convergence in probability is preserved under continuous transformations

Theorem 1.

Let $g\\colon\\mathbb{R}^{k}\\to\\mathbb{R}^{l}$ be a continuous function.If $\\{X_{n}\\}$ are $\\mathbb{R}^{k}$ -valued random variables converging to $X$ in probability, then $\\{g(X_{n})\\}$ converge in probability to $g(X)$ also.

Proof.

Suppose first that $g$ is uniformly continuous.Given $\\epsilon>0$ , there is $\\delta>0$ such that $\\lVert g(X_{n})-g(X)\\rVert<\\epsilon$ whenever $\\lVert X_{n}-X\\rVert<\\delta$ .Therefore,

\\mathbb{P}\\bigl{(}\\lVert g(X_{n})-g(X)\\rVert\\geq\\epsilon\\bigr{)}\\leq\\mathbb{P}%\\bigl{(}\\lVert X_{n}-X\\rVert\\geq\\delta\\bigr{)}\\to 0

as $n\\to\\infty$ .

Now suppose $g$ is not necessarily uniformly continuous on $\\mathbb{R}^{k}$ .But it will be uniformly continuous on any compact set $\\{x\\in\\mathbb{R}^{k}\\colon\\lVert x\\rVert\\leq m\\}$ for $m\\geq 0$ .Consequently, if $X_{n}$ and $X$ are bounded (by $m$ ), then the proof justgiven is applicable. Thus we attempt to reduce the general caseto the case that $X_{n}$ and $X$ are bounded.

Let

f_{m}(x)=\\begin{cases}x\\,,&\\lVert x\\rVert\\leq m\\\\mx/\\lVert x\\rVert\\,,&\\lVert x\\rVert\\geq m\\end{cases}

Clearly, $f_{m}\\colon\\mathbb{R}^{k}\\to\\mathbb{R}^{k}$ is continuous; in fact, it can be verified that $f_{m}$ is uniformly continuous on $\\mathbb{R}^{k}$ .(This is geometrically obvious in the one-dimensional case.)

Set $X_{n}^{m}=f_{m}(X_{n})$ and $X^{m}=f_{m}(X)$ ,so that $X_{n}^{m}$ converge to $X^{m}$ in probability for each $m\\geq 0$ .

We now show that $g(X_{n})$ converge to $g(X)$ in probability by a four-step estimate. Let $\\epsilon>0$ and $\\delta>0$ be given.For any $m\\geq 0$ (which we will later),

\\mathbb{P}\\bigl{(}\\lVert g(X_{n})-g(X)\\rVert\\geq\\delta\\bigr{)}\\leq\\mathbb{P}%\\bigl{(}\\lVert g(X_{n}^{m})-g(X^{m})\\rVert\\geq\\delta\\bigr{)}+\\mathbb{P}\\bigl{(%}\\lVert X_{n}\\rVert\\geq m\\bigr{)}+\\mathbb{P}\\bigl{(}\\lVert X\\rVert\\geq m\\bigr{%)}\\,.

Choose $M$ such that for $m\\geq M$ ,

\\mathbb{P}\\bigl{(}\\lVert X\\rVert\\geq m\\bigr{)}\\leq\\mathbb{P}\\bigl{(}\\lVert X%\\rVert\\geq M\\bigr{)}<\\frac{\\epsilon}{4}\\,.

(This is possible since $\\lim_{m\\to\\infty}\\mathbb{P}\\bigl{(}\\lVert X\\rVert\\geq m\\bigr{)}=\\mathbb{P}%\\bigl{(}\\bigcap_{m=0}^{\\infty}\\{\\lVert X\\rVert\\geq m\\}\\bigr{)}=\\mathbb{P}(%\\emptyset)=0$ .)

In particular, let $m=M+1$ .Since $X_{n}^{m}$ converge in probability to $X^{m}$ and $X_{n}^{m}$ , $X^{m}$ are bounded, $g(X_{n}^{m})$ converge in probability to $g(X^{m})$ .That means for $n$ large enough,

\\mathbb{P}\\bigl{(}\\lVert g(X_{n}^{m})-g(X^{m})\\rVert\\geq\\delta\\bigr{)}<\\frac{%\\epsilon}{4}\\,.

Finally, since $\\lVert X_{n}\\rVert\\leq\\lVert X_{n}-X\\rVert+\\lVert X\\rVert$ ,and $X_{n}$ converge to $X$ in probability,we have

\\mathbb{P}\\bigl{(}\\lVert X_{n}\\rVert\\geq m=M+1\\bigr{)}\\leq\\mathbb{P}\\bigl{(}%\\lVert X_{n}-X\\rVert\\geq 1\\bigr{)}+\\mathbb{P}\\bigl{(}\\lVert X\\rVert\\geq M\\bigr%{)}<\\frac{\\epsilon}{4}+\\frac{\\epsilon}{4}

for large enough $n$ .

Collecting the previous inequalities together,we have

\\mathbb{P}\\bigl{(}\\lVert g(X_{n})-g(X)\\rVert\\geq\\delta\\bigr{)}<\\epsilon

for large enough $n$ .∎