{"id":26769,"date":"2021-05-27T13:00:54","date_gmt":"2021-05-27T13:00:54","guid":{"rendered":"http:\/\/toposuranos.com\/material\/?p=26769"},"modified":"2024-05-22T01:36:11","modified_gmt":"2024-05-22T01:36:11","slug":"poisson-process-approximation-of-the-binomial-process","status":"publish","type":"post","link":"http:\/\/toposuranos.com\/material\/en\/poisson-process-approximation-of-the-binomial-process\/","title":{"rendered":"Poisson Process: Approximation of the Binomial Process"},"content":{"rendered":"<div style=\"background-color:#F3F3F3; padding:20px;\">\n<center><\/p>\n<h1>Poisson Process: Approximation of the Binomial Process<\/h1>\n<p><\/p>\n<p style=\"text-align:center;\"><strong>Summary<\/strong><br \/><em>This class focuses on the Poisson Process as an approximation to the Binomial Process, starting with the definition of the coefficients and the Poisson distribution, which is derived from a Bernoulli event with a large number of trials and a very small individual probability. The core of this class addresses the approximate Poisson processes, both spatial and temporal, using examples of tiny particles in a liquid and the emission of particles by a radioactive substance, respectively. Finally, it concludes with practical examples of the application of the Poisson distribution in different contexts, such as customer service in a supermarket and population density in a locality.<\/em><\/p>\n<p><\/center><br \/>\n<\/p>\n<p style=\"text-align:center;\"><strong>LEARNING OBJECTIVES:<\/strong><br \/>\nAt the end of this class, the student will be able to:\n<\/p>\n<ol>\n<li><strong>Understand<\/strong> the definition and coefficients of the Poisson distribution.<\/li>\n<li><strong>Understand<\/strong> the Poisson process as an approximation to the binomial process.<\/li>\n<li><strong>Understand<\/strong> the formal equivalence between spatial and temporal Poisson processes.<\/li>\n<li><strong>Use<\/strong> the Poisson distribution to solve practical problems.<\/li>\n<\/ol>\n<p><center><br \/>\n<strong><u>CONTENTS INDEX<\/u>:<\/strong><br \/>\n<a href=\"#1\">The Coefficients and the Poisson Distribution<\/a><br \/>\n<a href=\"#2\">Approximate Poisson Processes<\/a><br \/>\n<a href=\"#3\">Spatial Poisson Process<\/a><br \/>\n<a href=\"#4\">Temporal Poisson Process<\/a><br \/>\n<a href=\"#5\">Temporal and Spatial<\/a><br \/>\n<a href=\"#6\">Practical examples where the Poisson distribution is used<\/a><br \/>\n<\/center><br \/>\n<center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/mQ0j3FE8p2U\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>\n<\/div>\n<p><a name=\"1\"><\/a><\/br><\/br><\/p>\n<h2>The Coefficients and the Poisson Distribution<\/h2>\n<p style=\"text-align: justify; color: #000000;\"><a href=\"https:\/\/www.youtube.com\/watch?v=mQ0j3FE8p2U&amp;t=154s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">Now let&#8217;s consider an approximation<\/span><\/strong><\/a> for the <a href=\"https:\/\/toposuranos.com\/el-ensayo-de-bernoulli-para-n-intentos-independientes\/\" rel=\"noopener\" target=\"_blank\">binomial distribution<\/a>, where we consider a very large number of trials <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n<\/span><\/span> and all with a very small individual probability <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">p<\/span><\/span>. When we do this, we move from the typical Binomial process to a Poisson Process. To visualize this, let&#8217;s imagine a sequence of the form <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\{Bi(n;k;p_n)\\}_n,<\/span><\/span> where <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n\\to\\infty<\/span><\/span> and <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">p_n<\/span><\/span> satisfies the relationship <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">np_n=\\lambda \\gt 0<\/span><\/span>. From this, we will see that<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\\lim_{n\\to\\infty}P\\left(Bi(n;k;P_n) \\right) = \\frac{\\lambda^k}{k!