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5. Concepts of Stochastic Convergence 243
σ < ∞. Write for the sample mean. Then, as n →
∞.
Proof Consider arbitrary but otherwise fixed ε (> 0) and use Tchebysheffs
inequality (Theorem 3.9.3) to obtain
2
2
Thus, we have 0 ≤ P{| µ |≥ ε } ≤ σ /(nε ) → 0 as n → ∞. Hence, P
{| µ | ≥ε} → 0 as n → ∞ and by the Definition 5.2.1 one claims that
!
Intuitively speaking, the Weak WLLN helps us to conclude that the sample
mean of iid random variables may be expected to hang around the popu-
lation average µ with high probability, if the sample size n is large enough and
0 < σ < ∞.
2
Example 5.2.1 Let X , ..., X be iid N(µ, σ ), ∞ < µ < ∞, 0 < σ < ∞, and write
1 n
for n ≥ 2. Then, we show
that as n→∞. From (4.4.9) recall that we can express as
where Y ,..., Y are the Helmert variables. These Helmert variables
2 n
are iid N(0, σ ). Observe that ,
2
which is finite. In other words, the sample variance has the representation
of a sample mean of iid random variables with a finite variance. Thus, the
Weak WLLN immediately implies that as n→∞. !
Under very mild additional conditions, one can conclude that
as n → ∞, without the assumption of normality of the Xs.
Refer to Example 5.2.11.
Example 5.2.2 Let X , ..., X be iid Bernoulli(p), 0 < p < 1. We know that
1
n
E(X ) = p and V(X ) = p(1 p), and thus by the Weak WLLN, we conclude
1
1
that as n → ∞. !
In the following, we state a generalized version of the Weak WLLN. In
some problems, this result could come in handy.
Theorem 5.2.2 Let {T ; n ≥ 1} be a sequence of real valued random
n
variables such that with some r(> 0) and a ∈ℜ, one can claim that ξ = E{|T
r,n n
r
a | } → 0 as n → ∞. Then, as n → ∞.