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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?

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    $\begingroup$ Hint: consider the limits of $F_{X_n}(k+1/2)$ and $F_{X_n}(k-1/2)$ as $n\to\infty$ and apply a theorem about limits of differences of sequences. $\endgroup$ – whuber Dec 5 '20 at 18:27
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$F$ 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.

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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:

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

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    $\begingroup$ Since you seem interested in capturing the essence of the situation--which is always worthwhile--note that smoothness of these functions is of no import. What ultimately matters is that the support is discrete: that is, there is an open neighborhood around each point in the support that intersects the support at that point only. $\endgroup$ – whuber Dec 5 '20 at 20:50
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    $\begingroup$ @whuber: I have edited to specify continuity instead of smoothness, since continuity makes the limit results easier. Does this look better to you? $\endgroup$ – Ben Dec 5 '20 at 21:54
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    $\begingroup$ The point is that you don't need these functions to have much of any properties at all except to identify suitable neighborhoods of $k:$ specifying them just obscures that basic point. The problem is solved simply by restating $\Pr(X_n=x)$ in terms of the probability that $X_n$ lies between two defined values at which $F$ is continuous. We don't otherwise care about the properties of $F_n;$ it suffices that $F_n$ converges to $F$ pointwise and is discrete with the same support as $F.$ $\endgroup$ – whuber Dec 6 '20 at 2:17
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    $\begingroup$ Yes, fair enough. $\endgroup$ – Ben Dec 6 '20 at 4:29

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