Quantitative analyst Interview Questions in United States | Glassdoor

# Quantitative analyst Interview Questions in United States

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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?5 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.Show More ResponsesIt didn't mention linear. if it's linear, then ols. if not, CEF, conditional expectation function.It didn't mention linear. if it's linear, then ols. if not, CEF, conditional expectation function.

Jun 15, 2015
 You observe a sample of measurements coming from a fixed length ruler. If the object is shorter than the ruler you observe the actual measurement. Otherwise you observe the length of the ruler. What would be a good estimator of the ruler length?17 AnswersMy initial answer was to use the MAX of the sample. That however is a biased estimator. How can you account for the bias and come up with an unbiased estimator? I think this is where you need to start making assumptions on the distribution. A uniform distribution would allow to estimate the bias.what kind of distribution is it? what do we do with the data? what precision should we get? what happens if we "lose" the oversize data?I came up a solution: if we know the distribution of actual measurement and the value of actual measurement, then the expected probability of getting wrong measurement should equal to the probability of actual measurement greater than length of ruler. Not sure this is correct and will interview on friday. good luck to me.Show More ResponsesIf we now the distribution... I'd analyseIf we now the distribution... I'd analyse the tail of cumulative density function.round(central tendency) * 2Please ignore my previous two answers above. I misread the question and thought it was a regression problem, when it wasn't.get rid of measurements that are equal to the ruler length. then take the average of the rest of the measurements that are within the range (0, ruler_length), ruler_length is 2 times this average valueAssuming that the measurements are on a continuous scale, you would have a lot of mass on the point exactly corresponding to the ruler's length, so you could use something akin to a mode I'd imagine.The mode should work, right? The length of the ruler is likely to be the only specific value that shows up more than once in the data.Should first ask whether we have some prior knowledge about the object length distributionShow More ResponsesL^{hat} = N/(N+1) * max(X1,X2,X3,...XN) is an unbiased estimatorL^{hat} = 2*sum(X)/N is another unbiased estimatorThe length of the ruler would be the censored value in the data. If you draw the histogram of observed values, there should be a mass on the largest value, which is the length of the ruler. The more observation you have, the better the estimation.I think it should be (N+1)/N * max(X1,..XN). Is there anyone agreeing with me? One or more comments have been removed. Please see our Community Guidelines or Terms of Service for more information.

Jul 31, 2009

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

May 14, 2011
 You have 2 decks of cards (each deck contains both red and black cards). One deck has twice the number of cards in the other deck with the same color ration (so one deck has 52 cards and the other has 104, both half red and half black). I offer you to play a game. First you get to chose which deck of cards you want to play with. Second, you draw 2 cards at random from your deck of choice. If both are red, then I will give you a ferarri. Which deck of cards would you chose?13 AnswersThe unalert interviewee would answer "it doesn't matter, the probability is the same". While this is true for the first card, you have a higher probability of drawing a second red card with the big deck than the smaller one. So I chose the big deck (no homo) and I was right.yes, he would be rightMathematically: (26/52 * 25/51) vs (52/104 * 51/103) 51/206 > 25/102Show More ResponsesActually, the probabilities are the same for each deck. Consider than because you have a 50/50 chance of drawing your first card red, there's a 50% chance the numerator of the next fraction is reduced by one..so your probability is 26/52 * (25/51+26/51)/2 vs 52/104 * (51/103+52/103)/2 which are the same"G" did the right calculation. To calculate the probability of drawing two red cards in a row one needs to set up an equation where the first drawn was red, and the SECOND card was red as well. The question is asking what is the probability of drawing 2 red cards in a row, NOT what is the probability of drawing a red card then either a red or black card.Intuition tells us if you add in same number of red and black cards into the original problem with 52 cards, probability will go up. Just imagine when you add in 1 billion red cards and 1billion black cardsYou should definitely choose the larger deck if both are 50% red, 50% blue. Here's another explanation in addition to the other correct ones above. Each deck is naturally partitioned into maximal sub-stacks where each sub-stack consists of cards of a single color, either all red or all blue. If it is known ahead of time that half the cards are blue and half are red, then the expected size of the stack increases with the number of cards in the deck.Correction: expected size of the **sub-stacks** increases with the number of cards in the deck. Sorry about that.extreme case deck 1: 1 red 1 black deck 2: 2 red 2 black So more cards the better chance you get...2 out of 52 is the equivalent of 4 out of 104. For these chances to be equal then the problem should be 2 out of 52 or 4 out of 104. If you still get to draw only two cards to try to get two reds then the chances should be better with the smaller deck.Either one.If you don't like how the table is set, re arrange it. I would offer the dealer half of the reward if he/she let me draw 5 more times in the deck with more cards.It seems that the odds are greater with a lager number of choices. 2:1 ratio.

