# convergence in probability is preserved under continuous transformations

###### Theorem 1.

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

###### Proof.

Suppose first that $g$ is uniformly continuous^{}.
Given $\u03f5>0$, there is $\delta >0$ such that
$$ whenever
$$.
Therefore,

$$\mathbb{P}(\parallel g({X}_{n})-g(X)\parallel \ge \u03f5)\le \mathbb{P}(\parallel {X}_{n}-X\parallel \ge \delta )\to 0$$ |

as $n\to \mathrm{\infty}$.

Now suppose $g$ is not necessarily uniformly continuous on ${\mathbb{R}}^{k}$.
But it will be uniformly continuous on any compact set
$\{x\in {\mathbb{R}}^{k}:\parallel x\parallel \le m\}$ for $m\ge 0$.
Consequently, if ${X}_{n}$ 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 ${X}_{n}$ and $X$ are bounded.

Let

$${f}_{m}(x)=\{\begin{array}{cc}x,\hfill & \parallel x\parallel \le m\hfill \\ mx/\parallel x\parallel ,\hfill & \parallel x\parallel \ge m\hfill \end{array}$$ |

Clearly, ${f}_{m}:{\mathbb{R}}^{k}\to {\mathbb{R}}^{k}$ 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 $m\ge 0$.

We now show that $g({X}_{n})$ converge to $g(X)$ in probability by a four-step estimate. Let $\u03f5>0$ and $\delta >0$ be given. For any $m\ge 0$ (which we will later),

$$\mathbb{P}(\parallel g({X}_{n})-g(X)\parallel \ge \delta )\le \mathbb{P}(\parallel g({X}_{n}^{m})-g({X}^{m})\parallel \ge \delta )+\mathbb{P}(\parallel {X}_{n}\parallel \ge m)+\mathbb{P}(\parallel X\parallel \ge m).$$ |

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

$$ |

(This is possible since ${lim}_{m\to \mathrm{\infty}}\mathbb{P}(\parallel X\parallel \ge m)=\mathbb{P}\left({\bigcap}_{m=0}^{\mathrm{\infty}}\{\parallel X\parallel \ge m\}\right)=\mathbb{P}(\mathrm{\varnothing})=0$.)

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,

$$ |

Finally, since $\parallel {X}_{n}\parallel \le \parallel {X}_{n}-X\parallel +\parallel X\parallel $, and ${X}_{n}$ converge to $X$ in probability, we have

$$ |

for large enough $n$.

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 |