> [!definition] > > Let $X$ be a [[Topological Space|topological space]] and $\cb(X)$ be its [[Borel Sigma Algebra|Borel sigma-algebra]], then the space of [[Space of Bounded Continuous Functions|bounded and continuous]] functions $BC(X)$ induces a family of [[Seminorm|seminorms]] on the set of probability measures: > $ > \norm{\nu}_f = \abs{\int f d\nu} \quad f \in BC(X) > $ > This structure corresponds to the [[Weak Topology|weak topology]]. > [!definition] > > Let $\seq{X_n}$ be a family of $\real^n$-valued [[Random Variable|random variables]], and $\mu_n$ be the probability distribution of $X_n$. Then $X_n \to X$ [[Convergence in Distribution|in distribution]] if and only if $\mu_n \to \mu$ weakly. > [!theorem] > > Let $\seq{\mu_n} \subset \text{ca}(X)$. If for any $\seq{\mu_{n_k}} \subset \seq{\mu_n}$, there exists a further subsequence $\{\mu_{n_{k_l}}\}_1^\infty$ that converges to $\mu$ weakly, then $\mu_n \to \mu$ weakly. > [!theorem] > > Suppose that $X = \real^n$ and let $\seq{\mu_n} \ge 0$ such that $\mu_n \to \mu$, then for any [[Open Set|open]] set $U$, > $ > \liminf_{n \to \infty}\mu_n(U) \ge \mu(U) > $ > and for any closed set $K$, > $ > \limsup_{n \to \infty}\mu_n(K) \le \mu(K) > $ > *Proof*. Let $\seq{f_j} \subset C(X, [0, 1])$ such that $f_j \upto \one_{U}$ pointwise. By the [[Monotone Convergence Theorem for Sequences|monotone convergence theorem]], > $ > \begin{align*} > \mu(U) &= \limv{j}\mu(f_j) \\ > &= \limv{j}\limv{n}\mu_n(f_j) \\ > &\le \liminf_{n \to \infty}\mu_n(f) > \end{align*} > $ > Since $\mu_n(f_j) \le \mu_n(f)$ for all $j$, this provides a bound on the lower limit. > [!theorem] > > Let $\seq{\mu_n}$ be a family of probability measures on $\real^d$ such that $\mu_n \to \mu$ weakly, and $\seq{\xi_n} \subset \real^d$ such that $\xi_n \to \xi$. Then $\hat \mu_n(\xi_n) \to \hat \mu(\xi)$, and $\hat \mu_n \to \hat\mu$ [[Uniform Convergence on Compact Sets|uniformly on compact sets]]. > > *Proof*. Let $\varphi_n = e^{i\angles{\cdot, \xi_n}}$ and $\varphi = e^{i\angles{\cdot, \xi}}$, then these functions are uniformly bounded. > $ > \abs{\varphi_n(x) - \varphi(x)} = \abs{e^{i\angles{x, \xi_n}} - e^{i\angles{x, \psi}}} \le \abs{\angles{x, \xi_n - \xi}} \le \abs{x} \cdot \abs{\xi_n - \xi} > $ > Let $K$ be a compact set, then $\abs{x}$ is bounded on $K$, and $\varphi_n \to \varphi$ on uniformly on $K$. From here, we get that $\mu_n(\xi_n) = \mu_n(\varphi_n) \to \mu(\varphi)$. > > Suppose for contradiction such that the characteristic functions do not converge uniformly on compact sets. Then there exists $K$ compact such that > $ > \limv{n}\sup_{\xi \in K} \abs{\hat \mu_n(\xi) - \hat \mu(\xi)} > 0 > $ > yielding a subsequence $\seq{n_k}$ and $\varepsilon > 0$, such that > $ > \sup_{\xi \in K}\abs{\hat \mu_{n_k}(\xi) - \hat \mu(\xi)} \ge \varepsilon$ > for each $k$. Choose $\xi_k$ such that $\abs{\hat \mu_{n_k}(\xi_k) - \hat \mu(\xi)} \ge \varepsilon/2$. As $K$ is compact, there exists a subsequence $\seq{\xi_{k_l}}$ such that $\xi_{k_l} \to \xi_0$, while $\mu_{n_{k_l}} \to \mu$ still. By the previous part > $ > \abs{\hat \mu_{n_{k_l}}(\xi_{k_l}) - \hat \mu(\xi_{k_l})} \le \abs{\hat \mu_{n_{k_l}}(\xi_{k_l}) - \hat \mu(\xi_0)} + \abs{\hat \mu(\xi_0) - \hat \mu(\xi_{k_l})} > $ > the two parts here both converge to zero, which contradicts the lower bound on the subsequence. > [!theorem] > > Let $\seq{\mu_n}$ be a tight family of probability measures on $\real^d$. Suppose that there exists $f: \real^d \to \complex$ such that > $ > \limv{n}\hat \mu_n(\xi) = f(\xi) \quad \forall \xi \in \real^d > $ > Then there exists a probability measure $\mu$ such that $\mu_n \to \mu$ weakly. > > *Proof*. Let $\seq{n_k}$ be a subsequence. By Prokhorov's theorem, this sequence has a subsequence that converges weakly. From here, the characteristic functions of this subsequence converges to $f$ pointwise. Hence $f$ is the characteristic function of the subsequential limit, and thus the characteristic function of the sequence limit. > [!theorem] > > Let $\seq{\mu_n}$ and $\seq{\nu_n}$ be probability measures on $\real^d$. > 1. If both families are tight, then $\seq{\mu_n * \nu_n}$ are also tight. > 2. If $\mu_n \to \mu$ weakly and $\nu_n \to \nu$ weakly, then $\mu_n * \nu_n \to \mu * \nu$. >