In the negative binomial experiment, set $$k = 1$$ to get the geometric distribution. In the discrete case, as usual, the measure theory and topology are not really necessary. The random variables need not be defined on the same probability space. P. Billingsley, "Convergence of probability measures" , Wiley (1968), N. Bourbaki, "Elements of mathematics. The difference between the two kinds of convergence is that one might converge to a non probability while the other diverges. The probability of success $$p \in [0, 1]$$ is the same for each trial. Recall from analysis that since $$F_\infty^{-1}(u)$$ is increasing, the set $$D \subseteq (0, 1)$$ of discontinuities of $$F_\infty^{-1}$$ is countable. Warning: If $\mathcal{B}$ is the $\sigma$-algebra of Borel sets of a topological space $X$, we will then denote by $\mathcal{M}^b (X)$ the space of Radon signed measures, i.e. A sequence of probability measures converging in the narrow topology is often called a weakly converging sequence. If $X$ is compact, then the space of probability measures with the narrow (or wide) topology is also compact. For $$i \in \{1, 2, \ldots, n\}$$, the number of such permutations with $$i$$ in position $$i$$ is $$(n - 1)!$$. This is actually not a restriction in many cases, for instance if $X$ is the euclidean space. But $$\int_\R g_n \, d\mu = 0$$ so $$\int_\R g_n^+ \, d\mu = \int_\R g_n^- \, d\mu$$. Let $$n \to \infty$$ and $$\epsilon \downarrow 0$$ to conclude that $$\limsup_{n \to \infty} F_n^{-1}(u) \le F_\infty^{-1}(v)$$. However, if probability density functions of a fixed type converge then the distributions converge. For more information contact us at [email protected] or check out our status page at https://status.libretexts.org. Then $$P_n \Rightarrow P_\infty$$ as $$n \to \infty$$ if and only if $$P_n(A) \to P_\infty(A)$$ as $$n \to \infty$$ for every $$A \subseteq S$$. Note that the binomial distribution with parameters $$n$$ and $$p = r / m$$ is the distribution that governs the number of type 1 objects in a sample of size $$n$$, drawn with replacement from a population of $$m$$ objects with $$r$$ objects of type 1. It follows that $$F_\infty^{-1}(u) - \epsilon \lt x \lt F_n^{-1}(u)$$ for all but finitely many $$n \in \N_+$$. Therefore $$f_n(k) \to e^{-r} r^k / k!$$ as $$n \to \infty$$ for each $$k \in \N_+$$. The proof is finished, but let's look at the probability density functions to see that these are not the proper objects of study. For a specific construction, we could take $$\Omega = (0, 1)$$, $$\mathscr F$$ the $$\sigma$$-algebra of Borel measurable subsets of $$(0, 1)$$, and $$\P$$ Lebesgue measure on $$(\Omega, \mathscr F)$$ (the uniform distribution on $$(0, 1)$$). This convergence is sometimes called convergence in variation. Next, let $$v$$ satisfy $$0 \lt u \lt v \lt 1$$ and let $$\epsilon \gt 0$$. Hence $$F_n(x) \le F_\infty(x + \epsilon) + \P\left(\left|X_n - X_\infty\right| \gt \epsilon\right)$$. If $$P$$ is a probability measure on $$(\R^n, \mathscr R_n)$$, recall that the distribution function $$F$$ of $$P$$ is given by $F(x_1, x_2, \ldots, x_n) = P\left((-\infty, x_1] \times (-\infty, x_2] \times \cdots \times (-\infty, x_n]\right), \quad (x_1, x_2, \ldots, x_n) \in \R^n$. Again, the number of permutations of $$\{1, 2, \ldots, n\}$$ is $$n!$$. For $$n \in \N_+^*$$, let $$F_n$$ denote the distribution function of $$P_n$$ and define $$X_n = F_n^{-1}(U)$$ where $$F_n^{-1}$$ is the quantile functions of $$F_n$$. Let $$(\Omega, \mathscr F, \P)$$ be a probability space and $$U$$ a random variable defined on this space that is uniformly distributed on the interval $$(0, 1)$$. \int f\, \mathrm{d}\mu_n \to \int f\, \mathrm{d}\mu Note that the limiting condition on $$n$$ and $$p$$ in the last result is precisely the same as the condition for the convergence of the binomial distribution to the Poisson distribution. Recall next that Bernoulli trials are independent trials, each with two possible outcomes, generically called success and failure. This is often a useful result, again not computationally, but rather because the Poisson distribution has fewer parameters than the binomial distribution (and often in real problems, the parameters may only be known approximately). Recall that the Pareto distribution with shape parameter $$a \in (0, \infty)$$ has distribution function $$F$$ given by $F(x) = 1 - \frac{1}{x^a}, \quad 1 \le x \lt \infty$ The Pareto distribution, named for Vilfredo Pareto, is a heavy-tailed distribution sometimes used to model financial variables. We assume that $$(S, d)$$ is a complete, separable metric space and let $$\mathscr S$$ denote the Borel $$\sigma$$-algebra of subsets of $$S$$, that is, the $$\sigma$$-algebra generated by the topology. Missed the LibreFest?