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

# Quantitative Analyst Interview Questions in New York City, NY

290

Quantitative analyst interview questions shared by candidates

## Top Interview Questions

Sort: RelevancePopular Date

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

Jan 24, 2011
 expectation of max value of two dices4 Answers161/36if D1=D2: EV1 = sum_{i=1}^6 i/36=21/36 O.W.: EV2 = sum_{i=2}^6 i(i-1)/36=140/36 => EV=EV1+EV2=161/36;161/36. I did it by finding pdf of max(x1,x2).Show More ResponsesWhat is the expression for probability(max =n)? P(max=n)=P(1st=n, 2nd

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

Oct 2, 2010
 gambling question, whether it makes sense to gamble, if the chances are very small. Calculate the odds someone will take the grand prize if one million people goes, and odds are 1 in 1 million for each entry.4 Answers1/e = 0.3 1/1000000 is very small, so actual odd approaches the limiting case of 1/e.Odds are: 0.3678792572210609 that no one wins, i.e. 0.632120742778939 that someone wins. The number e is not related to this problem; it is the limit of (1+1/n)^n. Here we have odds of (1-1/n)^n that no one wins.poisson as limiting cxase if binomial distribut5ion. Large number of trials with small prob. Thus Lambda = 1. find prob n = 0 in poisson with lambda = 1 = 1-1/e.Show More ResponsesYou're all wrong. 1) Answer one makes no sense, where are you getting 1/e= 0.3? (I assume this is related to answer 3) 2) Lotteries are not binomially distributed, since lottery numbers are typically unique to each ticket. Therefore you are sampling without replacement. 3) Poisson is indeed the limiting case of the poisson, and using this approximation will give you the same answer as two, which since I was bored I calculated as p=0.3678794 that no one wins, but is wrong for the same reason in my view. If we are sampling without replacement, the problem becomes much easier, and you won't need to bring a slide-rule or pocket calculator to your interview. If the odds for each player are one in one million, then one million tickets must have been issued. If there are one million players, then all tickets have been issued, and thus with certainty someone has won.

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

Dec 20, 2010
 if you have 3 children one is boy and one is girl and another is unknown, you choose one from them what is the probability that that one is a boy.5 Answersmonty hall problem1/21/2Show More Responsesp(3rd == BOY) * 2/3 + P(3rd == GIRL) * 1/3 = (1/2) * (2/3) + (1/2) * (1/3) = 1/22/3

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

Dec 11, 2010
 There is a big line of people waiting outside a theatre for buying tickets. The theatre owner comes out and announces that the first person to have a birthday same as someone standing before him in the line gets a free ticket. Where wil you stand to maximise your chance?7 AnswersPlease post some answers guysyour are at nth position, the possibility that the people before you have different birthdays is (365/365)*(364/365)*- - - -*(366-(n-1)) then the possibility that your birthday is in one of these days is n-1/365 so total probablilty is (365/365)*(364/365)*- - - -*(366-(n-1))*(n-1/365) Clearly you should stand where f(n)>f(n-1) and f(n)>f(n+1) solvedLet there are n people before you. The fact that a person's birthday is the same as yours is a Bernoulli trial with probability of success p = 1/366. Probability of exactly 1 success in n trials is f(n) = n*p*(1-p)**(n-1) (1). To maximize (1), solve df/dn = 0; n = -1 / ln(1-p) ~ 1/p = 366; f(366) = (1-1/366)^365 ~1/eShow More ResponsesFirst in the lineSecond in lineAccoring to 2nd post, doing the calculation, I find n = 19?20th place

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

Mar 8, 2011
 A line in front of movie theater, a free ticket is to given to the first person whose birthday is the same as someone who has already bought a ticket. You choose the position in this line, which position have the largest chance of getting the free ticket?3 AnswersPossibility to win in he nth position: Pn=(n-1)*364!/365^n/(366-n)! It is easy to see that Pn/Pn+1 = 1, where n=19.61, so the best place is n=20. Note that the above calculation uses a 365 day year (no birthday on 29th Febr).this is not true for n=2 - if you are second in the line, then probability is 1/365 and formula above gives Pn|n=2 = P = 1/365^2 ...probability to get free ticket at position k is the product of -- the probability that none of the people at positions between 2 and k-1 got the free ticket and --the probability that the k-th person has the same birthday with the previous k-1 people. at a certain m-th position, given that there all the previous (m-1) people have different birthdays, the probability that the m-th person does not have the same birthday as one of the previous (m-1) people is (365 - (m-1)) / 365 . and the probability that he/she does have the same birthday as one of the previous (m-1) people is (m-1) / 365 --> probability to get free ticket at position k is : (n = 365) p(k) = (n-1) / n * (n-2)/n * .... * (n - (k-1))/n * (k-1)/n = (n-1) ! / ( n - (k-1)) ! * (k-1) / n^(k-1) --> p(k) maximum when k = 20.

