Quantitative research analyst Interview Questions

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How'd you get 0.98 and 0.46?

NND correlation matrix --> det(\Sigma)>=0 --> 0.98 and 0.46

minimum: 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) = 0.4585

How do you know this

12

sum of the series 66= n/2( 2*1 + (n-1)*1) n=12

You 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=12

You 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

4 is correct. If the question had asked to consecutive H (or T) then the answer would've been 6.

I would rather say 2. Consider each possible outcome of 2 consecutive draws : HH, TT, TH, HT so probability of getting 2 consecutive faces is 1/2 and therefore expected time to get 2 consecutive faces is 2. It would have been 4 if we were entitled to get either heads or tails.

expected number of coin tosses until 2 consecutive heads. Denote E(2H) = n = expected number of coin tosses until 2 consecutive heads E(2H|1=H) = expected number of coin tosses until 2 consecutive heads if the 1st toss is H E(2H|1=T) = expected number of coin tosses until 2 consecutive heads if the 1st toss is T = E(2H) + 1 //since we’ve already flipped 1 //and we have to start over whenever we get a tail —> E(2H)= n = E(2H|1=H) * P(1=H) + E(2H|1=T)*P(1=T) = (E(2H|1=H,2=H) * P(2=H) + E(2H|1=H,2=T)*P(2=T))*P(1=H) + (n+1)*0.5 = (2 * 0.5 + (n+2)*0.5)*0.5 + (n+1)*0.5 = 1.5 + 0.75n —> n = 6

The mean of a geometric distribution is (1/p) where p is the probability of success for any given trial. Group two tosses as a trial, and the probability to get two consecutive faces is p = (1/4). Hence answer is 4.

Your first flip happens, can't get a pair yet. Your next flip has an EV_pairing = .5, and if you miss, your next flip STILL has an EV of .5 for pairing. EV_pairing = 1 after 3 flips.

can 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=4/11

The 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

Dc

[ z * sigma_y^ (-2) ] / [ sigma_x^ (-2) + sigma_y^ (-2) ] From signal processing point of view, x is the signal, y is the noise, and z is the observation. We know X has a prior distribution X ~ N(0, sigma_x^ 2 ), noise Y has distribution Y ~ N(0, sigma_y^ 2 ) and the value Z = z, the questions is what is the MMSE estimate of X given Z, i.e., E(X|Z)? Using Bayesian theorem, or Gauss Markov Theorem, one can show that : E(X|Z) = [ z * sigma_y^ (-2) + 0 * sigma_x^ (-2) ] / [ sigma_x^ (-2) + sigma_y^ (-2) ] Comments: 1. This kind of problems are very common so please keep it in mind in Gaussian case the best estimate of X is a weighted linear combination of maximum likelihood estimate (z in this problem ) and the prior mean (0 in this problem). And the weights are the the inverse of variance. 2. In multi dimension cases where x, y, z are vectors, similar rules also apply. Check Gauss Markov Theorem 3. In tuition here is the larger variance of noise y, the less trust we will assign on ML estimate, which is sigma_y^ (-2) . Correspondingly, the more trust we put on the prior of X.

[ z * sigma_y^ (-2) ] / [ sigma_x^ (-2) + sigma_y^ (-2) ] From signal processing point of view, x is the signal, y is the noise, and z is the observation. We know X has a prior distribution X ~ N(0, sigma_x^ 2 ), noise Y has distribution Y ~ N(0, sigma_y^ 2 ) and the value Z = z, the questions is what is the MMSE estimate of X given Z, i.e., E(X|Z)? Using Bayesian theorem, or Gauss Markov Theorem, one can show that : E(X|Z) = [ z * sigma_y^ (-2) + 0 * sigma_x^ (-2) ] / [ sigma_x^ (-2) + sigma_y^ (-2) ] Comments: 1. This kind of problems are very common so please keep it in mind in Gaussian case the best estimate of X is a weighted linear combination of maximum likelihood estimate (z in this problem ) and the prior mean (0 in this problem). And the weights are the the inverse of variance. 2. In multi dimension cases where x, y, z are vectors, similar rules also apply. Check Gauss Markov Theorem 3. In tuition here is the larger variance of noise y, the less trust we will assign on ML estimate, which is sigma_y^ (-2) . Correspondingly, the more trust we put on the prior of X.

Z sigma_x^2/ (sigma_x^2 + sigma_y^2), similar to CAPM

If you don't know above theorems you can use good old bayes, P(X|Z) = P(Z|X)P(X)/P(Z) and set derivative=0, since you have pdfs of X and Z. But it's really messy and I don't wanna do it.

For one time evet, the probability is 0

I 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 0

nothing 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 = 0.5

try 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.

$1

Expectation under any measure is equivalent. That's why we change measures to calculate things as the result will be same. Answer is 1

It'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.

what 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 = 2/3

a stopped martingale is still a martingale. Ans: 2/3