Stock Exchanges interview questions | Glassdoor

# Stock Exchanges interview questions

Susquehanna International Group Interviews

www.sig.com /  HQ: Bala Cynwyd, PA

365 Interviews

3.1 Average

Optiver Interviews

www.optiver.com /  HQ: Amsterdam

143 Interviews

3.3 Difficult

Nasdaq Interviews

business.nasdaq.com /  HQ: New York, NY

49 Interviews

2.9 Average

## Interview Questions

Sort: RelevancePopular Date

Oct 5, 2009

Mar 2, 2010

Dec 28, 2011
 There is a 91% chance of seeing a shooting star in the next hour, what is the probability of seeing a shooting star in the next half hour?12 Answers70%How could this be the answer? surely it is 40.5%; half of the 91% probability?sorry I meant 45.5%Show More Responsesprobability shooting star in half hour = x probability of NOT seeing shooting star in 1 hour = (1-x)^2 probability of seeing shooting star in 1 hour = 1 - (1-x)^2 so we solve 1 - (1-x)^2 = 0.91 for xAccording to the question, suppose we have a random variable X that is the time when we see a shooting star, and let F(x) be the CDF of X. Hence, F(now+1h)-F(now)=0.91. And this is unfortunately the only thing we know about F(x). In other words, any valid F(x) satisfying such requirement could be the CDF of X. Therefore, we have no idea of the value of F(now+0.5h)-F(now) because F(now+0.5) could be any value between F(now) and F(now)+0,91. Answer: unknown result depending on the distribution.The second answer is incorrect. Take an extreme case as an example, if the probability of seeing a shooting star in 1 hour is 1, then the root of the function would be 1, which means that the probability of seeing a shooting star in the first 30 minutes is 1. An interesting question followed by that would be: what about in the first 15 (second 15) mins? According to this method, we can again conclude that the probability of having a shooting star in the first 15mins is still 1. And this can go on for ever, and result in the conclusion that at any time interval, P(shooting_Star) would be 1 which is obviously mistaken.let p= probability to see the star in half hour, (1-p)=won't see, (1-p)^2=can't see a star in an hour=1-0.91, solve we have p=0.7The answer should be 0.7 but most of the explanations here make no sense. It is reasonable to assume a exponential distribution here and solve the probability.P(no star in an hr) = 1 - 0.91 = 0.09 P(no star in half an hour)^2 = 0.09 P(no star in half an hour) = 0.3 P(star in half an hour) =1 - 0.3 = 0.7 = 70%Probability is 30%. The arrival of star follows a Poisson process. Seeing K arrivals in time T is P = (lambda*T)^K * exp(-lambda*T)/K!. Now we know there has been no arrival in 1 hour, T = 1, K = 0. P = exp(-lambda) = 1-0.91 = 0.09. Now T = 0.5 (half an hour), P = exp(-lambda*1/2) = sqrt(0.09) = 0.3. The probability of seeing a star is 1-0.3 = 0.7, or 70%.I would argue for the 91/2% = 45.5% chance of seeing the star. The wording of the problem implies this is not a poisson process. Rather, I would interpret as the weather channel telling residents in an area that there is a 91% chance of seeing the shooting star between 7pm and 8pm (this is an unusual occurrence). Hence, with 9% chance you can't see a star in the next half hour. Then, in the 91% of the time where there is a shooting star in the next hour, one could make the reasonable assumption that the time it comes is uniformly distributed in the hour. So, 45.5%.Like, think about it in the real world. It is kind of unreasonable to model seeing shooting stars as a Poisson process with parameter .91 stars per hour.

