Let $X,X_1,X_2,X_3,...$ be positive integer random variables. Show that $X_n \overset{d}{\to} X$ implies $\lim_{n\to\infty} P(X_n=k) = P(X=k)$

Question

Question

Let $X,X_1,X_2,X_3,...$ be positive integer random variables. Show that $X_n \overset{d}{\to} X$ implies $\lim_{n\to\infty} P(X_n=k) = P(X=k)$.

The $\overset{d}{\to}$ denotes convergence in distribution.

Attempt

Here I try to show

$$X_n \overset{d}{\to} X \implies \lim_{n\to\infty} P(X_n=k) = P(X=k)$$

Let $F(x)$, $F_{X_n}(x)$ be the cdfs of $X,X_n$ respectively. Let $k>0$ and suppose that $k$ is not in the support of $X$, i.e. $P(X=k)=0$.

$$ \begin{align} \lim_{n\to\infty}P(X_n=k) &= \lim_{n\to\infty}\left[ P(X_n\leq k) - P(X_n<k) \right] \\ &= \lim_{n\to\infty}\left[ F_{X_n}(k) - \lim_{x\to k^-}F_{X_n}(x) \right]\\ &= F(k) - \lim_{n\to\infty}\lim_{x\to k^-}F_{X_n}(x) \end{align} $$

I get stuck. Not sure where to go forward.

Now suppose that $k>0$ is in the support of $X$, so $P(X=k) >0$. I also get stuck here. Again, I can write

$$ \begin{align} \lim_{n\to\infty}P(X_n=k) &= \lim_{n\to\infty}\left( F_{X_n}(k) - \lim_{x\to k^-}F_{X_n}(x) \right)\\ \end{align} $$

But convergence in distribution only implies $F_{X_n}(t) \to F(t)$ when $F$ is continuous at $t$. Since $X$ is a discrete random variable, $F$ is not continuous at $k$ in the support. Again, not sure how to move forward?

Answer 1

$F$ actually is continuous at all non-integers (where, of course, its graph is horizontal, reflecting zero chance that $X$ will be nonintegral). Consequently, in particular, the convergence in distribution of the sequence implies convergence at (say) the values $k\pm 1/2,$ from which you may instantly conclude for any integral $k$ that

$$\begin{aligned} \lim_{n\to\infty}\Pr(X_n=k) &= \lim_{n\to\infty}\left(\Pr(X_n\le k+1/2) - \Pr(X_n\le k-1/2)\right) \\ &= \lim_{n\to\infty}\left(F_{X_n}(k+1/2) - F_{X_n}(k-1/2)\right)\\ &= \lim_{n\to\infty} F_{X_n}(k+1/2) - \lim_{n\to\infty}F_{X_n}(k-1/2)\\ &= F(k+1/2) - F(k-1/2)\\ &= \Pr(X=k), \end{aligned}$$

QED.

Answer 2

Following on from whuber's excellent answer, we can also see that this theory applies for any set of random variables with countable support, even if they are not positive integers. To see this, consider a sequence of random variables with distribution on a countable support $\mathscr{X} \subset \mathbb{R}$. For any such set, there exist continuous non-decreasing functions $u:\mathbb{R} \rightarrow \mathbb{R}$ and $r:\mathbb{R} \rightarrow \mathbb{R}$ with the property that:

$$x' < r(x) < x < u(x) < x'' \quad \quad \quad \text{for all } \ x' < x < x'' \ \text{ that are all in } \mathscr{X},$$

(Proving this result is a useful follow-up exercise you might like to try.) It follows that for all $x \in \mathscr{X}$ (and for any random variable in the sequence) we have:

$$\mathbb{P}(X_n = x) = \mathbb{P}(r(x) < X_n \leqslant u(x)) = F_{X_n}(u(x)) - F_{X_n}(r(x)).$$

By analogy to the solution for the case of positive integer random variables, we then have:

$$\begin{aligned} \lim_{n \rightarrow \infty} \mathbb{P}(X_n = x) &= \lim_{n \rightarrow \infty} \Big[ F_{X_n}(u(x)) - F_{X_n}(r(x)) \Big] \\[10pt] &= \lim_{n \rightarrow \infty} F_{X_n}(u(x)) - \lim_{n \rightarrow \infty} F_{X_n}(r(x)) \\[10pt] &= F_X(r(x)) - F_X(u(x)) \\[12pt] &= \mathbb{P}(X = x). \\[10pt] \end{aligned}$$