}e^{-\\lambda}<\/span><\/span><\/p>\n<p style=\"text-align: justify; color: #000000;\">This is actually not difficult to prove, if we take the probability of a Bernoulli event <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">Bi(n;k;p_n)<\/span><\/span> and multiply and divide by <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n^k<\/span><\/span>, we get the following reasoning:<\/p>\n<table style=\"text-align: justify; color: #000000;\">\n<tbody>\n<tr>\n<td width=\"80px\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">P(B(n;k;p_n))<\/span><\/span><\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle={{n}\\choose{k}}p^k(1-p)^{n-k}<\/span><span style=\"color: #f00000;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\cdot \\displaystyle \\frac{n^k}{n^k}<\/span><\/span><\/span><\/td>\n<\/tr>\n<tr>\n<td width=\"80px\"><\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle=\\frac{n!}{(n-k)!k!}p^k(1-p)^{n-k} \\cdot \\frac{n^k}{n^k}<\/span><\/span><\/td>\n<\/tr>\n<tr>\n<td width=\"80px\"><\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle=\\frac{n(n-1)\\cdots[n-(k-1)]}{n^k} \\cdot \\frac{(np_n)^k}{k!} (1-p_n)^{-k}(1-p_n)^n<\/span><\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: justify; color: #000000;\">So if we calculate the limit when <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n\\to\\infty<\/span><\/span>, we will have:<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><br \/>\n<span class=\"katex-eq\" data-katex-display=\"false\">\n\\begin{array}\n\n\\displaystyle \\lim_{n\\to\\infty} {{n}\\choose{k}}p_n^k(1-p_n)^{n-k} &amp;= \\lim_{n\\to\\infty} \\underbrace{\\frac{n(n-1)\\cdots[n-(k-1)]}{n^k}}_{\\to 1} \\cdot \\frac{\\overbrace{(np_n)^k}^{\\to\\lambda^k}}{k!} \\overbrace{(1-p_n)^{-k}}^{\\to 1} {(1-p_n)^n} \\\\ \\\\\n\n&amp;\\displaystyle = \\frac{\\lambda^k}{k!} \\lim_{n\\to\\infty}\\left(1 - \\frac{\\lambda}{n} \\right)^n \\\\ \\\\\n\n&amp; \\displaystyle = \\frac{\\lambda^k}{k!}e^{-\\lambda}\n\n\\end{array}\n\n<\/span>\n<\/span><\/p>\n<p style=\"text-align: justify; color: #000000;\">From this, the Poisson coefficients, <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">Po(k;\\lambda)<\/span><\/span>, are defined as follows:<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle Po(k;\\lambda) := \\lim_{n\\to\\infty} {{n}\\choose{k}}p^k(1-p_n)^{n-k} = \\frac{\\lambda^k}{k!}e^{-\\lambda} <\/span><\/span><\/p>\n<p style=\"text-align: justify; color: #000000;\">And a random variable <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">X<\/span><\/span> is said to have a Poisson distribution, <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">X\\sim Po(k,\\lambda),<\/span><\/span> if it holds that:<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\"> P(X=k) = Po(k;\\lambda) <\/span><\/span><\/p>\n<p><a name=\"2\"><\/a><\/br><\/br><\/p>\n<h2>Approximate Poisson Processes<\/h2>\n<p><a name=\"3\"><\/a><\/br><\/br><\/p>\n<h3>Spatial Poisson Process<\/h3>\n<p style=\"text-align: justify; color: #000000;\"><a href=\"https:\/\/www.youtube.com\/watch?v=mQ0j3FE8p2U&amp;t=665s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">Suppose we have a container of volume<\/span><\/strong><\/a> <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">V<\/span><\/span> with a liquid containing <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n<\/span><\/span> tiny particles uniformly mixed. Here we assume the liquid is well stirred and the particles do not interact with each other, neither attracting nor repelling. These are assumptions that can be formalized through the following statements:<\/p>\n<ul style=\"text-align: justify; color: #000000;\">\n<li><strong>Hypothesis of Spatial Homogeneity:<\/strong> The probability of finding a particle in a region <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">D<\/span><\/span> of the liquid depends solely on the volume of that region.<\/li>\n<li><strong>No-Interaction:<\/strong> The events \u00abthe j-th particle is in the region D,\u00bb with j=1,2,&#8230;,n are all n-independent.<\/li>\n<li><strong>No-Overlap:<\/strong> Two particles cannot occupy the same place in space.