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Suppose $X_n$ converges in distribution to $X$ , $x_n \rightarrow x$, also the cumulative distribution function for $X$ is continuous at $x$. Show that $ P(X_n \leq x_n) \rightarrow P(X \leq x)$.

PS: Since $F_n(x_n) \subseteq [0,1]$ and $[0,1]$ is compact, $F_n(x_n)$ has at least a subsequence converging to y. How to prove that $y = F(x)$ ?

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    $\begingroup$ Since this is a homework question, someone is bound to provide a hint. But, the best hints will come from knowing what exactly you are struggling with in this exercise. If you can edit your question to indicate this, that would be great. Cheers. $\endgroup$ – cardinal Jan 23 '13 at 17:20
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    $\begingroup$ Perhaps if you wrote out the exact definition of convergence in distribution, using $\epsilon$'s and $\delta$'s and "there exists a $N$ such that for all $n \geq N \cdots$, instead of $\to$ and $\lim_{n\to \infty}$ etc., you might see your way to figuring out what you need in the proof $\endgroup$ – Dilip Sarwate Jan 23 '13 at 19:47
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The interest of this result lies in the case where the $F_n$ are not continuous at $x$ for infinitely many $n$.

To illustrate the ideas, consider a sequence $(F_n)$ of standardized Binomial$(n,1/2)$ distributions converging to the standard Normal distribution (by virtue of the Central Limit Theorem) and let $x_n$ be a sequence of distinct points converging to $x = x_\infty= 1/2$ (I chose the points $x + 1/n^{1/3}$).

Plot of CDFs

To keep the plot from being too cluttered, only the points $x_1, x_{16}, x_{256},$ and $x_\infty$ are shown, along with graphs of the corresponding $F_n$. Thus $x_1$ lies on $F_1$ (shown in red; notice the big jump at $1$); $x_{16}$ lies on $F_{16}$ (gold, with moderate jumps), $x_{256}$ lies on $F_{256}$ (green with many small jumps), and $x_\infty = x$ lies on the limiting distribution $F$ (continuous blue curve). (Of course, because every $F_i$ must be càdlàg, we understand that each horizontal line on these graphs includes its left endpoint but not its right endpoint.)

It is not hard to see that that $F_{(4m)^2}$ has a finite jump at $1/2$ for all $m \ge 1$ (making this one of the interesting cases).

The question asks you to show that as the $x_i$ get closer to $x$, the corresponding points $(x_i, F_i(x_i))$ on the graphs--as exemplified by the black dots shown here--converge to $(x, F(x))$. What assures convergence of the second coordinates is that at the same time the $x_i$ approach $x$, the $F_i$ are all converging pointwise to $F$ and there is no jump in $F$ exactly at $x$.

The point of the question is to provide practice in translating these visual ideas into an epsilon-delta proof (or the equivalent, depending on what definitions of continuity you know and what theorems you have available). The technique doesn't appear to have much of a statistical interest--it's purely a matter of mathematical craft--; the result, however, is useful in statistics.

An interesting followup is to construct an explicit counterexample for the case where $F$ is not continuous at $x$ but otherwise the other conditions of this problem hold. When you can do that, you can be satisfied you understand these ideas.

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