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This is question 10.3.1 from Rosenthal's First Look at Rigorous Probability.

Suppose $\mathcal{L}(X_n) \Rightarrow \delta_c$ for some $c \in \mathbb{R}$. Prove that $\{X_n\}$ converges to $c$ in probability.

Here $\mathcal{L}(X_n)$ refers to the distribution of $X_n$. Is the following proof correct? Is a simpler proof available? Is there a simple proof using Skorohod's theorem?

I attempted to prove the contrapositive. Suppose $X_n \not \to c$ in probability. This means that there is an $\varepsilon > 0$ so that for all $N$, there exists an $m > N$ such that

\begin{equation} P(|X_m - c| \geq \varepsilon) > \eta, \text{ for some $\eta > 0$}. \end{equation}


\begin{equation} B := \{x \in \mathbb{R} : |x - c| \leq \varepsilon\} \in \mathcal{B} \text{ the Borel sets on $\mathbb{R}$}. \end{equation}

Then $B^c \in \mathcal{B}$ and denoting the distribution of $X_n$ as $\mu_n$ we have

\begin{equation} \mu_n(B^c) = P(X_n \in B^c) = P(|X_n - c| \geq \varepsilon) > \eta. \end{equation}

For all $N$, there exists an $m > N$ such that the cumulative distribution function of $X_m$ is not the same as the cumulative distribution function of $\delta_c$; in particular it's mass changes somewhere at least $\varepsilon$ distance from $c$. Finally because $\delta_c(\{x\}) = 0$ for all $x \neq c$ this completes the proof.

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There are various equivalent definitions of weak convergence. Which one are you using? – Nate Eldredge Oct 3 '12 at 2:34
I think proving the contrapositive here is unnecessarily tangled. As a start, try running your proof backwards and negating all the statements; I think it will be clearer. I would also avoid the language "even if you throw out $X_1, \dots, X_N$" and reword as "for all $N$ there exists $m > N$". – Nate Eldredge Oct 3 '12 at 2:36
The definition in Rosenthal is that $\int_{\mathbb{R}} f d \mu_n \to \int_{\mathbb{R}} f d \mu$ for all bounded continuous functions $f : \mathbb{R} \to \mathbb{R}$. However we proved that this is equivalent to $\mu_n(-\infty, x] \to \mu(-\infty, x]$ for all $x$ with $\mu\{x\} = 0$, which is what I used in the proof. – Alex Oct 3 '12 at 5:11
@NateEldredge thanks for the idea to run the proof in reverse with negations. Do you think I should post that as an answer? – Alex Oct 3 '12 at 5:13
up vote 2 down vote accepted

Here's another possible proof that is maybe more direct. Let $\mu_n$ denote the law of $X_n$.

Fix $\epsilon > 0$. We want to show $P(|X_n -c| \le \epsilon) \to 1$. Construct a continuous function $f$ such that:

  • $0 \le f \le 1$;
  • $f = 0$ outside $(c-\epsilon, c+\epsilon)$;
  • $f(c) = 1$.

Then we have $$P(|X_n - c| \le \epsilon) = \int 1_{[c-\epsilon, c+\epsilon]}\,d\mu_n \ge \int f\,d\mu_n \to \int f\,d \delta_c = f(c) = 1.$$

You could also use Skorohod's theorem if you want. Let $Y_n$ be the sequence of random variables produced by Skorohod, with $Y_n \sim \mu_n$. Then we have $Y_n \to c$ almost surely and in particular $Y_n \to c$ in probability. That is, $P(|Y_n - c| \le \epsilon) \to 1$. But $$P(|Y_n - c| \le \epsilon) = \mu_n([c-\epsilon, c+\epsilon]) = P(|X_n -c| \le \epsilon),$$ so $X_n \to c$ in probability as well. However, this seems like using too much machinery to me.

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