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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.

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Have you tried the wikipedia article:… ? 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

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.

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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.

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Or you could apply the bounded convergence theorem. – Calculon Mar 15 '15 at 11:33

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