# Convergence in distribution and CDF

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 $y = F(x)$ ?

• 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. Jan 23 '13 at 17:20
• 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 Jan 23 '13 at 19:47

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 $x + 1/n^{1/3}$). 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.