Let X and Y be two independent and identically distributed random variables that take only positive integer values. Their PMF is px (n) = py (n) = 2^-n for every n e N, where N is the set of positive integers. 1. Fix at E N. Find the probability P (min{X,Y} Y). Hint: Use your answer to the previous part, and symmetry. 4. Fix a positive integer k. Find the probability P (X KY). Your answer should be a function of k. Hint: You may find the following facts useful. • 2-k = 2-3+1 • For a geometric series, An = aj pn-1, we have m-1 an = a1/(1-r).

Question

Let X and Y be two independent and identically distributed random variables that take only positive integer values. Their PMF is px (n) = py (n) = 2^-n for every n e N, where N is the set of positive integers. 1. Fix at E N. Find the probability P (min{X,Y} <t). Your answer should be a function of t. 2. Find the probability P (X = Y). 3. Find the probability P(X>Y). Hint: Use your answer to the previous part, and symmetry. 4. Fix a positive integer k. Find the probability P (X > KY). Your answer should be a function of k. Hint: You may find the following facts useful. • 2-k = 2-3+1 • For a geometric series, An = aj pn-1, we have m-1 an = a1/(1-r).

Anonymous · Accepted Answer

1. 1-(1/(2^((2*t)-2))) 2. 1/3 3. 1/3 4. 2/((2^(k+1))-1)

a · Answer

1) 1-(1-CDFx)(1-CDFy) = 1- ((1-CDFx)^2) = 1- ((1-CDFx)^2) = 1-2^(-2t)

Anonymous · Answer

Integration of a^n is a^n/(ln(a)). So why doesn't the answer have ln in it ?

anon · Answer

integration is not needed as these are discrete RVs

Anonymous · Answer

How come I get? 2)1/2 3)1/4

Anonymous · Answer

2) Sum[P(Y=y)*P(X=y)] = Sum[1/2^(-2y)] = 1 - 1/2 = 1/2, not sure if this is correct.

Anonymous · Answer

Sum[P(Y=y)*P(X=y)] = Sum[2^(-2y)] = 1 - 1/2 = 1/2, not sure if this is correct.

Anonymous · Answer

why not the 1) (1-(1/2)^t)^2