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 random variable X is the sum of the product of each outcome and its probability.

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)

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

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, E(X) = p and var(X) = p/2 - p^2/2.