Does weak convergence imply convergence in probability? 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.
 A: 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. 
A: 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. 
Also: 
(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*}
