Convergence in distribution is the language for limits of probability laws when pointwise convergence of random variables is too strong or unavailable.

## The problem convergence in distribution solves

Many limiting arguments in probability give limiting laws without giving a limiting [random variable](/page/Random%20Variable) on the original [probability space](/page/Probability%20Space).

Statistics may be computed on sample spaces that change with $n$, while normalized counts may remain lattice-valued for every $n$ and still approach continuous laws.

Convergence in distribution records this law-level limiting behavior.

It is weaker than convergence in probability and almost sure convergence, but it is the right notion for limit theorems such as the [central limit theorem](/theorems/521).

The central object is the distribution, not the individual value of a random variable on a sample point.

If $X_n$ and $X$ are real-valued random variables, the question is not whether $X_n(\omega)$ is close to $X(\omega)$ for most $\omega$, but whether the law of $X_n$ assigns nearly the same mass as the law of $X$ to the sets that matter for weak limits.

For real-valued variables this can be stated with distribution functions.

For random elements in metric spaces it is more natural to use expectations of bounded continuous test functions.

Both views express the same idea: only the limiting law matters.

The notation used here is

\begin{align*} X_n \Rightarrow X \end{align*}

or

\begin{align*} \mathcal L(X_n) \Rightarrow \mathcal L(X). \end{align*}

The symbol $\mathcal L(X)$ denotes the law of $X$.

The arrow $\Rightarrow$ is reserved for convergence in distribution, not almost sure convergence or convergence in probability.

[example: Lattice Laws Approaching a Smooth Law] Let $Y_1,Y_2,\ldots$ be independent Bernoulli$(p)$ random variables with $0<p<1$, set $S_n=Y_1+\cdots+Y_n$, and define \begin{align*} X_n=\frac{S_n-np}{\sqrt{np(1-p)}}. \end{align*} For each $i$, the Bernoulli law gives $\mathbb P(Y_i=1)=p$ and $\mathbb P(Y_i=0)=1-p$, so \begin{align*} \mathbb E Y_i=1\cdot p+0\cdot(1-p)=p. \end{align*} Since $Y_i$ only takes the values $0$ and $1$, both possible values satisfy $y^2=y$, so $Y_i^2=Y_i$ almost surely. Therefore \begin{align*} \mathbb E(Y_i^2)=\mathbb E(Y_i)=p. \end{align*} Using $\operatorname{Var}(Y_i)=\mathbb E(Y_i^2)-(\mathbb E Y_i)^2$, we get \begin{align*} \operatorname{Var}(Y_i)=p-p^2=p(1-p). \end{align*} Because $0<p<1$, both $p>0$ and $1-p>0$, hence $p(1-p)>0$. Since each $Y_i$ is either $0$ or $1$, the sum $S_n$ can take only the integer values $0,1,\ldots,n$. If $S_n=k$, then substitution into the definition of $X_n$ gives \begin{align*} X_n=\frac{k-np}{\sqrt{np(1-p)}}. \end{align*} Thus every possible value of $X_n$ belongs to \begin{align*} \left\{\frac{k-np}{\sqrt{np(1-p)}}:k=0,1,\ldots,n\right\}. \end{align*} Conversely, whenever $S_n=k$, the displayed value of $X_n$ occurs by the same substitution. Hence these are exactly the possible values of $X_n$, so every finite-sample law of $X_n$ is supported on a finite lattice. For $k\in\{0,1,\ldots,n\}$, the event $S_n=k$ is the event that exactly $k$ of the variables $Y_1,\ldots,Y_n$ are equal to $1$ and the remaining $n-k$ are equal to $0$. For a fixed set $I\subset\{1,\ldots,n\}$ with $|I|=k$, the corresponding event is \begin{align*} \{Y_i=1\text{ for }i\in I,\ Y_j=0\text{ for }j\notin I\}. \end{align*} By independence, its probability is \begin{align*} \prod_{i\in I}\mathbb P(Y_i=1)\prod_{j\notin I}\mathbb P(Y_j=0)=\prod_{i\in I}p\prod_{j\notin I}(1-p). \end{align*} The set $I$ has $k$ elements and its complement has $n-k$ elements, so \begin{align*} \prod_{i\in I}p\prod_{j\notin I}(1-p)=p^k(1-p)^{n-k}. \end{align*} There are $\binom nk$ choices of such a set $I$, and the corresponding events are disjoint because a single outcome has one fixed set of positions where $Y_i=1$. Therefore \begin{align*} \mathbb P(S_n=k)=\binom nk p^k(1-p)^{n-k}. \end{align*} The denominator $\sqrt{np(1-p)}$ is positive, because $n\ge1$ and $p(1-p)>0$. Hence the map $k\mapsto (k-np)/\sqrt{np(1-p)}$ is one-to-one on $\{0,1,\ldots,n\}$, and \begin{align*} \left\{X_n=\frac{k-np}{\sqrt{np(1-p)}}\right\}=\{S_n=k\}. \end{align*} Therefore \begin{align*} \mathbb P\left(X_n=\frac{k-np}{\sqrt{np(1-p)}}\right)=\binom nk p^k(1-p)^{n-k}. \end{align*} The variables $Y_i$ are independent and identically distributed with mean $p$ and variance $p(1-p)>0$, so the [Central Limit Theorem](/theorems/532), applied with $\mu=p$ and $\sigma=\sqrt{p(1-p)}$, gives \begin{align*} \frac{S_n-np}{\sqrt{p(1-p)}\sqrt n}\Rightarrow Z, \end{align*} where $Z$ has the standard normal distribution. Since $n>0$, $p>0$, and $1-p>0$, both $\sqrt{p(1-p)}\sqrt n$ and $\sqrt{np(1-p)}$ are positive. Their squares agree: \begin{align*} \left(\sqrt{p(1-p)}\sqrt n\right)^2=p(1-p)n=np(1-p). \end{align*} Thus \begin{align*} \sqrt{p(1-p)}\sqrt n=\sqrt{np(1-p)}. \end{align*} Substituting this equality into the normalized sum gives \begin{align*} \frac{S_n-np}{\sqrt{p(1-p)}\sqrt n}=\frac{S_n-np}{\sqrt{np(1-p)}}=X_n. \end{align*} Hence \begin{align*} X_n\Rightarrow Z. \end{align*} Let $A$ be a Borel set whose boundary has normal probability zero: \begin{align*} \mathbb P(Z\in\partial A)=0. \end{align*} The law of $Z$ assigns mass $\mathbb P(Z\in B)$ to each Borel set $B$, so the displayed condition says that the law of $Z$ assigns zero mass to $\partial A$. Thus $A$ is a continuity set for the law of $Z$. By the continuity-set form of the *[Portmanteau Theorem](/theorems/1171)*, \begin{align*} \mathbb P(X_n\in A)\to \mathbb P(Z\in A). \end{align*} The finite laws never stop being lattice laws, but convergence in distribution records that their masses on normal-continuity sets approach the corresponding masses of the smooth limiting normal law. [/example]

## Definition

The formal definition captures convergence of the laws of real-valued random variables by checking their distribution functions only at continuity points of the limiting distribution function.