Nov 11, 2011

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

Mar 4, 2011
 3 random variables have equal pairwise correlation, what are the possible value of correlation.6 Answers0, 1.(-1/2, 1)let x be the correlation (1 x x) —> the correlation matrix A = (x 1 x) should be positive semi-definite (x x 1) —> |x| = 0 —> -1/2 why ? because intuitively, if x1 and x2 are negatively correlated, x2 and x3 are negatively correlated, then my naivety tends to tell me that x1 and x3 move in the same direction, i.e. positively correlated. but it seems like negative values for x are also possibleShow More Responseslet x be the correlation (1 x x) —> the correlation matrix A = (x 1 x) should be positive semi-definite (x x 1) —> |x| == 0 —> -1/2 = why ? because intuitively, if x1 and x2 are negatively correlated, x2 and x3 are negatively correlated, then my naivety tends to tell me that x1 and x3 move in the same direction, i.e. positively correlated. but it seems like negative values for x are also possiblefrom the requirement that the correlation matrix be positive semi-definite, we'd get correlation coefficient in [-1/2,1]Construct the correlation matrix ((1,x,x),(1,x,1),(x,1,1)) = C. For any vector v, we need dot(v, dot(C,v)) >= 0. Choose v= (1,1,1) to get x >= -1/2. To get the upper bound of 1, compute det (C) to get 2x^3 -3x^2 +1 >= 0. We know x=-1/2 is a zero, and we have 2 more. We can easily check that 1 is a zero. Thus x in [-1/2, 1].

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

Mar 4, 2011
 In 3-dim graph, how many paths to go from (0,0,0) to (3,3,3)4 Answers9!/(3!*3!*3!)1There are nine total steps to make. There are 9 choose 3 ways of picking the x-direction ones, say. After that, there are 6 choose 3 ways of picking the y-direction ones. This gives (9C3)*(6c3) = 1680.Show More Responses90

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

Feb 1, 2011
 3. There are 30 blue and 30 red balls and two urns. You play a game. Your opponent has the right to arrange the balls in the two urns as he pleases, without telling you what he did. You then must draw a ball from an urn of your choice. You win \$10 if you draw a blue ball, otherwise you get \$0. How much would you be willing to pay to play this game?2 AnswersI would pay less than 5\$ because the expected value of this game is 5\$Your opponent may put one red ball in a urn, 29 red balls and 30 blue ones in the other. So the probability you get a blue ball is approximately 1/4. So you should be willing to pay \$2.5.

### Quantitative Analyst at Knight Capital was asked...

Jul 20, 2012
 How much would you pay to play this game? You have a fair coin. You get heads you win \$1 and can continue to play. You get tails, you collect your winnings ans stop the game. Second question. Now what if the winnings double each time you get heads.4 Answerssimple expectation formula. For example, for 4 trows . E(Reward) = 0.5*0 + (0.5*1 + 0.5* ( 0.5 *2 + 0.5*(0.5*3 + 0.5 * 4)))) = ... The sum converges if you continue the formula, for example to 10 trows1) \$1 2) infinity (well, expectation is infinity)The fair amount to be paid for this game is the expected winnings from it. Let x be the expected winnings. Now, to calculate x, consider the first toss - With probability 1/2, we get a tail thus earning \$0. With probability 1/2, we get a head to get \$1 and make another \$x in expectation from further tosses. x = 1/2 * 0 + 1/2 * (x + 1) => x = 1 In the second scenario, if winnings double with every toss: y = 1/2 * 0 + 1/2 *( 2 * y + 1) => y = infinity (ie expectation diverges) where y is the expected winning.Show More ResponsesAfter simulating the second scenario a bunch of times on Python (each averaging 1,000,000 iterations), there was an average payoff roughly in the range of 4-12. Obviously, the expectation is infinite, but in practice it is reasonable to use monte carlo estimations for a more realistic answer. Indeed sometimes you can win massively, sometimes you can win nothing.
1120 of 290 Interview Questions