<\/li>\n<\/ul>\n<p style=\"text-align: justify; color: #000000;\">If we are given a region <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">D<\/span><\/span> with a volume <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">v<\/span><\/span>, the probability of the event \u00abthere are <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k<\/span><\/span> particles in <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">D<\/span>\u00bb depends solely on <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">v<\/span><\/span>; let&#8217;s call <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">g_k(v)<\/span><\/span> such an event. Let <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">h(v)<\/span><\/span> be the probability that the particle is inside a region of volume <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">v.<\/span><\/span> If <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">D_1<\/span><\/span> and <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">D_2<\/span><\/span> are two disjoint regions of volume <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">v_1<\/span><\/span> and <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">v_2<\/span><\/span> respectively, then if <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">D=D_1\\cup D_2,<\/span><\/span> has a volume <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">v,<\/span><\/span> then <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">v=v_1+v_2.<\/span><\/span> And since <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">D_1<\/span><\/span> and <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">D_2<\/span><\/span> are disjoint (<span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">D_1\\cap D_2 = \\emptyset <\/span><\/span>), it follows that<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\"> h(v) = h(v_1) + h(v_2) <\/span><\/span><\/p>\n<p style=\"text-align: justify; color: #000000;\">If <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">V<\/span><\/span> is the volume of the entire liquid, then<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\"> h(V) = 1 <\/span><\/span><\/p>\n<p style=\"text-align: justify; color: #000000;\">And consequently:<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\"> h(v) =\\displaystyle \\frac{v}{V} <\/span><\/span><\/p>\n<p style=\"text-align: justify; color: #000000;\">From here we have that the event <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">g_k(v)<\/span><\/span> is actually a Bernoulli-type event with <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">p=v\/V<\/span><\/span> and is given by:<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\"> g_k(v) =B(n;k;p=v\/V) <\/span><\/span><\/p>\n<p style=\"text-align: justify; color: #000000;\">However, most practical situations of this kind involve a large number of particles <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n<\/span><\/span> and the regions considered tend to be small relative to the size of the system, so that the conditions for applying the Poisson approximation are met and it holds that:<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle P(g_k(v)) = \\lim_{\\begin{matrix}n\\to\\infty\\\\ v\/V=c \\end{matrix}}P(B(n;k;p=v\/V)) =\\displaystyle \\frac{(cv)^k}{k!}e^{-cv}<\/span><\/span><\/p>\n<p><a name=\"4\"><\/a><\/br><\/br><\/p>\n<h3>Temporal Poisson Process<\/h3>\n<p style=\"text-align: justify; color: #000000;\"><a href=\"https:\/\/www.youtube.com\/watch?v=mQ0j3FE8p2U&amp;t=944s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">Suppose we are recording the number<\/span><\/strong><\/a> of particles emitted by a radioactive substance starting at time t=0, and from that, we will calculate the probability that exactly k particles are emitted in the interval [0,t[ under the following assumptions:<\/p>\n<ul style=\"text-align: justify; color: #000000;\">\n<li><strong>Invariance:<\/strong> The conditions of the experiment do not change over time.<\/li>\n<li><strong>No-Memory:<\/strong> What happened in [0,t[ does not affect what happens in [t,t'[.<\/li>\n<li><strong>Isolated Events:<\/strong> Particles are emitted one at a time.<\/li>\n<\/ul>\n<p style=\"text-align: justify; color: #000000;\">If we compare the assumptions of the temporal process with those of the spatial process, we will see that they are formally equivalent. Just as the probability of finding a particle in a region does not depend on the location of the region but only on its size, the probability of observing the emission of a particle does not depend on the moment chosen to measure, but only on the observation interval. The no-memory assumption is analogous to the no-interaction assumption of spatial processes: what happened at one time does not affect what happens at other times. And finally, isolated events imply that at any given moment, only one particle can be emitted, just as a place in space can only be occupied by one body at a time.