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Can we derive from weak convergence to convergence in probability? I know convergence in distribution is a special case of weak convergence, but it seems to be stronger than convergence in distribution.

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How do you define weak convergence and convergence in distribution? – Davide Giraudo Nov 11 '12 at 10:21
We can't answer your question until you won't give the definitions. Thanks in advance! – Davide Giraudo Nov 15 '12 at 22:58
Are you mentioning about the convergence in probability? By weak convergence, in probability theory, we usually mean the weak-* convergence of measures. – Sangchul Lee Mar 14 '13 at 4:42

Let for each $n$, $X_n=X$ a Bernoulli law of parameter $1/2$. Then $\{X_n\}$ converges in distribution to $1-X$, but $P(|X_n-(1-X)|\geqslant \delta)=P(2X-1\geqslant \delta)\neq 0$.

However, if $\{X_n\}$ converges in distribution to a constant $c$, we also have convergence in probability.

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Convergence in distribution is not the same as weak convergence. it's a special case. So I asked about weak convergence not in distribution. Could you please erase your answer? – peanut Nov 12 '12 at 1:54
Could you please reply the comment, in order to be clear about the definitions? – Davide Giraudo Nov 12 '12 at 11:00
+1. (Never saw a request like "Could you please erase your answer?" before...) – Did Apr 14 '13 at 18:05
so I guess the only difference is that in the case of convergence in measure (probability) we have to look on the original sample space......I guess that the two modes of convergence are the same if the random variables are identity mappings ($\Omega = S, S$ is separable metric space ) – user24367 Sep 19 '13 at 4:58

Convergence in distribution (which for real valued random variables is defined as pointwise convergence of the distribution functions in all points of continuity of the limit distribution)

IS the same as weak convergence (as defined more general for arbitrary metric spaces as: $E(f(\xi_i))\to E(f(\xi))$ for all bounded and continuous functions $f$).

This is a result of the portmanteau lemma.


(i): convergence in probability implies weak convergence, but the other direction is wrong in general.

However, (ii): weak convergence to a constant implies convergence in probability to this constant.

Proof: $(i)$ We verify the condition $Pf(\xi_n)\to Pf(\xi)$ for every uniformely continuous and bounded function $f\colon\to\mathbb{R}$ of the portmanteau lemma. Let $f\colon S\to\mathbb{R}$ be uniformely continuous and bounded, $|f|<M\in\mathbb{R}$. For every $\varepsilon>0$ there exists $\delta>0$, such that $d(x,y)\leq\delta\Rightarrow|f(x)-f(y)|<\epsilon/2$. By the assumption, there exists $n\in\mathbb{N}$, such that $P(d(\xi_n,\xi)\geq\delta)<\varepsilon/(4M)$. Now \begin{align*} |Pf(\xi_n)-Pf(\xi)|\leq& P|f(\xi_n)-f(\xi)|\\ &\leq P(\varepsilon/2\mathbb{1}_{d(\xi_n,\xi)<\delta)})+P(2M\mathbb{1}_{d(\xi_n,\xi)\geq\delta})\\ &\leq\varepsilon/2+\varepsilon/2=\varepsilon . \end{align*} The result follows if we let $\varepsilon\to 0$.\

$(ii)$ Let $\varepsilon>0$. We use the condition $P(\xi \in F) \geq \limsup_{n\to\infty} P(\xi_{n}\in F) $ for every closed set $F\subset S$ from the portmanteau lemma to calculate \begin{align*} \limsup P(d(\xi_n,a)>\varepsilon)\leq\limsup P(\xi_n\in B_\varepsilon(a)^C)\leq P(a\in B_{\varepsilon}(a)^C)=0. \end{align*}

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