Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If $X_i$ is a sequence of $\mathbb{R}^m$-valued random variables that converges either in probability or almost surely to $X$ and if $f$ is some measurable function from $\mathbb{R}^m$ into $\mathbb{R}$, does it follow that $f(X_i)$ converges to $f(X)$ as well?

I know about the continuous mapping theorem. Is there something similar for arbitrary measurable functions?

New Question: As the answer below suggested, let $\mathcal{C}$ be the set of functions $f$ that satisfy $f(X_i)$ converges in probability to $f(X)$ when $X_i$ converges in probability to $X$. What are the minimal properties of such a set?

share|improve this question
    
How about $X_i=1/i$ with probability one and $f(x)=1_{\{x\neq 0\}}$. Then $X_i\to X=0$ almost surely, but $f(X_i)=1 \nrightarrow 0=f(X)$. –  Stefan Hansen Nov 9 '12 at 8:58
2  
Note that you cannot avoid talking about continuity: If your statement holds for all random Variables (esp. for the constant ones), then $f$ is continuous. –  martini Nov 9 '12 at 9:17

1 Answer 1

up vote 3 down vote accepted

Let $\cal C_1$ the collection of measurable functions from $\Bbb R^m$ to $\Bbb R$ such that whenever $\{X_n\}$ is a sequence of random variables with values in $\Bbb R^m$ converging to $X$ almost surely, then $f(X_n)\to f(X)$ almost surely.

We define in the same way $\cal C_2$ replacing "almost everywhere" by "in probability".

Fix $(x_1,\dots,x_m)\in\Bbb R^n$, and $\{(x_1^{(k)},\dots,x_m^{(k)})\}_k$ an arbitrary sequence converging to $(x_1,\dots,x_m)$. Let $X=(x_1,\dots,x_m)$ and $X_k=(x_1^{(k)},\dots,x_m^{(k)})$ (constant random variables). Then $X_k\to X$ in probability and almost surely. If $f\in\cal C_i$, then $f$ is continuous at $(x_1,\dots,x_m)$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.