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5. Concepts of Stochastic Convergence 263
Notice that Ys are iid with mean = E(Y ) = σ and variance = E(Y ) E (Y )
2
2
2
1
1
1
= µ σ which is assumed finite and positive. Hence, by the CLT, we have
4
4
as n → ∞. Also, from the Example 5.2.10 it follows that
as n → ∞. Thus by Slutskys Theorem, part (i), we conclude that
as n → ∞. Next, we write
and reapply Slutskys Theorem. The result then follows since
and as n → ∞, so that as n
→ ∞. !
Remark 5.3.2 One can obtain the result mentioned in the Example 5.3.6
from this theorem by noting that µ = 3σ in the normal case.
4
4
Example 5.3.9 Under the setup of the Theorem 5.3.6, using Example
5.2.11 and Slutskys Theorem, we can immediately conclude the following
result when the Xs are iid but non-normal:
as n → ∞. In the literature, µ σ is traditionally denoted by
4
4
β , which is customarily referred to as a measure of the kurtosis in the parent
2
population. !
The following result is along the lines of the Theorem 5.2.5. It shows that
the property of the convergence in distribution is preserved under a continu-
ous transformation. We state it without giving its proof.
Theorem 5.3.7 Suppose that we have a sequence of real valued random
variables {U ; n ≥ 1} and another real valued random variable U. Suppose
n
also that as n → ∞. Let g(.) be a real valued continuous function.
Then, as n → ∞.
Example 5.3.10 Let X , ..., X be iid real valued random variables having
n
1
the common mean µ and variance σ , ∞ < µ < ∞ and 0 < σ < ∞. From the
2
CLT we know that as n → ∞. Thus, using the
2
Theorem 5.3.7 with g(x) = x , x ∈ ℜ, we can immediately conclude that
as n → ∞, since the square of a standard normal
variable has a Chi-square distribution with one degree of freedom.!
Example 5.3.11 (Example 5.3.10 Continued) Let X , ..., X be iid real
1
n
valued random variables having the common mean µ and variance σ , ∞
2
< µ < ∞ and 0 < σ < ∞. From the CLT we know that
N(0, 1) as n → ∞. Thus, using the Theorem 5.3.7 with g(x) = |x|, x ∈ ℜ,