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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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3 Answers 3

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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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You are right. I should have replaced (**) by saying that $x_1$ and $x_2$ are in $(n,n+1)$ for some $n$. Since there are uncountably many points in any open interval and only a countable set of points of discontinuity and since $F$ is non-decreasing, I think that if $F$ is not constant on the interval then it is possible to find two points of continuity where the value of $F$ differs. –  Jyotirmoy Bhattacharya Mar 10 '11 at 9:08
    
@Jyotirmoy Bhattacharya. Yes, it works. But to be clear, you prove by reduction to absurd that $F$ is constant on each interval $(n,n+1)$, where $n$ integer. $\textbf{After that}$, you deduce that $X$ is integer valued. Well done! Yuval Filmus'proof is faster –  Theta33 Mar 10 '11 at 18:59
    
@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. –  Theta33 Mar 10 '11 at 19:19

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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Please see my comment on @Sivaram's answer. –  Jyotirmoy Bhattacharya Mar 10 '11 at 7:01
    
@Jyotirmoy Bhattacharya. Since $F$ is constant on $(n,n+1)$, $F$ is continuous on such intervals. So the only points of discontinuity might be the integers. If $F$ is discontinuous at some $n$, it means $n$ is in the support of $X$. There's a jump in the graph of $F $ there. But this is what we wanted to prove. Tip: think about the problem graphically. –  Theta33 Mar 10 '11 at 8:08
    
@Mielu. $F$ being a constant on $(n,n+1)$ is precisely what we have to prove. –  Jyotirmoy Bhattacharya Mar 10 '11 at 9:04
    
@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. –  Yuval Filmus Mar 10 '11 at 16:11

$X$ 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. –  Jyotirmoy Bhattacharya Mar 10 '11 at 7:00

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