Quantitative research analyst Interview Questions

This is equivalent to Team A wins 3 out of the first 6 games. It's now a simple binomial problem.

to make it more details in the first answer: 20/64 = 5/16 = 31.25%

Yeah this is binomial distribution, you will need exactly 3 games from 6 to win , so assuming both teams are equally likely to win, 6C3 * .5^3 * .5^3 = .3125

14

can be found using markov chains and fundamental matrics matlab code A = [1/2 1/2 0; 1/6 1/2 1/3; 0 1/3 1/2]; E=[1 0 0; 0 1 0; 0 0 1]; X=(E-A)^(-1); P = [1; 1; 1]; C = X*P; PS = [1/8 3/8 3/8]; F = PS* C; display(C); display(F); so 14.5 average number of steps

14，2^(n+1)-2,with n denotes number of heads

100% in stock 2, with the smaller variance.

No, if you can short, the variance could be reduced even further since the stocks are correlated.

17/44 of the stock with variance of .3 and 27/44 from the stock with variance of .2

161/36

if D1=D2: EV1 = sum_{i=1}^6 i/36=21/36 O.W.: EV2 = sum_{i=2}^6 i(i-1)/36=140/36 => EV=EV1+EV2=161/36;

161/36. I did it by finding pdf of max(x1,x2).

Mapped the probability measure to a 3D polytope and calculated the center of mass.

0.75 by common sense

Let L and R be the left and right half of the broken stick (resp.) The probability that L is longer than R is equal to the probability that R is longer than L (these halves are symmetric, and so this probability is 1/2). This implies that the expected length of the longer side is 3/4.

No. You should be the second to play. The symmetry of round table guarantees the space for a 2nd coin for every coin placed.

be the first to play with the coin dead center Then the symmetry rules will apply for the the rest of the coins

be the first to play with the coin dead center Then the symmetry rules will apply for the the rest of the coins

In this line of work, it's harder to grab an intern position than a fulltimer

To give a sense of the "right" answer, I'm sure they're looking for you to explain what calculations you use to arrive at your answer and may also ask you for a 95% confidence window (note: people naturally are over-confident and are prone to giving confidence windows that are far too small). Google did a study in 2010 and estimated that 129,864,880 books have been published. Additionally, Wikipedia estimates that currently, an additional 2.2M titles are being published each year.

I love fermi problems. OK, step one: Lets say X is the fraction of people who are writers in a given population, and assume that this is constant over time. Lets say that each writer authors B books per year on average, and assume they all have distinct titles. Then Q=X*B is the average number of books per person per year under ideal conditions. Next we need to figure out how many people have lived since the advent of writing. We'll say that books started appearing 500 years ago with appreciable volume with the advent of the printing press. The current population is approximately 6 billion = 6* 10^9. 10000 years ago, the population was probably around 600,000. Populations grow exponentially in time so population ~ e^(k t), so we can solve for the growth rate as 6 10^9 =6* 10^5 e^(k (10^5)years) to get population growth rate = k~= 4 ln(10)/10^5 The population at the advent of the printing press was then P0~6*10^6.6 Since book growth rate is proportional to population growth rate, if there were B books 500 years ago, then there are B * \int_0^500 e^(k t)dt books now. The integral evaluates to 10^5*(10^0.2 -1)/4ln(10). We have to approximate 10^0.2 =(10^2)^(1/5) . The fifth root of 100 must be between 2 and 3, and 2.5 is decent enough. Then the total number of books now is B* (1.5 *10^5)/4ln(10) ~ B*10^4. We can say that during the first year after the printing press the number of books written should be about Q*P0. If we say that 1 in 100 people are writers and they publish 1 book per year on average we get B~ 10^4.6 Then the total number of titles ever written should be about 10^8.6. This is close to the actual result

By Ito W^n is a martingale for n=0,1

Agree, for any n>1 you get a drift term

Xt is a martingale you get f(Xt) is a martingale if f is linear otherwise convexity or concavity will increase or decrease the expectation, think of Jensen inequality.