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

Apr 17, 2011
 3) Poker. 26 red, 26 black. Take one every time, you can choose to guess whether it’s red. You have only one chance. If you are right, you get 1 dollar. What’s the strategy? And what’s the expected earn?13 Answersexpected earn is 25 cents. 1/2*1/2*1, prob of choosing to guess is 1/2, prob of guessing right is 1/2, and the pay is \$1I would start picking cards without making a decision to reduce the sample size. This is risky because I could just as easily reduce my chances of selecting red by taking more red cards to start, as I could increase my chances of selecting red by picking more black cards first. But I like my chances with 52 cards, that at some point, I will at least get back to 50% if I start off by picking red. Ultimately, I can keep picking cards until there is only 1 red left. But I obviously wouldn't want to find myself in that situation so I would do my best to avoid it, by making a decision earlier rather than later. Best case scenario, I pick more blacks out of the deck right off the bat. My strategy would be to first pick 3 cards without making a decision. If I start off by selecting more than 1 red, and thus the probability of guessing red correctly is below 50%, then I will look to make a decision once I get back to the 50% mark. (The risk here is that I never get back to 50%) However, if I pick more than 1 black card, then I will continue to pick cards without making a choice until I reach 51% - ultimately hoping that I get down to a much smaller sample size, and variance is reduced, while odds are in my favor that I choose correctly. The expected return, in my opinion, all depends on "when" you decide to guess. If you decide to guess when there is a 50% chance of selecting correctly, then your expected return is 50 cents (50% correct wins you \$1 ; 50% incorrect wins \$0 --- 0.5 + 0 = .5) If you decide to guess when there is a 51% chance of selecting red correctly, then the expected return adjusts to (0.51* \$1) + (0.49 * \$0) = 51 cents. So, in other words, your expected return would be a direct function of the percentage probability of selecting correctly. i.e. 50% = 50 cents, 51% = 51 cents, 75% equals 75 cents. Thoughts?There is symmetry between red and black. Each time you pull a card it is equally likely to be red or black (assuming you haven't looked at the previous cards you pulled). Thus no matter when you guess you odds are 50% and the expected return should be 50 cents.Show More Responsesscheme: guess when the first one is black, p(guess) x p(right) x 1=1/2 x 26/51=13/510.5, just turn the first card to see if it's red. I think it's more about trading psychology. If you don't know where the price is going, just get out of the market asap. Don't expect anything.The problem should be random draw card and dont put it back. Every draw you have one chance to guess. So the strategy is after first draw you random guess it's red. If correct you get one dollar, next draw you know there is less red than black. So you guess black on next draw. Else if first guess you are wrong, you guess red on next round. It's all about conditioning on the information you know from the previous drawingsThis should be similar to the brainteaser about "picking an optimal place in the queue; if you are the first person whose birthday is the same as anyone in front of you, you win a free ticket." So in this case we want to find n such that the probability P(first n cards are black)*P(n+1th card is red | first n cards are black) is maximized, and call the n+1th card?The problem statement is not very clear. What I understand is: you take one card at a time, you can choose to guess, or you can look at it. If you guess, then if it's red, you gain \$1. And whatever the result, after the guess, game over. The answer is then \$0.5, and under whatever strategy you use. Suppose there is x red y black, if you guess, your chance of winning is x/(x+y). If you don't, and look at the card, and flip the next one, your chance of winning is x/(x+y)*(x-1)/(x+y-1) + y/(x+y)*x/(x+y-1) = x/(x+y), which is the same. A rigorous proof should obviously done by induction and start from x,y=0,1.The answer above is not 100% correct, for second scenario, if you don't guess, and only look, the total probability of getting red is indeed the same. However, the fact that you look at the card means you know if the probability of getting red is x/(x+y)*(x-1)/(x+y-1) or y/(x+y)*x/(x+y-1). Therefore, this argument only holds if you don't get to look at the card, or have any knowledge of what card you passedDoesn't matter what strategy you use. The probability is 1/2. It's a consequence of the Optional Stopping Theorem. The percent of cards that are left in the deck at each time is a martingale. Choosing when to stop and guess red is a stopping time. The expected value of a martingale at a stopping time is equal to the initial value, which is 1/2.My strategy was to always pick that colour, which has been taken less time during the previous picks. Naturally, that colour has a higher probability, because there are more still in the deck. In the model, n = the number of cards which has already been chosen, k = the number of black cards out of n, and m = min(k, n-k) i.e. the number of black cards out of n, if less black cards have been taken and the number of red cards out n if red cards have been taken less times. After n takes, we can face n+1 different situations, i.e. k = 0,1,2, ..., n. To calculate the expected value of the whole game we are interested in the probability that we face the given situation which can be computed with combination and the probability of winning the next pick. Every situation has the probability (n over m)/2^n, since every outcome can happen in (n over m) ways, and the number of all of the possible outcomes is 2^n. Then in that given situation the probability of winning is (26-m)/(52-n), because there are 26-m cards of the chosen colour in the deck which has 52-n cards in it. So combining them [(n over m)/2^n]*[(26-m)/(52-n)]. Then we have to sum over k from 0 to n, and then sum again over n from 0 to 51. (After the 52. pick we don't have to choose therefore we only sum to 51) I hope it's not that messy without proper math signs. After all, this is a bit too much of computation, so I wrote it quickly in Python and got 37.2856419726 which is a significant improvement compared to a basic strategy when you always choose the same colour.dynamic programming, let E(R,B) means the expected gain for R red and B blue remain, and the strategy will be guess whichever is more in the rest. E(0,B)=B for all Bs, E(R,0)=R for all Rs. E(R,B)=[max(R,B)+R*E(R-1,B)+B*E(R,B-1)]/(R+B). I don't know how to estimate E(26,26) quickly.The question, to me, is not clear. Perhaps on purpose. If so, the best answers would involve asking for clarification.

