A communication channel accepts the input X ?{0,1, 2,3} and outputs Y=X+Z whereZ is a binary random variable taking values -1 and +1 with equal probability. AssumeX and Z are independent and all values of the input X have equal probability.a) Find the entropy of Y.b) Find the entropy of X given that Y=1.

To find the entropy of Y, we need to determine the probability distribution of Y and calculate the entropy using the formula:

H(Y) = -Σ P(Y=y) * log2(P(Y=y))

Since the output Y = X + Z, we can calculate the possible values of Y for each value of X:

When X = 0, Y can take two values: 0+1=1 and 0-1=-1.
When X = 1, Y can take two values: 1+1=2 and 1-1=0.
When X = 2, Y can take two values: 2+1=3 and 2-1=1.
When X = 3, Y can take two values: 3+1=4 and 3-1=2.

Now let's calculate the probabilities for each value of Y:

P(Y=1) = P(X = 0) * P(Z = 1) = 1/4 * 1/2 = 1/8
P(Y=-1) = P(X = 0) * P(Z = -1) = 1/4 * 1/2 = 1/8
P(Y=2) = P(X = 1) * P(Z = 1) = 1/4 * 1/2 = 1/8
P(Y=0) = P(X = 1) * P(Z = -1) + P(X = 2) * P(Z = 1) = 1/4 * 1/2 + 1/4 * 1/2 = 1/8
P(Y=3) = P(X = 2) * P(Z = -1) = 1/4 * 1/2 = 1/8
P(Y=4) = P(X = 3) * P(Z = 1) = 1/4 * 1/2 = 1/8

Now we can calculate the entropy of Y:

H(Y) = - [P(Y=1) * log2(P(Y=1)) + P(Y=-1) * log2(P(Y=-1))
+ P(Y=2) * log2(P(Y=2)) + P(Y=0) * log2(P(Y=0))
+ P(Y=3) * log2(P(Y=3)) + P(Y=4) * log2(P(Y=4))]

H(Y) = - [(1/8) * log2(1/8) + (1/8) * log2(1/8) + (1/8) * log2(1/8)
+ (1/8) * log2(1/8) + (1/8) * log2(1/8) + (1/8) * log2(1/8)]

H(Y) = - [(-3/8) * log2(1/8)]

H(Y) = 3/8 * log2(8)

H(Y) = 3/8 * 3 = 9/8

So, the entropy of Y is 9/8.

To find the entropy of X given that Y=1, we need to calculate the conditional probability distribution of X given Y=1 and then compute the entropy using the formula:

H(X|Y=1) = -Σ P(X=x|Y=1) * log2(P(X=x|Y=1))

Given that Y=1, we can determine the possible values of X:

When Y=1, X can be 0 or 2.

Now let's calculate the probabilities for each value of X given Y=1:

P(X=0|Y=1) = P(X=0 and Y=1) / P(Y=1)
= P(X=0) * P(Z=1) / P(Y=1)
= (1/4) * (1/2) / (1/8)
= 1

P(X=2|Y=1) = P(X=2 and Y=1) / P(Y=1)
= P(X=2) * P(Z=-1) / P(Y=1)
= (1/4) * (1/2) / (1/8)
= 1

Now we can calculate the entropy of X given Y=1:

H(X|Y=1) = - [P(X=0|Y=1) * log2(P(X=0|Y=1)) + P(X=2|Y=1) * log2(P(X=2|Y=1))]

H(X|Y=1) = - [1 * log2(1) + 1 * log2(1)]

H(X|Y=1) = - [0 + 0]

H(X|Y=1) = 0

So, the entropy of X given Y=1 is 0.