# A sequence of random variables which converges in distributon converges "to" some random variable

Let $(X_n)$ be a sequence of random variables on a probability space $\Omega$, with distribution functions $F_n$. Suppose $F_n \rightarrow F$ in distribution for some distribution function $F$. Must there be a random variable $X$ on $\Omega$ with $X_n \rightarrow X$ in distribution?

This seems like the sort of result one would want to be true, but I couldn't think of a way to prove it. To do so, one would need to somehow define $X$ "backwards" given its distribution $F$. Does anyone have any ideas?

• What do you mean by "Suppose $F_n \to F$ in distribution"? Did you mean "pointwise" instead? Commented Mar 16, 2013 at 2:35
• Yeah, $F_n \rightarrow F$ pointwise, so if $X_n \sim F_n$ and $X\sim F$ then $X_n \rightarrow X$ in distribution. Commented Mar 16, 2013 at 4:02
• I believe the answer is no; I've seen such random variables $X$ referred to as "fictitious" when there does not exist an $X$ defined on $\Omega$ with distribution function $F$. I don't have a reference, though, sorry.
– guy
Commented Mar 16, 2013 at 4:43
• Chow and Teicher's probability theory book uses the "fictitious random variable" terminology, so you might look there for a counterexample; I don't have access to the book though.
– guy
Commented Mar 16, 2013 at 4:58

For every distribution function $F$ there exists random variables $X$ such that $F=F_X$. A way to exhibit such a random variable is to set $X=G(U)$ where $U$ is uniform on $(0,1)$ and $G$ is an (extended) inverse of $F$. The function $$G:u\mapsto\inf\{x\mid F(x)\geqslant u\}$$ might do the job but you should check the inequality sign.
Edit: For a counterexample, one should consider $\Omega$ such that no uniform random variable $U$ as above exists on $\Omega$. Consider $\Omega=\mathbb N$ with the sigma-algebra $2^\Omega$ and $\mathbb P(\{n\})=p_n$. Call $S\subseteq\mathbb N$ the set $S=\{s(K)\mid K\subseteq\mathbb N\}$ where $s(K)=\sum\limits_{n\in K}p_n$ for $K\subseteq\mathbb N$. Choose $(s_n)\subseteq S$ such that $s_n\to s$. Then $X_n=\mathbb 1_{K_n}$, with $s_n=s(K_n)$, converges in distribution to a Bernoulli random variable $X$ with parameter $s$, hence if $s$ is not in $S$, $X$ cannot be defined on $\Omega$. To complete the proof, one must exhibit $(p_n)$ such that the set $S$ is not closed. At the moment, I do not know if this is possible...
• True, but in the situation I am interested in the probability space $\Omega$ is given. I am trying to show that $\Omega$ admits a random variable $X$ with the distribution $F$, given that it admits a sequence of random variables with the distributions $F_n$. Commented Mar 16, 2013 at 4:23