> [!definition] > > Let $\seq{(\Omega_n, \cf_n, \bp_n)}$ and $(\Omega_\infty, \cf_\infty, \bp_\infty)$ be [[Probability|probability spaces]], and $X_n: \Omega_n \to \real$ be a [[Sequence|sequence]] of [[Random Variable|random variables]]. The sequence $\seq{X_n}$ **converges in distribution** to $X_\infty$, denoted as $X_n \xrightarrow{\text{d}} X_\infty$, if > $ > \limv{n}\bp\bracs{X_n \le x} = \bp_\infty\bracs{X_n \le x} > $ > [!theorem] > > Let $\seq{X_n}$ be a sequence of random variables defined over a common probability space $(\Omega, \cf, \bp)$. If $X_n \to X$ [[Convergence in Measure|in probability]], then $X_n \to X$ in distribution. > > *Proof*. Let $F_X$ be the [[Cumulative Distribution Function|CDF]] of $X$, and let $x \in \real$ be a continuity point of $F_X$ meaning that > $ > \bp\bracs{X \le x} = \bp\bracs{X < x} > $ > If $x$ is not a continuity point, then $P\bracs{X = x} > 0$. > > Suppose that $x$ is a continuity point, then for any $\varepsilon > 0$, there exists $\delta > 0$ such that $F_X(x - \delta) > F_X(x) - \delta$. Note that > $ > \bracs{X \le n} \subset \bracs{X_n \le x + \delta} \cup \bracs{\abs{X_n - X} > \delta} > $ > Taking a limit on the right hand side yields > $ > \begin{align*} > \bp\bracs{X \le x - \delta} &\le \limsup_{n \to \infty}\bp\bracs{X_n \le x} + \underbrace{\limsup_{n \to \infty}\bp\bracs{\abs{X_n - X} > \delta}}_{0} \\ > \bp\bracs{X \le x} - \varepsilon &\le \limsup_{n \to \infty}\bp\bracs{X_n \le x} > \end{align*} > $ > Since the above holds for any $\varepsilon > 0$ and for $\liminf$, meaning that $\bp\bracs{X \le x} = \lim_{n \to \infty}\bp\bracs{X_n \le x}$. Since this is a continuity point, we can apply the argument to the other side as well. > > As there can only be countably many jumps, we can simply fill in these holes by continuity. > [!definition] > > Let $\seq{X_n}$ be a family of $\real^n$-valued random variables such that $X_n \to c$ in distribution, then $X_n \to c$ in probability. > [!definition] > > Let $\seq{X_n}$ be a family of $\real$-valued random variables. If each $X_n$ has density $f_n$ with respect to the Lebesgue measure, $X$ has density $f$, and $f_n \to f$ almost everywhere, then $X_n \to X$ in distribution.