# Basic Properties of Convergence in distribution

Suppose real-valued random variables $X_n$ converge to $X$ in distribution. Suppose $x_n$ be a sequence of numbers that for each $n$, $\{x_n \}$ is in the support of the distribution of $X_n$. Suppose $x$ is a limit of a sequence $\{x_n\}$. Will it be that $x$ is also in the support of the distribution of $X$?

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## 1 Answer

The answer is no. Recall that $x$ being in the support of $X$ means that $\mathrm P(X\in U)\ne0$ for every neighborhood $U$ of $x$ and that this definition implies that the support is always a closed set.

For a counterexample, assume that $X_n=Y+5B_n$ where $Y$ and $B_n$ are independent for every $n$, $Y$ is uniform on $(0,1)$ and $B_n$ is a Bernoulli random variable such that $\mathrm P(B_n=1)=b_n$ and $\mathrm P(B_n=0)=1-b_n$ for some parameter $b_n$ in $(0,1)$. Then the support of $X_n$ is $[0,1]\cup[5,6]$ for every $n$ and, if $b_n\to0$, then $X_n\to X$ in distribution, where $X$ is uniform on $(0,1)$. The support of $X$ is $[0,1]$ hence $x_n=x=5$ provides a counterexample.

Edit Here is an answer to the, quite different, question asked in comments. Let $x$ denote any point in the support of $X$, hence $\mathrm P(|X-x|\leqslant1/k)\ne0$ for every positive integer $k$. In particular, the liminf of $\mathrm P(|X_n-x|\leqslant1/k)$ is not zero, hence $\mathrm P(|X_n-x|\leqslant1/k)\ne0$ for every $n$ large enough, say every $n\geqslant n_k$.

Thus, for every $n\geqslant n_k$, the support of $X_n$ meets $[x-1/k,x+1/k]$, say at least at point $z_{n,k}$. Assume without loss of generality that $(n_k)_k$ is nondecreasing and define $x_n=z_{n,k}$ for every $n_k\leqslant n\lt n_{k+1}$. Then, for each $n$, $x_n$ is in the support of $X_n$, and $x_n\to x$ when $n\to\infty$.

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Then how about a weaker statement that for each $x$ that is in the support of $X$ there exists a sequence $x_n$ that converges to $x$ and each is in the support of $X_n$? – helloworld Dec 25 '11 at 16:40