Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

5.1. Convergence in Probability 323

Proof:Let>0 be given. Using the triangle inequality, we can write

|Xn−X|+|Yn−Y|≥|(Xn+Yn)−(X+Y)|≥.

SincePis monotone relative to set containment, we have

P[|(Xn+Yn)−(X+Y)|≥ ] ≤ P[|Xn−X|+|Yn−Y|≥ ]

≤ P[|Xn−X|≥/2] +P[|Yn−Y|≥/2].

By the hypothesis of the theorem, the last two terms converge to 0 asn→∞,
which gives us the desired result.

Theorem 5.1.3.SupposeXn→P Xandais a constant. ThenaXn→P aX.

Proof: Ifa= 0, the result is immediate. Supposea =0. Let>0. The result
follows from these equalities:

P[|aXn−aX|≥ ]=P[|a||Xn−X|≥ ]=P[|Xn−X|≥/|a|],

and by hypotheses the last term goes to 0 asn→∞

Theorem 5.1.4.SupposeXn

P

→aand the real functiongis continuous ata.Then

g(Xn)
P
→g(a).

Proof:Let>0. Then sincegis continuous ata,thereexistsaδ>0 such that if
|x−a|<δ,then|g(x)−g(a)|<.Thus

|g(x)−g(a)|≥ ⇒|x−a|≥δ.

SubstitutingXnforxin the above implication, we obtain

P[|g(Xn)−g(a)|≥ ]≤P[|Xn−a|≥δ].

By the hypothesis, the last term goes to 0 asn→∞, which gives us the result.

This theorem gives us many useful results. For instance, ifXn→P a,then

Xn^2

P

→ a^2

1 /Xn →P 1 /a, provideda =0

√

Xn →P

√

a, provideda≥ 0.

Actually, in a more advanced class, it is shown that ifXn
P
→X andgis a
continuous function, theng(Xn)→P g(X); see page 104 of Tucker (1967). We make
use of this in the next theorem.

Theorem 5.1.5.SupposeXn
P
→XandYn
P
→Y.ThenXnYn
P
→XY.