Quantitative research analyst Interview Questions | Glassdoor

# Quantitative research analyst Interview Questions

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

May 22, 2013
 If X, Y and Z are three random variables such that X and Y have a correlation of 0.9, and Y and Z have correlation of 0.8, what are the minimum and maximum correlation that X and Z can have?8 Answers0.9http://www.johndcook.com/blog/2010/06/17/covariance-and-law-of-cosines/0.98 & 0.46Show More Responseshttp://wolfr.am/1i1XT4PHow'd you get 0.98 and 0.46?NND correlation matrix --> det(\Sigma)>=0 --> 0.98 and 0.46minimum: 0.9*0.8+sqrt(1-0.9^2)*sqrt(1-0.8^2) = 0.9815 maximum: 0.9*0.8-sqrt(1-0.9^2)*sqrt(1-0.8^2) = 0.4585How do you know this

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

Aug 5, 2010
 At a party everyone shakes hands, 66 hand shakes occur, how many people are at the party? 9 Answers12requires explanationit's not 12, it's 11Show More Responses12sum of the series 66= n/2( 2*1 + (n-1)*1) n=12You can suppose there are n people in the room and think of them in a row. The first one has to shake hands with (n-1) people (because he doesn't have to shake hands with himself). The second one has already shaken hands with the first one, so he has (n-2) shakes remaining... and so on. So you have to sum: (n-1)+(n-2)+(n-3)+...+1= (n/2)*(n-1) Then you have to solve (n/2)*(n-1)=66 and you get n=12.n(n-1)/2=66 so n=12You can think of this problem as a combination problem: (N choose 2) = 66, and then solve for N. That is, N! / (2! (N-2)! ) = 66. Simplifying this equation leads to N(N-1)/2 = 66, and the integer solution to this problem is N = 12.66=x!/((x-2)!*2) which gives 12

Nov 11, 2015

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

Nov 11, 2011

Jan 2, 2010
 You flip four coins. At least two are tails. What is the probability that exactly three are tails? Do this in your head, you are not allowed to write anything down for this question.8 Answersprobability you get 3 tails is if you have 1 tail in the remaining 2 coins. This is just one halfits 4/11... it'd only be one half if they told you the first two coins were tails. Use bayes rule or actually write out all the possibilitiesbayes' theorem ftwShow More Responsescan you show me the math behind it? i got P(B|A) x P(A) / P(B) = (1x 1/4) / (5/8)=2/5...There are 16 (i.e., 2^4) possible outcomes. There is one way to get zero tails (i.e., HHHH), one way to get all tails (i.e., TTTT), and four ways to get only one tail (i.e., THHH, HTHH, HHTH, and HHHT). All other 10 (=16-6) outcomes have three tails, hence P=10/16.At least two are tails, no HHHH neither THHH, HTHH, HHTH, HHHT. 2^4-5=11 possible scenarios. Three tails, HTTT, THTT, TTHT, TTTH = 4 cases P=4/11The easy way to work this problem out is that knowing the sample space contains 2 coins (4 choose 2 = 6 ways), 3 coins (4 choose 3 = 4 ways) or 4 coins (4 choose 4 = 1 way). Hence, the probability that there are 2 tails provided the fact is 4 / (6 + 4 + 1) = 4/11.Three tails: 4 At least two tails: 6 (2 tails), 4 (3 tails), 1 (4 tails) 4/11

### Quantitative Research at Susquehanna International Group (SIG) was asked...

Mar 17, 2014

Aug 24, 2009
 What is the probability of breaking a stick into 3 pieces and forming a triangle?8 AnswersIts 1/4. Here is the key idea for my analysis of the problem: If we consider the original stick to be of unit length, then we can form a triangle whenever the longest stick is less than a half unit long.Suppose x is the length of the first piece and y is the length of the second piece (both must be nonnegative). Then y will be <= 1-x, and to be able to form a triangle, y must be <= .5 - x with x <= .5. The probability of being able to form a triangle is the area of the second set of (x,y) pairs divided by the area of the first set of (x,y) pairs, which is .125/.5 = .25.The probability is 0 given that it is a question about the probability of 2 breaking points falling on 1st thirdth and 2nd thirdth point. For any continuous variable, the probability of the variable equal to countable points (including indefinite countable numbers) equal to 0Show More ResponsesFor one time evet, the probability is 0I think, the probability is 1/2: Breaking a stick into three pieces corresponds to selecting three real positive numbers with a+b+c=1, and, w.l.o.g., a>=b>=c. The triangle inequation, that any two sides are longer than the third one (i.e., a= 0.5, then b+c a, so we cannot form a triangle. - If a 0.5 > a. The other two inequations a+b>c and a+c>b also hold because a>=b>=c: b is positive and so from a>=c we have a+b>c, similarly, as c>0, a+c>b holds. Hence, we can form a triangle iff a<0.5. Ultimately, selecting a number 0nothing in the question said it had to be equilateral triangle so the probability is 100%1.a+b>c 2.b+c>a 3.a+c>b A Triangle is formed when all three are true. As there are three pieces, so a>0, b>0 and c> 0 Only possibilities are: 1.T,T,F i.e one is bigger than sum of 2 2. T,T,T I.e all thee equations are valid Favorable possibility is T,T,T So answer = 1/2 = 0.5try this simulation in R: checkTri c & a+c>b & b+c>a, 1, 0) ) } mean(replicate(100, checkTri())) Theoretically, we have conditions that a+b>1/2, a<1/2, b<1/2. If you can draw this area in an coordinate axis, you can calc the probability.

### 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...

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...
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