Question

A subcarrier is one telecommunication signal carrier that is carried on top of another carrier so that effectively two signals are carried at the same time. Subcarriers can be used for an entirely different purpose than main carriers. For example, data subcarriers are used for data transmissions; pilot subcarriers are used for channel estimation and synchronization; and, null subcarriers are used for direct current and guard banks transmitting no signal. In the IEEE Journal of Oceanic Engineering (April, 2013), researchers studied the characteristics of subcarriers for underwater acoustic communications. Based on an experiment conducted off the coast of Martha’s Vineyard (MA), they estimated that 25% of subcarriers are pilot subcarriers, 15% are null subcarriers, and 60% are data subcarriers. Consider a sample of 60 subcarriers transmitted for underwater acoustic communications. a) What is the probability that you observe 15 pilot subcarriers, 20 null subcarriers, and 25 data subcarriers? b) How many of the 60 subcarriers do you expect to be pilot subcarriers? Null subcarriers? Data subcarriers? c) What is the variance of the number of pilot subcarriers?

Answer 1

We are given here that:
P( pilot subcarriers) = 0.25,
P( null subcarriers) = 0.15,
P( data subcarriers) = 0.6

a) The probability that we observe 15 pilot subcarriers, 20 null subcarriers, and 25 data subcarriers is computed here using the multinomial distribution as:

Therefore 0.000015 is the required probability here.

b) Expected number of 60 subcarriers who are the various types is computed here as:

pilot subcarriers: P( pilot subcarriers) * Total sample size = 0.25*60 = 15
null subcarriers: P( null subcarriers) * Total sample size = 0.15*60 = 9
data subcarriers: P( data subcarriers) * Total sample size = 0.6*60 = 36

These are the expected values here.

c) The variance of the number of pilot subcarriers is computed using the variance of a binomial variable Bin(n = 60, p = 0.25) as:

Variance = np(1-p) = 60*0.25*0.75 = 11.25

Therefore 11.25 is the required variance here.