clarification would be required before an answer of 1/3 or 1/2 could be givenShow More Responses1. 1/2 since it is just a probability of the second one=boy (well, statistics/biology says slightly higher than 1/2 actually) 2. No virtual constructor, but possibly was meant virtual copy constructor.It's 1 in 3. Of the possible combinations of two children, BB, BG, GB and GG, we have eliminated one possibility (GG). Of the three remaining possibilities, one of them is two boys. So the probability is 1/3.It depends on the prior..25 they try to trick you with the name being predominately a boy name, you have no confirmed Billy's gender in the riddle which is why they refer to Billy as "a kid" not as "a boy" hence all probabilities still remain the same .25

Feb 22, 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]=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.