Trading analyst Interview Questions | Glassdoor

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## Top Interview Questions

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Mar 20, 2014
 I was asked about if a machine failed and a frustrated production worker was on that machine when it failed, how do you deal with the production worker while you working to fix that machine as they tend to get more frustrated the longer the machine is down? 1 AnswerI said that I would give the production worker the knowledge of knowing what was wrong with the machine, and an approximate time as to how long it may take. I would then try to conversate and diffuse the situation with humor.

Mar 11, 2011

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?10 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.I think question was phrased as such: " You can take 3.50 before seeing the value of the card, or pay an amount to see the value of the card". We have established that the EV of seeing the card is 4.5, but why would we pay 4.5 to get an EV of 4.5? That nets 0 profit per play on average as opposed to taking the risk free 3.5 Instead, we can see that we would at most bet 1 dollar per play. This way, the total EV of the game with the bet is 3.5 = the arbitrage free price. We can show it in terms of "plays" as follows, with first number as taking 3.5 and second as seeing the game Play 1: 3.5, 4.5-B Play 2: 7, 2(4.5-B) ... We can see that we would at most pay 1 dollar for this game

Oct 4, 2011
 Say I take a rubber band and randomly cut it into three pieces. What's the probability that one of the pieces has length greater than 1/2 of the original circumference of the rubber band.9 Answers3/4Suppose you have two cuts on the rubber band placed randomly. The probability of having one segment greater than half the circumference is the probability that the third cut will be inside the combined range of 90* to either side of the cuts. Since the average distance between the first two cuts is also 90*, the combined range is 270*, or 3/4 of the circle.You need 3 cuts to end up with 3 pieces. The first cut doesn't matter. The second cut can also be anywhere and the largest piece will still be at least half the circumference. What matters is the third cut, which should lie in the same half as the second cut. So the probability is actually 1/2.Show More ResponsesThe correct answer is 3/4, as this problem is equivalent to the famous 3-points-on-semicircle problem. Why? If one of the pieces has length greater than 1/2 the circumference, then the three points of cutting must lie in the same semicircle. On other hand, if the three points of cutting lie on the same semicircle, then the longest piece must be at least 1/2 of the circumference.For reference to the 3-points-on-same-semicircle problem, see e.g., http://godplaysdice.blogspot.com/2007/10/probabilities-on-circle.html1/4 1 -3/4suppose I have two points whose minor arc distance is t <= 1/2. Then the range of semicircles covering both points gives an arc length of (1/2+1/2)-t = 1-t. say we fix the first point, tracing the second point around gives minor arc lengths from 0 to 1/2 and then 1/2 to 0. Therefore the answer is 2*integral (1-t) from 0 to 1/2, which is 2(1/2-1/8) = 3/4It's 3/4. Cut it into 1 piece make a line. Cut it close. Pretend the length is 100. If you cut the first at x=1, as long as it isn't between x=50-51, it will have a length greater than 50% so there's 99% chance. You can imagine that if the cut was infinitely close to the end it would be about 100%. Now cut at x=2 you can't do between x=50-52. For x=3 it's 50-53 etc. So when you get to right to infinitely close to 50 it is pretty much between x=50-100 so there is a 50% chance you hit your spot. (obviously 50-50 is 100%, but since this length is continuous there's little chance it lands on that point). Obviously since this is symmetrical you can see this pattern going from 50% to 100% at the other end. Since each point on the continuous line has the same probability of happening the answer is clearly 75%.This problem is also equivalent to the probability that, if you have a line segment from 0 to 1 and you make 2 random cuts on that line segment, what is the probability that the three resulting pieces do NOT make a triangle?

Feb 21, 2012
 You have three barrels. One barrel is filled with apples, one with oranges, and one with both apples and oranges. Each barrel is mislabeled. You can take out as many fruit as you like form each barrel and look at it. What is the minimum number of fruits you need to remove to correctly label the barrels?8 AnswersJust one. Take a fruit from the barrel labeled apples and oranges. You know this barrel is not the apples and oranges barrel, so whatever fruit you take must be the only fruit in that barrel. The other two barrels are mislabeled, so you can figure out the other two as well.Two. Take a fruit from A+O, if its an O then this barrel is O ...Now take out a fruit from A, if its an O then this barrel is A+O...and the 3rd is by default barrel AOne. Start of from A+O, the fruit u pick say O should be the barrel name. now that means that the barrel labelled A can be A+O or A... But since we know that barrels are mislabelled it cant be A. hence, we concluded it to be A+O and the last remaining barrel is A.Show More Responses1Trick question, just look inside the barrel...One. Since it's mislabeled, A can be labeled as O or A+O. Similarily, O = {A, A+O} and A+O = {O, A}. Case 1: Take 1 from A+O label, it turns out to be A. Relabel it. We have A, O mislabels left, and A+O, O real labels left. Mislabeled ones can ot be the same as real labels. So the mislabeled O should be A+O and A should be O. Case 2: also choose from A+O, but it turns out to be O. Solving the same as case 1.None, you don't need to move any of the fruit to correctly label the barrels. You need to change the labels.All three are mislabeled and you cannot look inside the barrels. If you take one out of the apples+oranges barrel, whatever fruit you pull out is the fruit of the barrel. And Since you know the other two are mislabeled, you would switch the labels. So you need to take out only one fruit

Jun 12, 2012

Aug 17, 2011
 We play a game. You flip two coins, if you get both heads I give you \$1. If you get both tails you flip again. If you get one of each, I pay you nothing. What's the expected value of this game?6 AnswersDismiss TT since it is not part of the sample space (ie there is no outcome from it) You are twice as likely to get one of each (H-T or T-H) versus getting both heads answer: (2/3)(\$1) + (1/3)(0) = 66.7 centsE=0.25x1+0.5x0+0.25xE 0.75xE=0.25 E=0.33interviewer told me .667 is correctShow More ResponsesHT and TH add up to 1/2 of the probability with zero payout. HH is 1/4 probability with \$1 payout. TT results in a "reset" so we can disregard. After we normalize the probabilities, we get: E[payout]= (1/3)*\$1 + (2/3)*\$0 = \$0.33You can also see it a geometric series. By writing E=0.25x1+0.5x0+0.25xE, Kerzane could as well have written E?0.25x1 + 0.5x0 + 0.25x(0.25x1 + 0.5x0 + 0.25x(0.25x1 + 0.5x0 + 0.25x(....))) This is quite simply the geometric sum: (n-> infinity) E=0.25 + 0.25^2 + 0.25^3....0.25^n which is equal to 0.25/(1-0.25) = 0.25/0.75 = 1/3 or 0.333Sample Space = H/H, H/T, T/H, T/T, except T/T isn't part of the sample space due to the fact that T/T loops back into the game. An outcome, part of the sample space, is only counted if you "win" or "lose". So the sample space is: H/H, H/T, T/H. H/H = 1/3 probability of getting; T/H, H/T = 2/3 probability of getting. Expected value, or E, is the theoretical value attained from the law of large numbers which is arrived at as the number of trails approaches infinity. Therefore, in probability theory we may say that E(X) = 1/3 for winning or 2/3 for losing. 1/3 = .33 chance of winning, and 2/3 =.667 chance of losing.

Aug 17, 2011