Page 269 - Probability and Statistical Inference

246 5. Concepts of Stochastic Convergence

Here again, it easily follows that as n → ∞, but E(V ) = 1 – 1/n+n →
n
∞ as n → ∞. The relationship between the two notions involving the conver-
gence in probability (that is, as n → ∞) and the convergence in mean
(that is, E(U ) → u as n → ∞) is complicated. The details are out of scope for
n
this book. Among other sources, one may refer to Sen and Singer (1993,
Chapter 2) and Serfling (1980).

{U ; n ≥ 1} may converge to u in probability, but this fact by
n
itself may not imply that E(U ) → u. Also, a fact that E(U )
n n
→ u alone may not imply that . Carefully study the
random variables given in (5.2.6) and (5.2.7).

The Exercise 5.2.22 gives some sufficient conditions under which
one claims: as n → ∞ ⇒ E[g(U )] → g(u) as n → ∞.
n
The Exercise 5.2.23 gives some applications.

In textbooks at this level, the next theorem’s proof is normally left out as
an exercise. The proof is not hard, but if one is mindful to look into all the
cases and sub-cases, it is not that simple either. We include the proof with the
hope that the readers will think through the bits and pieces analytically, thereby
checking each step with the care it deserves.
Theorem 5.2.4 Suppose that we have two sequences of real valued ran-
dom variables {U , V ; n = 1} such that . Then,
n
n
we have:
(i)
(ii) as n → ∞;
(iii) as n → ∞ if P(V = 0) = 0 for all n = 1 and v ≠ 0.
n
Proof (i) We start with an arbitrary but otherwise fixed ε(> 0) and write

The second inequality follows because
implies that [| U – u | + | V – v | < ε], so that [| U – u | + | V – v | ≥ ε]
n n n n