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Given a probability space $(\Omega,\cal F, \Bbb P)$, the distribution function of a random variable $X$ is defined as $F(x)=\Bbb P\{X \le x\}$. Now if $F_1,F_2,...,F_{\infty}$ are distribution functions, then the question is

Is $F_n \xrightarrow{w} F_{\infty}$ equivalent to $\lim_{n\uparrow\infty}\int\phi dF_n=\int\phi dF_{\infty}$ for every $\phi \in C(\Bbb R)$?

Here ${F_n}\xrightarrow{w}{F_\infty }$ means weak convergence, and the integral involved are Riemann-Stieltjes integrals.

Someone has pointed out that this is the Helly-Bray theorem, which says the above claim is true when $\phi$ is bounded. I have searched the Internet but was not able to find a proof of the theorem. Can anyone help provide a proof? Thanks!

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    $\begingroup$ Helly-Bray Theorem $\endgroup$
    – A.Γ.
    Aug 31, 2015 at 5:12
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    $\begingroup$ Please specify more precisely what you mean. Are all $F_i$ meant to be CDFs of probability measures? Does $F_n \to F_\infty$ mean pointwise convergence or convergence in distribution? If it's the latter, what is your definition of convergence in distribution? In any case, you should restrict $\phi$ so some subfamily, so that all occuring integrals are well-defined. If the $F_i$ are distribution functions of probability measures, then it is enough to consider bounded functions. If the corresponding measures are only locally finite, you need compactly supported functions. $\endgroup$
    – Dominik
    Sep 2, 2015 at 11:17
  • $\begingroup$ @Dominik Hi. Thank you for your suggestion. I have made the refinement. $\endgroup$
    – Tony
    Sep 2, 2015 at 23:13
  • $\begingroup$ I'm unclear what you're asking here. Are you asking for a proof of the Helly-Bray theorem, or an answer to your boxed question about the equivalence of the conditions? The answer to your boxed question is no, they are not equivalent. $\endgroup$ Sep 3, 2015 at 0:07
  • $\begingroup$ @NateEldredge Thank you for your comment. This question comes from my teacher's lecture note. I am not sure but I myself is confused by the question. Can I ask a question first. If in the boxed question the "convergence in distribution" is replaced by "weak convergence", is the claim true? $\endgroup$
    – Tony
    Sep 3, 2015 at 0:56

2 Answers 2

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I will try to give an extensive answer to this problem. Instead of working with distribution functions, I will work with the corresponding random variables $X_1, X_2, \ldots, X_\infty$ to make notation easier.

Is the claim correct if we allow $\phi$ to be any continuous function?
Unfortunately, the answer is no. The first problem is, that in general $E[f(x_i)]$ doesn't need to exist for every continuous funciton $f$. Take for example the Cauchy distribution. This distribution doesn't have finite expectation, so we can construct a counterexample by assuming $X$ to be Cauchy distributed and setting $X_n = \frac{X}{n}$. Then clearly $X_n$ converges almost surely to $X_\infty = 0$, and therefore also weakly. But for $f(x) = x$ the value of $E[f(X_n)]$ is undefined, except for $E[f(X_\infty)] = E[0] = 0$.

What if we restrict $\phi$ to the class of functions so that all $E[f(X_n)]$ exist?
Unfortunately, this still won't work. Assume $P(X_n = \frac{1}{n}) = 1 - \frac{1}{n}$ and $P(X_n = n + \frac{1}{n} - 1) = \frac{1}{n}$. It's easy to see that $X_n$ converges in probability to $X_\infty = 0$, and therefore weakly. Choosing $f$ to be the identity, we can see that $E[f(x_n)] = 1$ doesn't converge to $E[f(x_\infty)] = 0$.

The problem in this case is, that a general continuous function can "blow up" a neglibible part of the distribution.

What if we restrict $\phi$ to be a continuous, bounded function?
In this case, the equivalence is correct.

To see that the convergence of the functions implies weak convergence, let $x$ be a continuity point of the distribution of $X_\infty$. Now we can approximate the Indicator $f(t) = I\{t \le x\}$ from below and from the top by continuous functions. Namely we choose $\underline{f_m}$ to be one on $(-\infty, x - \frac{1}{m}]$, zero on $[x, \infty)$ and linear inbetween. Similarly, we choose $\overline{f_m}$ to be one on $(-\infty, x]$, one on $[x+ \frac{1}{m}, \infty)$ and linear inbetween. I recommend doing a quick sketch of these functions.

We can now easily see that $\underline{f_m} \le f \le \overline{f_m}$ and therefore $$\lim \limits_{n \to \infty} E[\underline{f_m}(X_n)] \le \lim \limits_{n \to \infty}E[f(X_n)] \le \lim \limits_{n \to \infty} E[\overline{f_m}(X_n)]$$ which by assumption implies $$E[\underline{f_m}(X_\infty)] \le \lim \limits_{n \to \infty}P(X_n \le x) \le E[\overline{f_m}(X_\infty)].$$

Now observe that $P(X_\infty \le x - \frac{1}{m}) \le E[\underline{f_m}(X_\infty)]$ and a similar inequality for the upper bound yields $$P\left(X_\infty \le x - \frac{1}{m}\right) \le \lim \limits_{n \to \infty}P(X_n \le x) \le P\left(X_\infty \le x + \frac{1}{m}\right).$$ The assertion now follows from observing that $x$ is a continuity point of $F_\infty$ and letting $m \to \infty$.

The other direction of the equivalence is a bit more tedious work. This proof works by simple approximations, I will only sketch how it works.

First we note that weakly convergent measures are tight, so the first step of the approximation is to restrict our continuous function to some compact interval $K$. Since $F_\infty$ is monotone, it has at most countably many points of discontinuity. Therefore we can choose a sequence of partitions $S_m$ of $K$ that consist only of continuity points of $F_\infty$ and whose mesh converges to zero. If $S_m = \{x_1, \ldots, x_m\}$, we choose to approximate our continuous, bounded function $f$ by $f_m(t) = \sum \limits_{i = 1}^{m - 1} f(x_i) I\{x_i < t \le x_{i + 1}\}$. Observe that $E[I\{x_i < X_\infty \le x_{i + 1}\}] = P_\infty(x_{i + 1}) - P_\infty(x_i)$.

Now the rest of this proof consists in showing that our approximation is actually good enough to ensure $E[f(X_n)] \to E[f(X_\infty)]$ actually holds.

Can we expand or further restrict our class of functions, so that the equivalence still holds?
The answer is yes. A list of possible characterizations of weak convergence can be found here.

Can we restrict our functions to continuous functions that are compactly supported?
The answer is no, take for example $X_n \sim \mathcal{U}[n, n + 1]$. Then $X_n$ doesn't converge weakly [as it isn't tight], but for every compactly supported function $f$ we have $E[f(X_n)] \to 0$.

However, the notion of convergence that is induced by only considering compactly supported functions is called vague convergence. It can be shown [Thm 5.20 in Kallenberg] that a sequence of random variables converges weakly iff it converges vaguely and is tight.

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I can refer you this link (http://www.ma.utexas.edu/users/gordanz/notes/weak.pdf). Look for the portmanteau theorem here.

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