Page 270 - Probability and Statistical Inference

5. Concepts of Stochastic Convergence 247

implies [| U – u | ≥ ½ ε ∪ | V – v | = 1/2ε]. The third inequality follows
n
n
because P(A ∪ B) ≤ P(A) + P(B) for any two events A and B. From the fact
that as n → ∞, both the probabilities in the last step in (5.2.8)
converge to zero as n → ∞, and hence the lhs in (5.2.8), which is non-
negative, converges to zero as n → ∞, for all fixed but arbitrary ε(> 0). Thus,
as n → ∞. The case of U – V can be tackled similarly. It
n
n
is left as the Exercise 5.2.6. This completes the proof of part (i). "
(ii) We start with arbitrary but otherwise fixed ε(> 0) and write

Case 1: u = 0, v = 0
Observe that implies that [| U V | < ε], that
n n
is [| U V | ≥ ε] implies . Hence, we have
n n
,
and both these probabilities converge to zero as n → ∞, and
thus P{| U V | ≥ ε} → 0 as n → ∞.
n n
Case 2: u ≠ 0, v = 0
Observe by triangular inequality that | U | = | (U – u) + u | ≤ | U – u | +
n
n
n
| u | and hence P{| U | ≥ 2 | u |} ≤ P{| U – u | + | u | ≥ 2 | u |} = P{| U – u
n n n
| ≥ | u |} which converges to zero as n → ∞, since as n → ∞. Thus,
we claim that P{| U | ≥ 2 | u |} → 0 as n → ∞. Let us now write
n

But, we had verified earlier that the first term in the last step converges to zero
as n → ∞, while the second term also converges to zero as n → ∞ simply
because as n → ∞. Thus the lhs converges to zero as n → ∞.
Case 3: u = 0, v ≠ 0