convergence in probability is preserved under continuous transformations
Theorem 1.
Let g:Rk→Rl be a continuous function.
If {Xn} are Rk-valued random variables
converging to X
in probability, then {g(Xn)} converge in probability to
g(X) also.
Proof.
Suppose first that g is uniformly continuous.
Given ϵ>0, there is δ>0 such that
∥g(Xn)-g(X)∥<ϵ whenever
∥Xn-X∥<δ.
Therefore,
ℙ(∥g(Xn)-g(X)∥≥ϵ)≤ℙ(∥Xn-X∥≥δ)→0 |
as n→∞.
Now suppose g is not necessarily uniformly continuous on ℝk.
But it will be uniformly continuous on any compact set
{x∈ℝk:∥x∥≤m} for m≥0.
Consequently, if Xn and X are bounded (by m), then the proof just
given is applicable. Thus we attempt to reduce the general case
to the case that Xn and X are bounded.
Let
fm(x)={x,∥x∥≤mmx/∥x∥,∥x∥≥m |
Clearly, fm:ℝk→ℝk is continuous; in fact, it can be verified that fm is uniformly continuous on ℝk. (This is geometrically obvious in the one-dimensional case.)
Set Xmn=fm(Xn) and Xm=fm(X),
so that Xmn converge to Xm in probability for each m≥0.
We now show that g(Xn) converge to g(X) in probability by a four-step estimate. Let ϵ>0 and δ>0 be given. For any m≥0 (which we will later),
ℙ(∥g(Xn)-g(X)∥≥δ)≤ℙ(∥g(Xmn)-g(Xm)∥≥δ)+ℙ(∥Xn∥≥m)+ℙ(∥X∥≥m). |
Choose M such that for m≥M,
ℙ(∥X∥≥m)≤ℙ(∥X∥≥M)<ϵ4. |
(This is possible since lim.)
In particular, let . Since converge in probability to and , are bounded, converge in probability to . That means for large enough,
Finally, since , and converge to in probability, we have
for large enough .
Title | convergence in probability is preserved under continuous transformations |
---|---|
Canonical name | ConvergenceInProbabilityIsPreservedUnderContinuousTransformations |
Date of creation | 2013-03-22 16:15:05 |
Last modified on | 2013-03-22 16:15:05 |
Owner | stevecheng (10074) |
Last modified by | stevecheng (10074) |
Numerical id | 10 |
Author | stevecheng (10074) |
Entry type | Theorem |
Classification | msc 60A10 |