Let X and Y be independent Bernoulli random variables with parameter 1/2, and let Z be the random variable that returns the remainder of the division of X + Y by 2.

(a) Prove that Z is also a Bernoulli random variable, also with parameter 1/2.
(b) Prove that X; Y;Z are pairwise independent but not mutually independent.
(c) By computing Var[X+Y +Z] according to the alternative formula for variance and using the variance of Bernoulli r. v.'s, verify that Var[X +Y +Z] =Var[X]+Var[Y ]+Var[Z]

** Kindly use the properties of Bernoulli random variables for solving questions. Thanks!

a) Since Z is a remainder of a divition by 2, then it can only take the values 0 and 1, thus it is a bernoulli random variable. Z is 1 when either X or Y is 1 and the other is 0, therefore

P(Z=1) = P(X = 1 and Y = 0) + P(X = 0 and Y=1)

For independence of X and Y , that sum is equal to

P(X=1)*P(Y=0) + P(X=0) *P(Y=1) = 1/2*1/2 + 1/2*1/2 0 1/4 + 1/4 = 1/2.

The probability for success is 1/2, so effectively Z is a bernoulli with parameter 1/2.

b) X and Y are independent because we are given that information.

If we prove that X and Z are independent, then we can prove with a similar argument that Y and Z are.

Lets compute every possible probability. Note that since both X and Z are bernoullis with parameter 1/2, then P(X=k)*P(Z = j) = 1/2*1/2 = 1/4 for k,j in {0,1}.

P(X = 0 and Z = 0): if X is 0, then Z is 0 only if Y is 0 (otherwise it would be 1), thus

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

P(X=0, Z=1) = (for Z to be 1 given than X is 0, we need Y to be 1) P(X=0 and Y=1) = P(X=0) *P(Y=1) = 1/2*1/2 = 1/4

P(X=1, Z=0) = P(X=1, Y=1) = P(X=1)*P(Y=1) = 1/4 (If X is 1, then Z is 0 when Y is also 1)

P(X=1,Z=1) = P(X=1,Y=0) = P(X=1)*P(Y=0) = 1/4 (If X is 1, then Z is 1 only if Y is 0)

This show that X and Z are independent (hence so are Y and Z)

However, Z is heavily dependant on X and Y, for example, If X and Y are both 1, then Z should be 0, this means that

P(X=1,Y=1,Z=1) = 0

but P(X=1)*P(Y=1)*P(Z=1) = 1/2*1/2*1/2 = 1/8 ≠ 0

Thus, X, Y and Z are pairwise independent but not mutually independent.

c)

Since X is bernoully with parameter 1/2, then

V(X) = 1/2-(1/2)² = 1/4

Therefore,

V(X) + V(Y) + V(Z) = 1/4*3 = 3/4

On the other hand,

V(X+Y+Z) = E( ( X + Y + Z )² ) + E(X+Y+Z)²

X+Y+Z could take any value from 0 to 3. since X, Y and Z cant take the value 1 at the same time, then X+Y+Z cant be 3. It cant also be 1, because if X and Y are both 0, then so is Z, and if X is 1 and Y is 0 (or otherwise), then Z is 1.

Thus, X+Y+Z can take the values 0 and 2. It only takes the value 0 when X, Y and Z is 0, and that happens only if X and Y are 0. Therefore

P(X+Y+Z = 0) = P(X = 0, Y=0, Z=0) = P(X=0,Y=0) = P(X=0)*P(Y=0) (independence) = 1/2*1/2 = 1/4.

and, as a result,

P(X+Y+Z=2) = 1-1/4 = 3/4.

As a consequence,

E(X+Y+Z)²= (0*1/4+3/4*2)² = (3/2)² = 9/4

On the other hand, (X+Y+Z)² can take the values 0, with probability 1/4 (When X+Y is 0), and 2² = 4, with probability 3/4 (When X+Y is 2)

Thus, E((X+Y+Z)²) = 0*1/4+4*3/4 = 3

We conclude that

V(X+Y+Z) = 3-9/4 = 3/4 = V(X)+V(Y)+V(Z)

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