convergence in probability is preserved under continuous transformations
Suppose first that is uniformly continuous. Given , there is such that whenever . Therefore,
Now suppose is not necessarily uniformly continuous on . But it will be uniformly continuous on any compact set for . Consequently, if and are bounded (by ), then the proof just given is applicable. Thus we attempt to reduce the general case to the case that and are bounded.
Clearly, is continuous; in fact, it can be verified that is uniformly continuous on . (This is geometrically obvious in the one-dimensional case.)
Set and , so that converge to in probability for each .
We now show that converge to in probability by a four-step estimate. Let and be given. For any (which we will later),
Choose such that for ,
(This is possible since .)
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 .
Collecting the previous inequalities together, we have
for large enough . ∎
|Title||convergence in probability is preserved under continuous transformations|
|Date of creation||2013-03-22 16:15:05|
|Last modified on||2013-03-22 16:15:05|
|Last modified by||stevecheng (10074)|