<\/p>\n<p style=\"text-align: justify; color: #000000;\">Thus, if we define the event \u00abk particles are emitted in a time interval <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">t<\/span><\/span>,\u00bb its probability of occurring will be an event of the form <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">g_k(t)<\/span><\/span>, that is:<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">P(g_k(t)) =\\displaystyle \\frac{(ct)^k}{k!} e^{-ct}<\/span><\/span><\/p>\n<p><a name=\"5\"><\/a><\/br><\/br><\/p>\n<h3>Temporal and Spatial<\/h3>\n<p style=\"text-align: justify; color: #000000;\"><a href=\"https:\/\/www.youtube.com\/watch?v=mQ0j3FE8p2U&amp;t=1102s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">Both processes, spatial and temporal, are formally equivalent.<\/span><\/strong><\/a> They only vary in how they are interpreted for practical purposes. A quick way to make this distinction clearer is by observing the role played by the constant \u00abc\u00bb that appears in both cases. For the exponential function to be well-defined, its argument must be dimensionless; however, it contains units of time or space depending on whether we are dealing with temporal or spatial processes. This problem is fixed precisely by the constant c. We have:<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">Po(k;\\lambda)=\\displaystyle \\frac{\\lambda^k}{k!}e^{-\\lambda}=\\left\\{\\begin{matrix} {Taking\\,\\lambda = \\rho v } &amp; \\longmapsto &amp;\\displaystyle \\frac{(\\rho v)^k}{k!}e^{-\\rho v} &amp; {Spatial\\,Process} \\\\ {Taking\\,\\lambda = \\nu t } &amp; \\longmapsto &amp;\\displaystyle \\frac{(\\nu t)^k}{k!}e^{-\\nu t} &amp; {Temporal\\,Process} \\end{matrix} \\right.<\/span><\/span><\/p>\n<ul style=\"text-align: justify; color: #000000;\">\n<li>If <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">c=\\rho<\/span><\/span>, it is a spatial density (number of things per unit space), therefore it defines a spatial Poisson process.<\/li>\n<li>If <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">c=\\nu<\/span><\/span>, it is a temporal density (or frequency, number of occurrences per unit time), therefore it defines a temporal Poisson process.<\/li>\n<\/ul>\n<p><center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/8qvHRoEckSc\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center><br \/>\n<a name=\"6\"><\/a><\/br><\/br><\/p>\n<h2>Practical examples where the Poisson distribution is used<\/h2>\n<ol style=\"text-align: justify; color: #000000;\">\n<li>The checkout at a supermarket serves an average of 2 customers every 9 minutes. Create a table showing the probabilities of serving between 1, 2, 3, and so on, up to 5 people in a 5-minute time frame.<\/li>\n<li>A veterinary clinic has the capacity to serve a maximum of 12 clients per day. If they receive an average of 9 clients each day, what is the probability that on any given day, the clinic&#8217;s capacity will be exceeded?<\/li>\n<li>A certain locality has a population density of 10 people per 1000 square meters. What is the probability that in an area of 60 square meters, we find less than 15 people?<\/li>\n<li>A chicken wants to cross the road. Walking in a straight line, it takes 58 seconds. If the road has a traffic flow of 3 vehicles per minute, and if a vehicle passes while the chicken is attempting to cross, it will certainly be run over with fatal results. What is the probability that the chicken will make it to the other side alive?<\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"<p>Poisson Process: Approximation of the Binomial Process SummaryThis class focuses on the Poisson Process as an approximation to the Binomial Process, starting with the definition of the coefficients and the Poisson distribution, which is derived from a Bernoulli event with a large number of trials and a very small individual probability. The core of this [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":26422,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"iawp_total_views":7,"footnotes":""},"categories":[567,670],"tags":[],"class_list":["post-26769","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-mathematics","category-probabilities-and-statistics"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.4 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>Poisson Process: Approximation of the Binomial Process - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"Understand the Poisson distribution as a result obtained from binomial processes taken to the limit.