Answer:
the expected value of Xn , E(Xn) = 0 and the variance σ²(Xn) = n*(1-2n)
Step-by-step explanation:
If X1= number of tails when n fair coins are flipped , then X1 follows a binomial distribution with E(X1) = n*p , p=0,5 and the number of heads obtained is X2=n-X1
therefore
Xn =X1-X2 = X1- (n-X1) = 2X1-n
thus
E(Xn) =∑ (2*X1-n) p(X1) = 2*∑[X1 p(X1)] -n∑p(X1) = 2*E(X1)-n = 2*n*p--n= 2*n*1/2 -n = n-n =0
the variance will be
σ²(Xn) = ∑ [Xn - E(Xn)]² p(Xn) = ∑ [(2X1-n) - 0 ]² p(X1) = ∑ (4*X1²-4*X1*n+n²) p(X1) = = 4*∑ X1²p(X1) - 4n ∑X1 p(X1) - n²∑p(X1) = 2*E(X1²) -4n*E(X1)- n²
since
σ²(X1) = n*p*(1-p) = n*0,5*0,5=n/4
and
σ²(X1) = E(X1²) - [E(X1)]²
n/4 = E(X1²) - (n/2)²
E(X1²) = n(n+1)/4
therefore
σ²(Xn) = 4*E(X1²) -4n*E(X1)- n² = 4*n(n+1)/4 - 4*n*n/2 - n² = n(n+1) - 2n² - n²
= n - 2n² = n(1-2n)
σ²(Xn) = n(1-2n)
