# 48K

Marketing Research Analyst interview questions shared by candidates

## Top Interview Questions

Sort: Relevance|Popular|Date
Research Analyst was asked...5 August 2014

### If you had a machine that produced \$100 dollars for life what would you be willing to pay for it today?

I already have the machine, why would I pay for it??

I would not pay anything. Only the Federal Reserve can legally produce \$100 dollar bills. Less

Why is this considered a top 10 odd interview question? It's a basic accounting question that applies to any applicant at a financial institution. Let's assume the proper phrasing of the question is "If you had a machine that produced a free \$100 dollars per year for life, what would you be willing to pay for it today?" Given that Aksia is a financial firm, they're basically asking what is the present value of a perpetuity with a \$100 annual payment. PV=pmt/r where: PV=PResent value PMT= payment per period r= discount rate Given current US fed reserve discount rate is 0.75%, the Present value of such a device would be \$13,333.33 Answer varies obviously if discount rate changes or if proper phrasing was meant to be \$100 for a different time period. Less

### 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?

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. Less

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 drawings Less

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. Less

Yes

Yes

Yes

### Given log X ~ N(0,1). Compute the expectation of X.

exp(mu + (sigma^2)/2) = exp(0+1/2) = 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). Less

Complete the square in the integral

### What are the last 4 digits of 2015^(2013^2014]? The three distinct entries of a 2 x 2 symmetric matrix are drawn from the uniform distribution [-60, 60]. What is the expected determinant of the matrix? How many times a day do the hour and minute hands of an analog clock form a right angle? If WHITE=000, RED-101, BLUE-110, and PURPLE-100, then what three-digit string corresponds to YELLOW? There are 4 green and 50 red apples in a basket. They are removed one-by-one, without replacement, until all 4 green ones are extracted. What is the expected number of apples that will be left in the basket? Given two assets that have expected excess returns of 7 and 4, and given their expected covariance matrix: {1,1}{1,2} What is the maximum expected Sharpe ratio that you can achieve by combining the two assets into a portfolio? Consider a polynomial f(x) and its derivative f (x) that are related according to: f(X) - f‘(X) = X"3 + 3*X"2 + 3*X + 1 What is f(9)? A pedestrian starts walking from town A to town B. At the same time, another pedestrian starts walking from town B to town A. They pass each other at noon and continue on their paths. One of them arrives at 4 PM, the other at 9 PM. How many hours had each walked before passing each other? Seven people are in an argument, but potentially some or all of them are liars. They give the following statements: Bob: "No one lies." Jennifer: "No one tells the truth." Conrad.: "Jennifer is not a liar." Tom: "Conrad and Sherry always lie at the same time." Sherry: "Danny never lies." Danny: "Sherry is a liar." Adam: "Danny sometimes lies.” How many of them are lying? A city is composed of three parallel east-west streets and four parallel north-south streets: Note there are 12 intersections and 17 street segments. A policeman needs to visit every street segment, but he wants to take the shortest path. The policeman can start at any intersection, and he can only traverse streets, going from one intersection to another. How many street segments are there in the shortest path that visits each street segment at least once? Three riflemen A, B, and C take turns shooting at a target. The first rifleman to hit the target gets 2002 dollars. A shoots first, B second, and C third, after which the cycle repeats again with A, until one of the riflemen hits the target. Each hits the target with probability 0.5. What is rifleman A's expected winnings in dollars? Nine boys and seven girls are seated randomly around a circular table with 16 seats. Find the expected number of girl-boy neighbors. For example, in the seating below there are four such pairs. GBBBBB G B G B GGGBBG

Yellow = 001 it`s my calculation but i`ve also found the same answer in one chinese forum Less

E = sum(k=1, 50) (53-k)(52-k)(51-k)x4xk/(54x53x52x51) = 10

### Game: I throw 1 die 4 times, trying to reach at least one 6, you throw 2 dice 24 times and try to reach at least one double 6 (6,6). Who has greater chance of winning

To estimate, compare (5/6)^4 and (35/36)^24 this is 5/6 and (35/36)^6 this is 30/36 and (35/36)^6 notice that 30/36 is missing six 1/6th from 1 (36/36) and taking powers of (35/36)^6 will reduce the number by nearly 1/36th each time, but less than than, so that (35/36)^6 is greater than 1-6/36=30/36. Therefore the probability of not getting any double six is greater than probability of not getting any 6, and you should choose to roll one die. To understand the reasoning, think about taking powers of 0.90, 0.90^1 = 1-1x0.10 0.90^2 &gt; 1-2x0.10 0.90^3 &gt; 1-3x0.10 and so on Less

It all comes to which is greater: 1-(5/6)^4 or 1-(35/36)^24. They will expect you to calculate this (which is greater, not actual numbers) without a calculator. Less

First comment is correct, second comment is wrong since it asks for at least one 6 or at least one (6,6). This also includes the outcomes 2 or more sixes or double (6,6). Hence the easiest way of calculating this is by calculating the complementary probabilities P(no six) and P(no double six), respectively, to get P(at least one six) = 1 - P(no six) and P(at least one double 6) = 1 - P(no double six), which gives the result in the first comment. Less

### There is a solar system with three planets orbiting around the sun. One of them has a translation period of 60 years, another one of 84 years and another one of 140 years. Today, the three planets are aligned with the sun. When is the next time the three planets will be aligned with the sun?

Should be 210 years

These planets can be aligned on either the same side of the sun or opposite sides. So the answer is a number x that is the least common multiplier of 30, 42 and 70, which is 210. Less

sorry i was wrong

### 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?

http://wolfr.am/1i1XT4P

http://www.johndcook.com/blog/2010/06/17/covariance-and-law-of-cosines/

0.98 &amp; 0.46

### 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?

I remember it being mostly probability and finance. Nothing too terrible.

1 is correct, 2 is wrong. Try again. :)

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. Less