Question

Suppose X1,X2, ..., Xn be a sequence of idpendently identically distributed random variables with mean 4 and variance 16. Let S =X1+X2+...+Xn, Z1=sqrt(n)S/4, Z2=(S-4)/16, and Z3=sqrt(n)(S/n-4)/4. Then Select one:

a. Z1 converges in prpbability to Normal(0,1) as n increases to infinity.
b. Z3 converges in prpbability to Normal(0,1) as n increases to infinity.
c. Z2 converges in prpbability to Normal(0,1) as n increases to infinity.
d. None of the other alternatives.

Answer 1

Answer :-

a. Z1 converges in prpbability to Normal(0,1) as n increases to infinity.

Answer 2

Step-by-step explanation:

Convergence in probability is stronger than convergence in distribution. In particular, for a sequence X1, X2, X3, ⋯ to converge to a random variable X, we must have that P(|Xn−X|≥ϵ) goes to 0 as n→∞, for any ϵ>0. To say that Xn converges in probability to X, we write

Xn

p

→

X.

Here is the formal definition of convergence in probability: