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Define a subprobability measure to be a measure on Borel sigma algebra of the real line $\mathbb{R}$ with the measure for the whole real line less or equal than 1.

I was wondering about the definition of vague convergence of a sequence of subprobability measures $\{ \mu_n, n\geq 1 \}$ to another subprobability measure $\mu$. The convergence can be defined in slightly two different ways as in Chung's probability theory book p85 and p90:

(1) if there exists a dense subset D of the real line $\mathbb{R}$ so that $ \forall a \text{ and } b \in D \text{ with } a .

(2) if there exists a dense subset D of the real line $\mathbb{R}$ so that $ \forall a \text{ and } b \in D \text{ with } a (the original text just says converges, not mention that converges to $ \mu((a,b))$ which I guess can be added?).

How to show these two definitions are equivalent?

A side question: is there a definition of vague convergence for general measures on more general sigma algebra with more general underlying space?

Thank you so much!

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    When I wrote my previous comments, I wasn't aware that the book pages are available online. This changes the situation completely, and I delete my previous comments. It is now clear that you shouldn't specify the limit in definition (2): you are only given that the sequence $\mu_n ((a,b))$ converges for any $a,b \in D$.2010-11-17

4 Answers 4

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Try $(a,b] = \bigcap_{c \in D, c > b} (a,c)$ and $(a,b) = \bigcup_{c \in D, c < b} (a,c]$.

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    Thanks, Yuval! Can you be more specific how to apply these to the proof?2010-11-17
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Here is the general notion of vague convergence, taken from Olav Kallenberg's Foundations of Modern Probability (2nd edition).

Let $S$ be a locally compact, second countable Hausdorff space equipped with its Borel $\sigma$-field ${\cal S}$. Let $\hat{\cal S}$ be the ring of relatively compact Borel sets, and ${\cal M}(S)$ the space of locally finite non-negative measures. Locally finite means that $\mu(B)<\infty$ when $B\in \hat{\cal S}$.

The vague topology on ${\cal M}(S)$ is the topology generated by the mappings $\mu\mapsto \int f\ d\mu$ for every $f$ a non-negative continuous function with compact support.

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    Thanks, B$y$ron! Nice to $k$now the generalization of the definition.2010-11-17
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In my previous answer, I have shown that definition (1) implies definition (2). Now I prove the converse implication. Suppose that the condition in definition (2) holds. That is, $\mu_n ((a,b))$ converges for any $a,b \in D$, $a < b$, where $D$ is a dense subset of $\mathbb{R}$. We are now going to specify the limit. For any $a,b \in D$, $a < b$, define $ \mu ((a,b)) = \mathop {\lim }\limits_{n \to \infty } \mu _n ((a,b)). $ Since $\mu_n$ is a sequence of subprobability measures (s.p.m.), $\mu ((a,b)) \in [0,1]$. Next, fix any $a,b \in {\mathbb R}$, $a < b$. If $a_m$ is a decreasing sequence converging to $a$, and $b_m$ is an increasing sequence converging to $b$, $a_m,b_m \in D$, then, by the last definition, $\mu ((a_m,b_m))$ is a monotone increasing sequence, bounded from above by $1$. Hence, this sequence converges, and we can define $ \mu ((a,b)) = \mathop {\lim }\limits_{m \to \infty } \mu ((a_m ,b_m )) \in [0,1]. $ This actually determines a s.p.m. $\mu$ on $\mathbb{R}$. Now, define $\tilde D = \lbrace x \in {\mathbb R}:\mu (\lbrace x \rbrace ) = 0 \rbrace$ and proceed similarly to my previous answer, based on \mu _n ((a',b')) \le \mu _n ((a,b]) \le \mu _n ((a'',b'')), to conclude that for any $a < b$ in (the dense set) $\tilde D$, $\mu _n ((a,b]) \to \mu ((a,b])$.

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We will only prove here that definition (1) implies definition (2) (this is the easy part).

Suppose that the condition in definition (1) holds. That is, $\mu_n ((a,b]) \to \mu ((a,b])$ for any $a,b \in D$, $a < b$, where $D$ is a dense subset of $R$. Define $\tilde D = \lbrace x \in R:\mu (\lbrace x \rbrace ) = 0 \rbrace$. Then $\tilde D$ is dense in $R$, since its complement is at most countable. For any $a,b \in \tilde D$, $a < b$, and a',a'',b',b'' \in D, a' < b', a'' < b'', such that a'' < a < a' and b' < b < b'', we have \mu _n ((a',b']) \le \mu _n ((a,b)) \le \mu _n ((a'',b'']). By assumption, \mu _n ((a',b']) \to \mu ((a',b']) and \mu _n ((a'',b''] \to \mu ((a'',b'']) as $n \to \infty$. Then, since a',a'' and b',b'' can be chosen arbitrarily close to $a$ and $b$, respectively, we can conclude that $ \mu ((a,b)) \le \mathop {\lim \inf }\limits_{n \to \infty } \mu _n ((a,b)) \le \mathop {\lim \sup }\limits_{n \to \infty } \mu _n ((a,b)) \le \mu ([a,b]). $ However, by assumption $\mu (\lbrace a \rbrace) = \mu (\lbrace b \rbrace) = 0$, and hence $\lim _{n \to \infty } \mu _n ((a,b)) = \mu ((a,b))$. Thus, the condition in definition (2) is satisfied (with $\tilde D$ playing the role of $D$ in that definition).

It is instructive to consider here the following simple example. As is customary, denote by $\delta_x$ the probability measure concentrated at $x$. Define $\mu_n = \delta_{-1/n}$, $n \geq 1$, and $\mu = \delta_0$. Then, for any $a,b \in R$, $a < b$, we have $\mu _n ((a,b]) \to \mu ((a,b])$, and hence, by definition (1), $\mu_n$ converges vaguely to $\mu$. On the other hand, $\mu _n ((-2,0)) = 1$ for any $n \geq 1$, whereas $\mu ((-2,0)) = 0$. However, if we define $\tilde D$ as above, then $\tilde D = R - \lbrace 0 \rbrace$, and for any $a,b \in \tilde D$, we do have $\mu _n ((a,b)) \to \mu ((a,b))$; hence, $\mu_n$ converges vaguely to $\mu$ also according to definition (2).

Finally, as for the OP's remark (concerning definition (2)) that "the original text just says converges, not mention that converges to $ \mu((a,b))$ which I guess can be added?", it is clear that we are deliberately only given that $\mu_n ((a,b))$ converges -- this is what makes the converse implication (from definition (2) to (1)) a substantially more difficult problem.

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    In any case, a direct proof is in order for the OP's question.2010-11-18