Convergence in probability implies convergence in distribution

Solution 1:

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.

Solution 2:

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.

Solution 3:

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.