Quantitative research analyst Interview Questions | Glassdoor

# Quantitative research analyst Interview Questions

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## Top Interview Questions

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Oct 26, 2017
 What is the probability that a best of seven series goes to the seventh game?5 AnswersThis is equivalent to Team A wins 3 out of the first 6 games. It's now a simple binomial problem.to make it more details in the first answer: 20/64 = 5/16 = 31.25%Yeah this is binomial distribution, you will need exactly 3 games from 6 to win , so assuming both teams are equally likely to win, 6C3 * .5^3 * .5^3 = .3125Show More Responses6C3 * p^3 * (1-p)^3 * p One or more comments have been removed. Please see our Community Guidelines or Terms of Service for more information.

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

Aug 4, 2010
 What is the expected value of tosses to get 3 heads in a row?4 Answers14can be found using markov chains and fundamental matrics matlab code A = [1/2 1/2 0; 1/6 1/2 1/3; 0 1/3 1/2]; E=[1 0 0; 0 1 0; 0 0 1]; X=(E-A)^(-1); P = [1; 1; 1]; C = X*P; PS = [1/8 3/8 3/8]; F = PS* C; display(C); display(F); so 14.5 average number of steps14，2^(n+1)-2,with n denotes number of headsShow More ResponsesDenote H as an expected number of tosses. The key is to solve an appropriate equation. First toss, if you get tails than the expected number of tosses is (H+1), since you have to start from the beginning.. If you get heads than you move to the next throw with the same logic: H=0.5*(H+1) + 0.5*(0.5(H+2)+0.5(0.5(H+3)+0.5*3)) H=14

### 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 Finance Summer Associate at Morgan Stanley was asked...

Aug 5, 2010
 There are two stocks with the same expected value, with variance 0.3 and 0.2 and correlation 0.5. What proportion of each stock do you invest in to yield a minimum risk portfolio? 4 Answers100% in stock 2, with the smaller variance.No, if you can short, the variance could be reduced even further since the stocks are correlated.17/44 of the stock with variance of .3 and 27/44 from the stock with variance of .2Show More ResponsesCovariance = 0.3 * 0.2 * 0.5 = 0.03 Cov Matrix = c = [0.3 0.03; 0.03 0.2] Inv Cov Matrix = c_inv = [2000/591 -100/197; -100/197 1000/197] Answer = (c_inv * [1; 1])/([1 1] * c_inv * [1; 1]) = [17/44; 27/44]

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

Oct 1, 2016
 Expected length of the longest segment of a unit-length stick broken in 2 places.4 AnswersMapped the probability measure to a 3D polytope and calculated the center of mass.0.75 by common senseLet L and R be the left and right half of the broken stick (resp.) The probability that L is longer than R is equal to the probability that R is longer than L (these halves are symmetric, and so this probability is 1/2). This implies that the expected length of the longer side is 3/4.Show More Responses11/18, also verified by simulation, but I don't know how to do it without 15minutes of calculus

### 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

Oct 15, 2015
 Approximate the number of book titles ever published3 AnswersIn this line of work, it's harder to grab an intern position than a fulltimerTo give a sense of the "right" answer, I'm sure they're looking for you to explain what calculations you use to arrive at your answer and may also ask you for a 95% confidence window (note: people naturally are over-confident and are prone to giving confidence windows that are far too small). Google did a study in 2010 and estimated that 129,864,880 books have been published. Additionally, Wikipedia estimates that currently, an additional 2.2M titles are being published each year.I love fermi problems. OK, step one: Lets say X is the fraction of people who are writers in a given population, and assume that this is constant over time. Lets say that each writer authors B books per year on average, and assume they all have distinct titles. Then Q=X*B is the average number of books per person per year under ideal conditions. Next we need to figure out how many people have lived since the advent of writing. We'll say that books started appearing 500 years ago with appreciable volume with the advent of the printing press. The current population is approximately 6 billion = 6* 10^9. 10000 years ago, the population was probably around 600,000. Populations grow exponentially in time so population ~ e^(k t), so we can solve for the growth rate as 6 10^9 =6* 10^5 e^(k (10^5)years) to get population growth rate = k~= 4 ln(10)/10^5 The population at the advent of the printing press was then P0~6*10^6.6 Since book growth rate is proportional to population growth rate, if there were B books 500 years ago, then there are B * \int_0^500 e^(k t)dt books now. The integral evaluates to 10^5*(10^0.2 -1)/4ln(10). We have to approximate 10^0.2 =(10^2)^(1/5) . The fifth root of 100 must be between 2 and 3, and 2.5 is decent enough. Then the total number of books now is B* (1.5 *10^5)/4ln(10) ~ B*10^4. We can say that during the first year after the printing press the number of books written should be about Q*P0. If we say that 1 in 100 people are writers and they publish 1 book per year on average we get B~ 10^4.6 Then the total number of titles ever written should be about 10^8.6. This is close to the actual result

### Quantitative Finance Summer Associate at Morgan Stanley was asked...

Aug 5, 2010
 If W is standard Brownian Motion, for what values of n is W^n still a martingale? 3 AnswersBy Ito W^n is a martingale for n=0,1Agree, for any n>1 you get a drift termXt is a martingale you get f(Xt) is a martingale if f is linear otherwise convexity or concavity will increase or decrease the expectation, think of Jensen inequality.

### Quantitative Finance Program at Morgan Stanley was asked...

Sep 13, 2010
 In a set of cereals, there are 4 kinds of coupons. A child wants to collect all the 4 kinds of coupons. How many boxes in average should he buy? 3 Answers25/3169/16sorry, should be 25/3
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