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text: '[summary: "Normalization" obtains a set of [1rf probabilities] summing to 1, in [1rd cases where they ought to sum to 1]. We do this by dividing each pre-normalized number by the sum of all pre-normalized numbers.\n\nSuppose the [1rb odds] of Alexander Hamilton winning an election are 3 : 2. We think the proportions are right (Alexander is 1.5 times as likely to win as not win) but we want *probabilities*. To say that Hamilton has probability 3 of winning the election would be very strange indeed. But if we divide each of the terms by the sum of all the terms, they'll end up summing to one: $3:2 \\cong \\frac{3}{3+2} : \\frac{2}{3+2} = 0.6 : 0.4.$ Thus, the probability that Hamilton wins is 60%.]\n\n"Normalization" is an arithmetical procedure carried out to obtain a set of [1rf probabilities] summing to exactly 1, in cases where we believe that [1rd exactly one of the corresponding possibilities is true], and we already know the [1rb relative probabilities].\n\nFor example, suppose that the [1rb odds] of Alexander Hamilton winning a presidential election are 3 : 2. But Alexander Hamilton must either win or not win, so the *probabilities* of him winning *or* not winning should sum to 1. If we just add 3 and 2, however, we get 5, which is an unreasonably large probability.\n\nIf we rewrite the odds as 0.6 : 0.4, we've preserved the same proportions, but made the terms sum to 1. We therefore calculate that Hamilton has a 60% probability of winning the election.\n\nWe normalized those odds by dividing each of the terms by the sum of terms, i.e., went from 3 : 2 to $\\frac{3}{3+2} : \\frac{2}{3+2} = 0.6 : 0.4.$\n\nIn converting the odds $m : n$ to $\\frac{m}{m+n} : \\frac{n}{m+n},$ the factor $\\frac{1}{m+n}$ by which we multiply all elements of the ratio is called a [https://en.wikipedia.org/wiki/Normalizing_constant normalizing constant].\n\nMore generally, if we have a relative-odds function $\\mathbb{O}(H)$ where $H$ has many components, and we want to convert this to a probability function $\\mathbb{P}(H)$ that sums to 1, we divide every element of $\\mathbb{O}(H)$ by the sum of all elements in $\\mathbb{O}(H).$ That is:\n\n$\\mathbb{P}(H_i) = \\frac{\\mathbb{O}(H_i)}{\\sum_i \\mathbb{O}(H_i)}$\n\nAnalogously, if $\\mathbb{O}(x)$ is a continuous distribution on $X$, we would normalize it (create a proportional probability function $\\mathbb{P}(x)$ whose integral is equal to 1) by dividing $\\mathbb{O}(x)$ by its own integral:\n\n$\\mathbb{P}(x) = \\frac{\\mathbb{O}(x)}{\\int \\mathbb{O}(x) \\operatorname{d}x}$\n\nIn general, whenever a probability function on a variable is *proportional* to some other function, we can obtain the probability function by *normalizing* that function:\n\n$\\mathbb{P}(H) \\propto \\mathbb{O}(H) \\implies \\mathbb{P}(H) = \\frac{\\mathbb{O}(H)}{\\sum \\mathbb{O}(H)}$',
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