Suppose X is a 0/1 random variable (i.e. a Bernoulli random variable) such that  P ( X = 1 ) = p,   and   P ( X = 0 ) = 1 − p . (a) Derive E( X ).(b) Derive var( X ).Now suppose X = ( X1 + X2 ) / 2, where X1 and X2 are independent and  P ( Xi = 1 ) = p,   and   P ( Xi = 0 ) = 1 − p  where i can be 1 or 2. (a) Derive E( X ).(b) Derive var( X ).

(a) Derive E(X):

The expected value E(X) of a Bernoulli random variable X is given by the probability of success (X = 1), which is p. So, E(X) = p.

(b) Derive var(X):

The variance var(X) of a Bernoulli random variable X is given by the product of the probability of success (p) and the probability of failure (1 - p). So, var(X) = p(1 - p).

Now suppose X = (X1 + X2) / 2, where X1 and X2 are independent and P(Xi = 1) = p, and P(Xi = 0) = 1 - p where i can be 1 or 2.

(a) Derive E(X):

The expected value E(X) is the average of the expected values of X1 and X2. Since X1 and X2 are independent and identically distributed, their expected values are both p. So, E(X) = (E(X1) + E(X2)) / 2 = (p + p) / 2 = p.

(b) Derive var(X):

The variance var(X) is the average of the variances of X1 and X2, because the variance of the average of independent random variables is the average of their variances. Since X1 and X2 are independent and identically distributed, their variances are both p(1 - p). So, var(X) = (var(X1) + var(X2)) / 4 = (p(1 - p) + p(1 - p)) / 4 = p(1 - p) / 2.

First, let's derive E(X) and var(X) for a Bernoulli random variable X.

(a) Derive E(X)

The expected value E(X) of a random variable X is the sum of the product of each outcome and its probability.

E(X) = 1p + 0(1-p) = p

(b) Derive var(X)

The variance var(X) of a random variable X measures how spread out the numbers are from their expected value.

var(X) = E(X^2) - (E(X))^2 = [1^2p + 0^2(1-p)] - p^2 = p - p^2

Now, let's derive E(X) and var(X) for X = (X1 + X2) / 2, where X1 and X2 are independent.

(a) Derive E(X)

Given X = (X1 + X2) / 2, we can find E(X) by finding the expected values of X1 and X2 and then averaging them, since X1 and X2 are independent.

E(X1) = 1p + 0(1-p) = p E(X2) = 1p + 0(1-p) = p

So, E(X) = (E(X1) + E(X2)) / 2 = (p + p) / 2 = p

(b) Derive var(X)

Given X = (X1 + X2) / 2, we can find var(X) by finding the variances of X1 and X2 and then averaging them, since X1 and X2 are independent.

var(X1) = E(X1^2) - (E(X1))^2 = [1^2p + 0^2(1-p)] - p^2 = p - p^2 var(X2) = E(X2^2) - (E(X2))^2 = [1^2p + 0^2(1-p)] - p^2 = p - p^2

So, var(X) = (var(X1) + var(X2)) / 2^2 = (p - p^2 + p - p^2) / 4 = (2p - 2p^2) / 4 = p/2 - p^2/2

Therefore, for a Bernoulli random variable X, E(X) = p and var(X) = p - p^2. For X = (X1 + X2) / 2, where X1 and X2 are independent, E(X) = p and var(X) = p/2 - p^2/2.