Say that we have a sequence of discrete random variables, $\left\{X_n\right\}_{n \in \mathbb{N}}$, which converges to a random variable, $X$, with a continuous distribution, e.g., the Normal (Gaussian) distribution.

Denote the corresponding sequence of PMFs for $\left\{X_n\right\}_{n \in \mathbb{N}}$ by $\left\{g_n\right\}_{n \in \mathbb{N}}$. Is this sequence of PMFs, $\left\{g_n\right\}_{n \in \mathbb{N}}$, uniformly bounded?

Based on the definition I have seen for uniform boundedness, the sequence of PMFs, $\left\{g_n\right\}_{n \in \mathbb{N}}$, would be uniformly bounded when

$$\sup_n \left\lvert g_n(x) \right\rvert \leq M(x) < \infty \text{ for each } x \in \mathbb{R}.$$

Even if the limiting random variable has a continuous distribution, it would seem that uniform boundedness of the sequence of PMFs trivially holds: For each $n = 1, 2, 3, \dots$ in the sequence, $\left\{g_n\right\}_{n \in \mathbb{N}}$, the largest possible value of $g_n(x)$ is $1$ since $g_n$ is a probability mass function. Hence, wouldn't letting $M(x) = M \geq 1$ for every $n \in \mathbb{N}$ ensure that $\left\{g_n\right\}_{n \in \mathbb{N}}$ satisfies uniform boundedness as defined above? Or is my reasoning incorrect?

Many thanks to all for any help.

  • $\begingroup$ Thank you. I think your answer clarifies. My concern (based on I think faulty reasoning) was as follows:A discrete variable converges in CDF to a continuous variable. The corresponding PDF of the limiting continuous variable may be unbounded. So if the sequence of PMFs limits to this PDF, then the limiting PDF may be unbounded. Yet this reasoning is wrong, I suppose, because (1) the limiting PDF of a sequence is irrelevant for uniform boundedness of the sequence of PMFs and (2) convergence in CDFs does not imply convergence of PMFs to PDF of the limiting distribution? $\endgroup$ Jun 18, 2022 at 21:43
  • $\begingroup$ You're right: axiomatically, no value of a pmf may exceed $1.$ But so what? $\endgroup$
    – whuber
    Jun 18, 2022 at 22:54

1 Answer 1


Note that convergence to a random variable is convergence of the cumulative distribution function (CDF) and not of the probability density function (PDF).

Possibly that is why you might think that your reasoning is not correct? Like the following:

  • If the discrete variable convergences to a continuous distribution
  • and the continuous distribution can be unbounded
  • then how can the discrete distribution be bounded?.

Well, the reason is because the convergence relates to the CDF and not to the PMF and PDF. The values of the PMF can be a lot different from the values of the PDF.

Below is an example of a construction with discrete distributions that converges to the exponential distribution, even when the PMF has a constant value. (The idea is that we pick n evenly spaced quantiles) Possibly that construction can help you see the difference between the PMF and the PDF. The latter relates to a density and this can exceed 1.

pmf and pdf

The discrete distribution used here $f_n$ follows the function:

$$f_n(X = x) = \begin{cases} \frac{1}{n} & \text{if $x = -\frac{\ln\left(1-\frac{0}{n}\right)}{\lambda}$}\\ \frac{1}{n} & \text{if $x = -\frac{\ln\left(1-\frac{1}{n}\right)}{\lambda}$}\\ \frac{1}{n} & \text{if $x = -\frac{\ln\left(1-\frac{2}{n}\right)}{\lambda}$}\\ \vdots&\vdots\\ \frac{1}{n} & \text{if $x = -\frac{\ln\left(1-\frac{n-1}{n}\right)}{\lambda}$}\\ 0 & \text{else} \end{cases}$$ So the probability is $\frac{1}{n}$ when $X$ is equal to the $\frac{k}{n}$-th quantile or else it is zero.

Related question: How can a probability density function (pdf) be greater than 1 ?


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