AKUNA CAPITAL Interview Question: Let X and Y be random variabl... | Glassdoor.co.in

Interview Question

Quantitative Research Interview Boston, MA (US)

Let X and Y be random variables with standard deviation a

  and b. What are the bounds of Var(X + Y)? Pile of 100 coins, 99 fair 1 double heads. Pick random coin, flip 10 times, get 10 heads. What is the probability the coin is double headed coin? Dice sum games.
Answer

Interview Answer

2 Answers

0

1. Var(X+Y) = Var(X) + Var(Y) + 2Cov(X, Y) <= Var(X) + Var(Y) + 2*sqrt[Var(X)*Var(Y)] = a^2 + b^2 + 2ab = (a+b)^2

2. Use bayesian theorem:
event A is pick double heads coin
event B is flip 10 times and get 10 heads
Given event A then probability of event B is 1^10, i.e. P(B|A) = 1
P(A|B) = P(B|A)P(A)/P(B) = 1*(1/100) / [1*(1/100) + (1/2)^10 * (99/100)] = 1024/(1024+99) = 1024/1123 = 0.9118

Di Yang on 08-Jan-2020
0

Refinement on the first answer to Q1. Actually it also has a lower bound, (a-b)^2. Upper bound is correct, (a+b)^2.

Hess on 05-Feb-2020

Add Answers or Comments

To comment on this, Sign In or Sign Up.