Let X1 X2 X3 be aniid sequence of Bernoulli random variable

Let X1, X2, X3, ... be ani.i.d. sequence of Bernoulli random variables where Xi = 0 with probability 0.5 and xi = 1 with probability 0.5. For a fixed value of N, define X = X1 + X2 = X3 + ... + Xn / N Find mux = E[X) and sigma^2 X = Var(X). Determine an N that guarantees that Pr(|X - mu X| GE 0.01) LE 0.0001.

Solution

Given X1,X2, ......., Xn be a i.i.d sequence of bernoulli random variables where Xi= 0, with p=0.5 and Xi=1, with p=0.5, then E(Xi)=0.5, E(Xi²)=0.5 and V(Xi)=0.25.

(a):_xbar=E[Xbar]=E[(X1+X2+X3+.........+XN)/N]=[E(X1)+E(X2)+.......+E(XN)]/N=[(0.5+.5+.5+.......+0.5)/N]=[N(0.5)]/N=0.5. V(Xbar)=E(Xbar²)-[E(Xbar)]²=0.5-(0.5)²=0.25.

(b):By the theorem of Bernoulli\'s law of large numbers, for any €(epslon)>0, then P{|(X/n)-p|>=€}<=1/(4n£²)

P{|Xbar-_xbar|>=0.01}<=0.0001=>1/(4*N*(0.01)²)=0.0001=>N=25000000.