Jan 8, 2012
 I have 10 cards face-down numbered 1 through 10. We play a game in which you choose a card and I give you the corresponding dollar amount. a) What is the fair price of this game? b) Now, after picking a card you can either take the dollar value on the card or \$3.50. Also, cards worth less than 5 are now valued at \$0. What is the maximum price you are now willing to pay for the game?9 AnswersFair value is 5.5 2nd game: FV = (3.5 x 5 + 6 + 7 + 8 + 9 + 10)/10 = 4.75You decide on taking the dollar value on the card or \$3.50 *before* seeing the number on the card. And you get \$5, if you open that card, not \$0. So: 2nd game: (1/2)*3.5 + (1.2)*(1/10*[5+6+7+8+9+10]) = 43.5(.4) + (4.5)*.6 = \$4.10 3.5 because this is what 40% will select, 4.5 as this is the avg of 5-10*1/10. There is not a 50% chance of selecting 3.5 @Your are WRong @ EASY. There is not a 10% chance of receiving \$3.5 as numbers under 5 are worthless this will /should change your perspective.Show More ResponsesFJM is correct. "Cards worth less than 5 are now valued at \$0" - that does not include card number 5.how does 4.10 make sense if 4.5 is the ev of taking the card..ii think you are incorrect. can you explain your reasoning. using 1/10 to take your average and then multiplying by .6 seems like you are double countingTotal amount=3.5*4+(5+6+~+10) Expected value=total/10=5.9 @FJM no idea where your 4.5 coming fromOK agreed with xinzhuo. Should be 7.5 instead of 4.5 in FJM's calculation, which gives 5.9.Agreed. I double counted. 3.5*.4 + 7.5*.6 3.5 was given 7.5 =(5+6+7+8+9+10)/6"You decide on taking the dollar value on the card or \$3.50 *before* seeing the number on the card." 2nd game: Obviously, the game price should be no less than 3.5, otherwise there is an opportunity for arbitrage. Now that the price is higher than 3.5, it makes no sense for a rational player to choose just take 3.5 dollar and exist the game. The player should always choose to bet. And the expectation for betting is (5+6+...+10)/10 = 4.5.

Apr 22, 2011
 If there is an 84% chance of an event happening in an hour, what is the probability that it happens in half an hour?7 Answers1-(1-p)^1/2How do you get that answer? We don't even know the distribution of event, to say, uniformly or normally, etc.1-(1-p)^1/2 is simply wrongShow More Responses.5 X .84 = .42 or 42% chanceSuppose the event meets Poisson distribution. lambda is the rate P(event happens in one hour)=1-e^(-lambda)=0.84 e^(-lambda)=0.16 P(event happens in .5 hour)=1-e^(-lambda/2)=1-sqrt(0.16)=0.6cb is right.If cb is right then Interview Candidate:is right too. Just observe 1-p=e(lambda).

Sep 13, 2011

### Quant Research Intern at Susquehanna International Group was asked...

Mar 17, 2013
 Suppose you have two covariance matrices A and B. Is AB also a covariance matrix? Suppose that, by plain dumb luck, we also have that AB=BA. Is AB a covariance matrix under this additional condition?10 AnswersI suppose it's clear from how I wrote the question that the answer to the first question is no (for you: why?). For the second question, this is a little bit harder if you aren't experienced in linear algebra. I actually have a PhD in algebra, and the interviewer also had a PhD in algebra, so on some level this question might have been specifically targeting my background. There is a standard result that applies here; see if you can figure it out.1. no 2. no, give an example where diagonal positivity is not true.1 is correct, 2 is wrong. Try again. :)Show More Responses2. Just checked Wikipedia, AB is cov matrix because it is symmetric semi positive definite. Is it right?If AB=BA then yes it is symmetric. So the question boils down to: Is it in fact positive semi-definite? Why or why not? Work through that and you have your answer. My hint: Take a look at some results related to diagonalizing matrices that commute with each other.Hi for your telephone interview with doug, what was the focus for technical interview parts, everything from math, finance, programming? or mainly probability type of question? thanksI remember it being mostly probability and finance. Nothing too terrible.Hi Mr.interview candidate, what were the types of finance questions Doug asked? Thanks!Hi, Mr. interview. Do you remember what kind of data analysis example was for on-site interview? If so, can you share a little bit? What was the level of difficulty? Thanks!1, no 2, yes

Jan 27, 2010
 square root of .165 Answers.4 -- it's basically the square root of 16 and you move back a decimal point.+.4 or -.4nign on the -.4 square root by convention must be positive.Show More Responsesthe answer above is incorrect. indeed the answer can be -0.4. the square itself may not be negative howeverIf x^2 = y, Then x = + or - square root of y square root of any real number is positive