# Convergence in probability implies convergence in distribution

A sequence of random variables $\{X_n\}$ converges to $X$ in probability if for any $\varepsilon > 0$, $$P(|X_n-X| \geq \varepsilon) \rightarrow 0$$

They converge in distribution if $$F_{X_n} \rightarrow F_X$$ at points where $F_X$ is continuous.

(There is another equivalent definition of converge in distribution in terms of weak convergence.)

It seems like a very simple result, but I cannot think of a clever proof.

• Have you tried the wikipedia article: en.wikipedia.org/wiki/… ? Most books on probability theory include a proof. – Gautam Shenoy Nov 14 '12 at 4:54
• Oh, how come I didn't find it! It looks like something I have in mind. Thank you so much! – Hawii Nov 14 '12 at 5:34

## 3 Answers

A slicker proof (and more importantly one that generalizes) than the one in the wikipedia article is to observe that $X_n \Longrightarrow X$ if and only if for all bounded continuous functions $f$ we have $E f(X_n) \to E f(X)$. If you have convergence in probability then you can apply the dominated convergence theorem (recalling that $f$ is bounded and that for continuous functions $X_n \to X$ in probability implies $f(X_n) \to f(X)$ in probability) to conclude that $E |f(X_n) - f(X)| \to 0$, which implies the result.

Here is an answer that does not rely on dominated convergence.

To prove convergence in distribution, we only need to establish that $E[f(X_n)]$ converges to $E[f(X)]$ for bounded continuous functions $f$. By definition of the limit, we need to prove that for any $\epsilon>0$, there some $n_0=n_0(\epsilon)$ such that for all $n>n_0$ the inequality $| E[f(X_n)] - E[f(X)]| < \epsilon$ holds.

1. As suggested in another answer, the first step is to show that if $X_n$ converge to $X$ in probability then $f(X_n)$ also converges in probability to $f(X)$ for any continuous $f$.
2. Let $f$ be any continuous function bounded by $K$. Take any $\epsilon>0$ and show that $$| E[f(X_n)] - E[f(X)]| \le E[|f(X_n)] - E[f(X)|] \le (\epsilon/2) \; P(A_n^c) + K \; P(A_n)$$ where $A_n$ is the event $\{ |f(X_n)] - E[f(X)| > \epsilon /2 \}$.
3. It remains to show that $P(A_n^c)\le 1$ (obvious) and that for $n$ large enough, one has $P(A_n^c)\le \epsilon/(2 K)$ thanks to the convergence in probability established in 1.

Chris J.'s answer more or less is correct, but you require almost sure convergence to be able to apply dominated convergence. Fortunately, convergence in probability implies almost sure convergence along a subsequence, and the proof more or less can proceed as desired.

For more details, Kallenberg's Foundations of Modern Probability, First Edition, Lemma 3.7 is useful.

• Or you could apply the bounded convergence theorem. – Calculon Mar 15 '15 at 11:33
• Dominated convergence theorem also applies with convergence in probability. – perlman Oct 29 '17 at 0:26