A binary communication system is used to send one of two messages:
(i) message A is sent with probability 2/3, and consists of an infinite sequence of zeroes,
(ii) message B is sent with probability 1/3, and consists of an infinite sequence of ones.
The ith received bit is “correct" (i.e., the same as the transmitted bit) with probability 3/4, and is “incorrect" (i.e., a transmitted 0 is received as a 1, and vice versa), with probability 1/4. We assume that conditioned on any specific message sent, the received bits, denoted by Y1,Y2,… are independent.
Is Y2+Y3 independent of Y1?
Is Y2−Y3 independent of Y1?
Is Y2+Y3 independent of Y1? NO!
Is Y2−Y3 independent of Y1? YES!
1. Is Y2+Y3 independent of Y1?
NO
2.Is Y2−Y3 independent of Y1?
YES
To determine if Y2+Y3 or Y2−Y3 is independent of Y1, we need to verify if the conditional probability distributions satisfy the definition of independence.
Let's start by calculating the conditional probability distribution for Y2+Y3 given Y1.
P(Y2+Y3 | Y1 = 0) = P(Y2+Y3 = 0 | Y1 = 0) * P(Y1 = 0) + P(Y2+Y3 = 1 | Y1 = 0) * P(Y1 = 1)
= P(Y2 = 0, Y3 = 0 | Y1 = 0) * P(Y1 = 0) + P(Y2 = 1, Y3 = 0 | Y1 = 0) * P(Y1 = 1)
= (P(Y2 = 0 | Y1 = 0) * P(Y3 = 0 | Y1 = 0)) * P(Y1 = 0) + (P(Y2 = 1 | Y1 = 0) * P(Y3 = 0 | Y1 = 0)) * P(Y1 = 1)
Using the given probabilities, we have:
P(Y2+Y3 | Y1 = 0) = (3/4 * 3/4) * (2/3) + (1/4 * 3/4) * (1/3)
= 9/16 * 2/3 + 3/16 * 1/3
= 27/48 + 3/48
= 30/48
= 5/8
Now, let's calculate the unconditional probability of Y2+Y3:
P(Y2+Y3) = P(Y2+Y3 = 0) * P(Y1 = 0) + P(Y2+Y3 = 1) * P(Y1 = 1)
= P(Y2 = 0, Y3 = 0) * P(Y1 = 0) + P(Y2 = 1, Y3 = 0) * P(Y1 = 1)
= (P(Y2 = 0) * P(Y3 = 0)) * (2/3) + (P(Y2 = 1) * P(Y3 = 0)) * (1/3)
Using the given probabilities, we have:
P(Y2+Y3) = (3/4 * 3/4) * (2/3) + (1/4 * 3/4) * (1/3)
= 9/16 * 2/3 + 3/16 * 1/3
= 27/48 + 3/48
= 30/48
= 5/8
Since P(Y2+Y3 | Y1) = P(Y2+Y3), we can conclude that Y2+Y3 is independent of Y1.
Now let's check if Y2−Y3 is independent of Y1 in a similar manner.
P(Y2−Y3 | Y1 = 0) = P(Y2−Y3 = 0 | Y1 = 0) * P(Y1 = 0) + P(Y2−Y3 = 1 | Y1 = 0) * P(Y1 = 1)
= P(Y2 = Y3 | Y1 = 0) * P(Y1 = 0) + P(Y2 ≠ Y3 | Y1 = 0) * P(Y1 = 1)
Using the given probabilities, we have:
P(Y2−Y3 | Y1 = 0) = (3/4 * 1/4) * (2/3) + (1/4 * 3/4) * (1/3)
= 3/16 * 2/3 + 3/16 * 1/3
= 6/48 + 3/48
= 9/48
= 3/16
Now let's calculate the unconditional probability of Y2−Y3:
P(Y2−Y3) = P(Y2−Y3 = 0) * P(Y1 = 0) + P(Y2−Y3 = 1) * P(Y1 = 1)
= P(Y2 = Y3) * (2/3) + P(Y2 ≠ Y3) * (1/3)
Using the given probabilities, we have:
P(Y2−Y3) = (3/4 * 1/4) * (2/3) + (1/4 * 3/4) * (1/3)
= 3/16 * 2/3 + 3/16 * 1/3
= 6/48 + 3/48
= 9/48
= 3/16
Since P(Y2−Y3 | Y1) = P(Y2−Y3), we can conclude that Y2−Y3 is independent of Y1.