Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $X$?

share|improve this question
add comment

1 Answer

up vote 3 down vote accepted

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 $n\to\infty$.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.