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Let $X_1,X_2,...$ be a sequence of integer-valued random variables that converge in distribution to some random variable $X$. Am I right in thinking that we can always pick $X$ to be integer valued?

I thought like this: Let $F_n$ be the distribution function of $X_n$ and $F$ be the distribution function of $X$. Then $X$ cannot be integer valued only if there are two points $x_1$ and $x_2$ such that $F(x_1) \neq F(x_2)$ and $|x_1-x_2|<1$

Now since a distribution function cannot have more than a countable number of points of discontinuity, we can pick $x_1$ and $x_2$ such that they are points of continuity of $F$. But then for large enough $n$ $F_n(x_1) \neq F_n(x_2)$ But this contradicts the fact that $X_n$ is integer-valued.

Am I right?

[This is motivated by exercise 5.12 of Wasserman's All of Statistics where he assumes that $X$ is integer-valued]

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    This has been answered by Byron before.Please have a look here: http://math.stackexchange.com/questions/571025/the-limit-of-integer-valued-random-variables-must-be-integer-valued2015-04-21

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$X$ has to be an integer valued random variable.

$X_n$ converges to $X$ in distribution if $F_n$ is the cdf of $X_n$ and if $F$ is the cdf of $X$, then $\lim_{n \rightarrow \infty} F_n(x) = F(x)$ Note that if $X_n$ is an integer valued random variable then $F_n(x) = F_n(\lfloor x \rfloor)$.

Consider $|F(x) - F(\lfloor x \rfloor)|$ and argue that this can be made smaller than $\epsilon$, $\forall$ $\epsilon >0$, by adding and subtracting $F_n(x)$ and making use of the fact that $F_n(x) = F_n(\lfloor x \rfloor)$ and then use triangle inequality and choose $N(\epsilon)$ such that both the $\frac{\epsilon}{2}$ are satisfied.

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    In the definition of convergence in distribution $\lim_{n \to \infty} F_n(x)=F(x)$ needs to be true only on points of continuity of $F$. So we need to somehow tackle potential points of discontinuity.2011-03-10
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The CDFs $F_i$ all have the property that $F_i$ is constant on $(n,n+1)$ for every integer $n$. This property is conserved by the limiting function, and is in fact equivalent to the random variable being supported on the integers.

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    @Jyotirmoy. This is trivial since if $x_m = y_m$ and $\lim_{m\rightarrow\infty} x_m$ exists then $\lim_{m\rightarrow\infty y_m}$ exists and the limits are equal.2011-03-10
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There are problems with your proof.

I thought like this: Let $F_n$ be the distribution function of $X_n$ and $F$ be the distribution function of $X$. Then $X$ cannot be integer valued only if there are two points $x_1$ and $x_2$ such that $F(x_1) \neq F(x_2)$ and $|x_1-x_2|<1$ (**)

Not necessarily. What if the range of $X$ are non-integers with spaces between them bigger or equal to 1? For example , all the points $n+1/2$, $n\in \mathbb{Z}$

Now since a distribution function cannot have more than a countable number of points of discontinuity, we can pick $x_1$ and $x_2$ such that they are points of continuity of $F$.

You pick such points of continuity, but at the same time you pick them with the property (**) from the first part. You have to justify that.

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    @Jyotirmoy Bhattacharya. Actually, on second thought, I agree with you-Yuval Filmus'proof is incomplete because he does not consider that the limit in distribution is for the points where $F$ is continuous.2011-03-10