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"http:\/\/toposuranos.com\/material\/en\/poisson-process-approximation-of-the-binomial-process\/\" \/>\n<meta property=\"og:locale\" content=\"es_ES\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Poisson Process: Approximation of the Binomial Process\" \/>\n<meta property=\"og:description\" content=\"Understand the Poisson distribution as a result obtained from binomial processes taken to the limit.\" \/>\n<meta property=\"og:url\" content=\"http:\/\/toposuranos.com\/material\/en\/poisson-process-approximation-of-the-binomial-process\/\" \/>\n<meta property=\"og:site_name\" content=\"toposuranos.com\/material\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/groups\/toposuranos\" \/>\n<meta property=\"article:published_time\" content=\"2021-05-27T13:00:54+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2024-05-22T01:36:11+00:00\" \/>\n<meta property=\"og:image\" content=\"http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2024\/04\/poisson-e1712962118751-1024x285.jpg\" \/>\n<meta name=\"author\" content=\"giorgio.reveco\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:title\" content=\"Poisson Process: Approximation of the Binomial Process\" \/>\n<meta name=\"twitter:description\" content=\"Understand the Poisson distribution as a result obtained from binomial processes taken to the limit.\" \/>\n<meta name=\"twitter:image\" content=\"http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2024\/04\/poisson-e1712962118751.jpg\" \/>\n<meta name=\"twitter:creator\" content=\"@topuranos\" \/>\n<meta name=\"twitter:site\" content=\"@topuranos\" \/>\n<meta name=\"twitter:label1\" content=\"Escrito por\" \/>\n\t<meta name=\"twitter:data1\" content=\"giorgio.reveco\" \/>\n\t<meta name=\"twitter:label2\" content=\"Tiempo de lectura\" \/>\n\t<meta name=\"twitter:data2\" content=\"6 minutos\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\\\/\\\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"http:\\\/\\\/toposuranos.com\\\/material\\\/en\\\/poisson-process-approximation-of-the-binomial-process\\\/#article\",\"isPartOf\":{\"@id\":\"http:\\\/\\\/toposuranos.com\\\/material\\\/en\\\/poisson-process-approximation-of-the-binomial-process\\\/\"},\"author\":{\"name\":\"giorgio.reveco\",\"@id\":\"http:\\\/\\\/toposuranos.com\\\/material\\\/#\\\/schema\\\/person\\\/e15164361c3f9a2a02cf6c234cf7fdc1\"},\"headline\":\"Poisson Process: Approximation of the Binomial Process\",\"datePublished\":\"2021-05-27T13:00:54+00:00\",\"dateModified\":\"2024-05-22T01:36:11+00:00\",\"mainEntityOfPage\":{\"@id\":\"http:\\\/\\\/toposuranos.com\\\/material\\\/en\\\/poisson-process-approximation-of-the-binomial-process\\\/\"},\"wordCount\":1589,\"commentCount\":0,\"publisher\":{\"@id\":\"http:\\\/\\\/toposuranos.com\\\/material\\\/#organization\"},\"image\":{\"@id\":\"http:\\\/\\\/toposuranos.com\\\/material\\\/en\\\/poisson-process-approximation-of-the-binomial-process\\\/#primaryimage\"},\"thumbnailUrl\":\"http:\\\/\\\/toposuranos.com\\\/material\\\/wp-content\\\/uploads\\\/2024\\\/04\\\/poisson-e1712962118751.jpg\",\"articleSection\":[\"Mathematics\",\"Probabilities and Statistics\"],\"inLanguage\":\"es\",\"potentialAction\":[{\"@type\":\"CommentAction\",\"name\":\"Comment\",\"target\":[\"http:\\\/\\\/toposuranos.com\\\/material\\\/en\\\/poisson-process-approximation-of-the-binomial-process\\\/#respond\"]}]},{\"@type\":\"WebPage\",\"@id\":\"http:\\\/\\\/toposuranos.com\\\/material\\\/en\\\/poisson-process-approximation-of-the-binomial-process\\\/\",\"url\":\"http:\\\/\\\/toposuranos.com\\\/material\\\/en\\\/poisson-process-approximation-of-the-binomial-process\\\/\",\"name\":\"Poisson Process: Approximation of the Binomial Process - 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