For z in R, the floor function is denoted by [z] = floor(z) and defined as follows: [z] = k, where k is an integer and k...

Question

For z in R, the floor function is denoted by [z] = floor(z) and defined as follows: [z] = k, where k is an integer and k <= z < k+1.

Let Y = [X], where X is an exponentially distributed random variable with mean 1/ln2, where ln denotes natural logarithm. For any positive integer l, one can write the probability of the event Y = l as:

P(Y = l) = q^l * (1 - q)

The value of q is

A. 0.1
B. 0.01
C. ln3
D. 0.5

Accepted Answer

q = 0.5 (answer choice A). X ~ Exp(lambda) with mean 1/ln2, so lambda = ln2. Y = floor(X) takes integer values 0, 1, 2, ... The event Y = l is equivalent to l <= X < l+1. Using the exponential survival function, P(Y = l) = P(l <= X < l+1) = e^{-lambda*l} - e^{-lambda*(l+1)} = e^{-lambda*l}*(1 - e^{-lambda}). With lambda = ln2, e^{-lambda} = 1/2, so P(Y = l) = (1/2)^l * (1 - 1/2) = (1/2)^l * (1/2). For l >= 1 this matches q^l * (1-q) with q = 1/2 = 0.5.