Quantitative Analyst Interview Questions in New York City, NY | Glassdoor

# Quantitative Analyst Interview Questions in New York City, NY

289

Quantitative analyst interview questions shared by candidates

## Top Interview Questions

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

Jan 21, 2014
 What's the best unbiased estimator for a series random variables?3 AnswersI guess it is just a Gaussian distribution (Normal dist.). Since it has the smallest uncertainty (from quantum point of view) or variance.I guess it is just a Gaussian distribution (Normal dist.), since it has the smallest uncertainty (from quantum point of view) or variance.It is the OLS estimator (with Gauss-Markov approximations and normality), by Fisher's theorem on Maximum Likelihood Estimators.

May 14, 2011

### Quantitative Analyst at Citi was asked...

Oct 4, 2010
 how many combinations that have at least one five from zero to 10008 Answersit's 271xyz x=5, y(0..9) z(0..9) => 10x10 = 100 y=5 x(0..4,6..9) z(0..9) => 9x10 = 90 z=5 x(0..4,6..9) y(0..4,6..9) => 9x9 =81 100+90+81 = 27110! - 9!Show More Responsestypo...10^3-9^3Isn't it 111 since he said combination and not permutation? Don't 252 and 225 and 522 count as one?271.295. 1 digit: 1 2 digits: 10+9=19 3 digits: 100+90+81=271 total: 295271

### 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?6 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.

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

Feb 1, 2011
 2. In an urn you have red and blue balls (same number of balls in each of the two colors). You extract 3 balls what is the probability that the number of balls you have extracted is odd. Now you extract 10 balls, what is the probability that the number of balls you have extracted is odd? 6 Answerswhat? 3 is an odd number. so the probability you have drawn an odd number of balls after drawing 3 balls is 1 similarly, 10 is an not an odd number. so the the probability you have drawn an odd number of balls after drawing 10 balls is 0I'm guessing that this question was written incorrectly and it actually should be "odd number of blue balls" or whatever. In which case, the answer, I believe, is always 1/2. For 3, we can work this out P(odd) = P(exactly 1) + P(exactly 3) = 3C1 * (1/2)^3 + 3C3 * (1/2)^3 = 3/8 + 1/8 = 1/2 working this out for 10 is exactly the same. P(odd) = P(1) + P(3) + P(5) + P(7) + P(9) = 10C1/1024 + 10C3/1024 + 10C5 / 1024 +10C7/1024 + 10C9 / 1024 = 10/1024 + 120/1024 + 252/1024 + 120/1024 + 10/1024 = 512/1024it is always one halfShow More ResponsesFor ten balls situation, the sample space is BBBBBBBBBB BBBBBBBBBR BBBBBBBBRR BBBBBBBRRR .... BBRRRRRRRR BRRRRRRRRR RRRRRRRRRR There are 11 samples, and 5 of which give you odd number of blue balss. So it is 5/11The question is equivalent to (1/2 + 1/2 x)^3, what are the sum of the coefficients at has x^n, where n is odd. Plug x = 1, get 1; plug x = -1; get 0. So sum is (1-0)/2 = 2. For (1/2+1/2x)^10, it's the same.For n = 3, answer is 1/2 by symmetry. For n =10, the original question doesn't assume choose with replacement. If choose with replacement, then it is 1/2. Otherwise, it is slightly smaller than 1/2, depending on the total number of red and blue balls.

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

Dec 19, 2010
 The price of a stock is \$10 now. It has .6 prob. increasing to 12 and .4 prob. going down to 8. Interest rate is 0. What's the value of a call option with strike \$10?8 AnswersFirst compute risk neutral prob.10.40Calculate first the risk neutral probability. (I.e you are assuming the current stock price is efficiently priced in) With the stock price tree, risk neutral probability Pi: Its \$10 = pi * \$12 + (1-pi)*\$8 Solving for Pi = 0.5 Then to calculate the value of call: Substitute Pi with the payoffs (\$2 = \$12 - 10 and \$0 as option is worthless at <\$10) C(0) = (pi*(\$2) + (1pi)*\$0 )/ r = 0.5 * 2 = \$1 (Since r = 1, 0 interest rate) A normal expected value of the option (non risk neutral) will just be 0.6*\$2 + (0.4)*\$0 = \$1.2.Show More Responses\$1Expectation under any measure is equivalent. That's why we change measures to calculate things as the result will be same. Answer is 1It's easy to get an answer 1.2, but it's wrong. Answer is 1.Can someone explain why the answer is 1 and not 1.2:?Jim, because you get 1.2 with physical probabilities, and 1 with risk neutral probabilities. To use the binomial tree pricing approach you should use the risk neutral probabilities. Th calculations are provided above.

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

Feb 22, 2011
 66 handshakes occur at a party how many people at a party if everyone shakes hands with everyone else.9 AnswersAns 12; 12 *11 =132 132/2 = 66Solve: 'n'choose'2' =66 n=12yuuurp... http://brainteaserbible.com/interview-brainteaser-66-handshakes-at-the-partyShow More Responses11another interesting way of thinking about this one is as a complete graph (where each person is a node, each undirected egde is a handshake). A complete graph has n*(n-1) edges where n is the number of nodes12It has to be 11simple sum formula: 1+2+3+…+(n-2)+(n-1)+n = n(n+1)/212

### 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 Analyst at D. E. Shaw & Co. - Investment Firm was asked...

Feb 12, 2012
 If 3 of my friends have phone numbers ending (last four digits) in some permutation of 0,1,4, and 9 and I have 150 friends, then is that just a chance occurance?7 AnswersChances of only 3 in 150 having some permutation of 0, 1, 4, 9 is quite low...many more would be expected to have such a permutation in a sample size of 150.My guess: There are 4! = 24 ways in which we can pick a 4 digit number that is a permutation of 0,1,4,9. There are 10^4 = 10000 ways in which we can pick 4 digit number(each of the digits can be between 0 to 9). Hence probability of getting a permutation of 0,1,4,9 is 24/10000 = 3/1250 < 3/150 So having 3 in 150 with the given permutation seems to be more of a chance occurrence. is that correct?As G wrote p=3/1250, q=1-p. We have Bernoulli random variables. 3 friends out of 150 are three successes out of 150 with p=3/1250. Using the corresponding formula, probability of 3 successes out of 150 = 0.0054, 2 out 150 = 0.0451, 1 out of 150 = 0.2517, 0 out of 150 = 0.6974. Looks like chances of just 3 such successes are pretty low.Show More Responses@G: It's incorrect to compare 3/1250 with 3/150 as the latter doesn't tell us the probability of the first event (i.e., phone numbers ending in 0,1,4 and 9) for 3 times out of 150. The correct way of calculating this probability is C(150,3)p^3(1-p)^147 where C(150,3) is 150 choose 3 and p = 3/1250.In fact, the p-value defined by fisher exact test should be P( x<=3) = C(150,0) p^150 + C(150, 1) (1-p)p^149 + C(150,2).... My guess based on (24*150/10000)/3 = 0.36/3 = 0.12 is not significant(statistically )Am I over simplifying this? The odds of getting any one of the numbers at first is 4/10, then 3/10, 2/10 and 1/10. Leaves with .0024 or 1/416 chance of getting those four digits in any order. The odds of three being 1/(416^3).@S, Actually, I think we don't need to calculate C(150,3)p^3(1-p)^147 as @G mentioned, the probability that one has a phone number ending with such permutation is p = 4!/10^4 = 3/1250, so the expected number of such friends is E = np = 150 * 3/1250 = 0.36, which is much smaller than 3, so it's not just a chance occurrence.

### 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
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