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could I possibly have feedback on my attempt to prove the following statement?

If $g$ is a continuous function and $X_{n}\rightarrow X$, where $X_{n}$ is a sequence of random variable, then $g\left( X_{n} \right) \rightarrow g\left(X\right)$

Attempt on Proof. $Start$

We want to prove that for any $\epsilon>0$, $P\left(\left|g(X_{n})-g(X)\right|>\epsilon\right)\rightarrow0$ as $n\rightarrow\infty$. For some $\delta>0$, by Law of Total Probability, we have $$ P\left(\left|g(X_{n})-g(X)\right|>\epsilon\right)=P\left(\left|g(X_{n})-g(X)\right|>\epsilon,\left|X_{n}-X\right|<\delta\right)+P\left(\left|g(X_{n})-g(X)\right|>\epsilon,\left|X_{n}-X\right|\geq\delta\right). $$

As $\delta\rightarrow0$, the first term on the RHS converget to $0$ by the continuity of $g$. Meanwhile, the latter term also converge to $0$ by convergence of $X_{n}$ to $X$.

Consequently, $P\left(\left|g(X_{n})-g(X)\right|>\epsilon\right)\rightarrow0$, and we have shown that the statement holds. $End$

Thanks in advance for the help.

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  • $\begingroup$ I don't see any problem with this proof. For an alternate proof you can see theorem 20.5 and the discussion below it of Billingsley's Probability and measure. $\endgroup$
    – Kumara
    Jan 28, 2013 at 7:15
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    $\begingroup$ Assuming your hypothesis is the convergence in probability of $X_n$ to $X$, the problem with your argument to deal with the first term on the RHS is that one needs the function $g$ to be uniformly continuous. If one assumes only that $g$ is continuous, the proof is incomplete. $\endgroup$
    – Did
    Jan 30, 2013 at 11:43
  • $\begingroup$ @Did The argument by the OP works with $g$ just continuous, but of course you are right, the details are missing. $\endgroup$
    – air
    Nov 25, 2016 at 13:22
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    $\begingroup$ @BeerR Do you know that a sequence $X_n$ converges in probability to $X$ if and only if every subsequence of $X_n$ has a further subsequence that converges almost surely to $X$? Using this characterization of convergence the proof is very easy. $\endgroup$
    – Dominik
    Nov 25, 2016 at 13:33

1 Answer 1

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BeerR's proof

The proof is correct, although justifying convergence of your first term is not completely trivial.

Pick a sequence $\delta_m \searrow 0$ and $\varepsilon >0$ fixed. Define

$$ A_m := \{\omega: |g(X_n(\omega))-g(X(\omega))| > \varepsilon, |X_n(\omega)-X(\omega)| <\delta_m \text{ for some } n\} $$

Clearly $A_{m'} \subset A_m$ for $m' > m$.

We will show that $A_m \searrow \emptyset$ as $m \to \infty$. Then by continuity of probability measures we will get $\Pr(A_m) \searrow 0$ as $m \to \infty$.

For this fix $\omega$:

By continuity of $g$ at $X(\omega)$, there exists a $\delta$ such that $|g(y) - g(X(\omega))| < \varepsilon$ for all $y$ with $|y-X(\omega)| < \delta$.

In particular, pick $M$ such that $\delta_M < \delta$, then $\omega \notin A_M$. Since $\omega$ was arbitrary it follows that:

$$ \bigcap_{m=1}^{\infty} A_m = \emptyset $$

This shows that the first term is negligible, i.e. for all $n$:

$$ \Pr(\{\omega: |g(X_n(\omega))-g(X(\omega))| > \varepsilon, |X_n(\omega)-X(\omega)| <\delta_m \}) \leq \Pr(A_m)$$

Now, in BeerR's expression, one can first take $n \to \infty$ and then $m \to \infty$ to get the wanted result.

Did's comment

If $g$ is uniformly continuous, then things simplify quite a bit:

For fixed $\varepsilon > 0$ choose $\delta >0$ such that $|g(x)-g(y)| < \varepsilon$ for all $x,y$ with $|x-y| < \delta$.

Then: $ |g(X_n(\omega))-g(X(\omega)| \geq \varepsilon$ implies $|X_n(\omega) - X(\omega)| \geq \delta$.

Therefore:

$$ \Pr(|g(X_n)-g(X)| \geq \varepsilon) \leq \Pr( |X_n - X| \geq \delta) $$

The RHS now goes to $0$ since $X_n$ converges to $X$ in probability ($\delta >0$) .

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    $\begingroup$ I don't think this supplements OPs proof sufficiently. You prove that for any fixed $n$ the sequence $P(A_m)$ goes to zero for $m \to \infty$. But you need to fix $m$ at some point and then left $n$ go to infinity. $\endgroup$
    – Dominik
    Nov 25, 2016 at 13:27
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    $\begingroup$ @Dominik Thanks for the remark. I fixed it now: One can just take $A_m$ to be the union over all $n$ of the sets I had defined before. All other arguments still go through and there is no $n$ dependence in $A_m$. $\endgroup$
    – air
    Nov 25, 2016 at 13:35
  • $\begingroup$ @air I just posted a question where I include a proof of a bivariate version of this result here: math.stackexchange.com/questions/2572526/… Would you be willing/able to take a look at my proof and let me know if it is correct or not? Thanks. $\endgroup$
    – user100463
    Dec 19, 2017 at 0:50

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