Jan 2, 2010

### Software Engineering and Quantitative Research at D. E. Shaw & Co. - Investment Firm was asked...

Jan 9, 2014

Nov 18, 2015
 Given log X ~ N(0,1). Compute the expectation of X.13 AnswersThis is a basic probability question.Exp[1/4]exp(mu + (sigma^2)/2) = exp(0+1/2) = exp(1/2)Show More ResponsesLet Y = log(X), then X = exp(Y) = r(Y), if we call the pdf of X f(X), then E[X] = integral(Xf(X)dX). By variable transformation, f(x) = g(r^-1(X))r^-1(X))', plug this into E[X] = integral(Xf(X)dX), we get integral( f(y)dy ), which equals to 1Suppose the density function of Y is P(y) and the one for X is F(x), it obeys that P(y)*dy = F(x)*dx; then the expectation of X is E(x) = Integral( x*F(x)*dx ) = Integral( Exp(y) * P(y) * dy ); if you plug the gaussian function and standard deviation in, you will find E(x) = Integral( Exp(1/2) * P(y-1/2)*d(y-1/2) ) = Exp(1/2) So, mojo's ans is correct.I m not that sure, as I got E(x) = 4 I substituted log X = y e^y = X ;and e^2y = t and plz do not forget to change the integration limitsDo they care if you explain the theory or not? I just looked at it, it's standard normal, therefore x=50%P(logX P(XSorry misread the problem. ignore.X has a log-normal distribution, so yes the mean is exp(mu+sigma^2/2)=exp(1/2)Expanding on the correct answers above: E[X] = E[exp(logX)], and logX is normally distributed. So: E[X} is the moment-generating-function (mgf) of a standard normal distribution, evaluated at 1. The mgf of a normal distribution with mean mu, SD sigma is exp(mu*t + (1/2) * sigma^2 * t^2), now set mu = 0, sigma = 1, t = 1 to get exp(1/2).Complete the square in the integral One or more comments have been removed. Please see our Community Guidelines or Terms of Service for more information.