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Portmanteau:

Let $(E,\text{d})$ be a metric space, and $\mathbb{P},\mathbb{P}_1,\mathbb{P}_2\dotsc$ be probability measures on $E$. Then the sequence $(\mathbb{P}_n)_{n\in\mathbb{N}}$ converges weakly against $\mathbb{P}$ if and only if $$\limsup\limits_{n\to\infty}\mathbb{P}_n(F)\leq \mathbb{P}(F)$$ for every closed set $F\subset E$.

My question is, does this assertion also hold for every compact set instead of closed? i.e.

Let $\mathbb{P},\mathbb{P}_1,\mathbb{P}_2\dotsc$ be probability measures on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$. Then the sequence $(\mathbb{P}_n)_{n\in\mathbb{N}}$ converges weakly against $\mathbb{P}$ if and only if $$\limsup\limits_{n\to\infty}\mathbb{P}_n(K)\leq \mathbb{P}(K)$$ for every compact set $K\subset \mathbb{R}$.

I would say yes, because compact implies closed, not?

Thanks.

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In a Hausdorff space every compact set is closed, however in general a compact set does not have to be closed. –  Ilya Jan 10 '13 at 17:48
    
Is it your question that can we replace "closed" with "compact" in this theorem? Or, that if $(\mathbb{P}_{n})$ converges weakly to $\mathbb{P}$, then does the inequality hold for compact sets also? Note that in a metric space compact sets are closed, or in general as Ilya noted, in any Hausdorff space. –  Thomas E. Jan 10 '13 at 17:53
    
Yes, I meant replacing "closed" with "compact" in this theorem. Maybe I should also mention that $\mathbb{P},\mathbb{P}_1,\mathbb{P}_2,\dotsc$ are probability measures on $(\mathbb{R},\mathcal{B}(\mathbb{R})$. –  Ichigo Jan 10 '13 at 18:33
    
Has anybody found a counter-example to this? I.e., a metric space and a sequence of weakly convergent probability measures such that there is a compact set K not fulfilling the lim-sup condition? –  user58449 Jan 16 '13 at 6:24

1 Answer 1

If we replace the word "closed" by "compact" in the theorem, it won't be true.

Since in a metric space, a compact set is closed, the condition remains necessary. However, it's not sufficient. Consider $E=\Bbb R$ with the usual metric, $\Bbb P_n:=\delta_n$ and $\Bbb P$ any probability measure. Then for each compact $K$, $\Bbb P_n(K)=0$ for $n$ eventually large, so the condition is satisfied.

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I don't understand it well. So you said if I define $\mathbb{P}_n:=\delta_n$, then the assertion with the compact set is true, otherwise not? –  Ichigo Jan 10 '13 at 19:48
    
Yes, this one is true. But not the assertion about convergence in law. –  Davide Giraudo Jan 10 '13 at 21:06

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