{"id":23136,"date":"2023-03-03T06:02:12","date_gmt":"2023-03-03T06:02:12","guid":{"rendered":"https:\/\/www.booksofall.com\/id\/?post_type=product&p=23136"},"modified":"2023-03-03T06:02:12","modified_gmt":"2023-03-03T06:02:12","slug":"introduction-to-statistical-thought","status":"publish","type":"product","link":"https:\/\/www.booksofall.com\/id\/introduction-to-statistical-thought\/","title":{"rendered":"Introduction to Statistical Thought"},"content":{"rendered":"

Chapter 1 – Probability<\/h2>\n

1.1 Basic Probability<\/h3>\n

Let X be a set and F a collection of subsets<\/a> of X. A probability measure, or just a probability, on (X,F ) is a function \u00b5<\/a> : F \u2192 [0, 1]. In other words, to every set in F , \u00b5 assigns a probability between 0 and 1. We call \u00b5 a set function because its domain is a collection of sets. But not just any set function will do. To be a probability \u00b5 must satisfy<\/p>\n

    \n
  1. \u00b5(\u2205) = 0 (\u2205 is the empty set<\/a>.),<\/li>\n
  2. \u00b5(X) = 1, and<\/li>\n
  3. if A1 and A2 are disjoint then \u00b5(A1 \u222a A2) = \u00b5(A1) + \u00b5(A2).<\/li>\n<\/ol>\n

    One can show that property 3 holds for any finite collection of disjoint sets, not just two; see Exercise 1. It is common practice, which we adopt in this text, to assume more \u2014 that property 3 also holds for any countable collection of disjoint sets.<\/p>\n

    When X is a finite or countably infinite<\/a> set (usually integers) then \u00b5 is said to be a discrete probability. When X is an interval, either finite or infinite, then \u00b5 is said to be a continuous probability. In the discrete case, F usually contains all possible subsets of X. But in the continuous case, technical complications prohibit F from containing all possible subsets of X. See Casella and Berger [2002] or Schervish [1995] for details. In this text we deemphasize the role of F and speak of probability measures<\/a> on X without mentioning F .<\/p>\n

    In practical examples X is the set of outcomes of an \u201cexperiment\u201d and \u00b5 is determined by experience, logic or judgement. For example, consider rolling a six-sided die. The set of outcomes is {1, 2, 3, 4, 5, 6} so we would assign X \u2261 {1, 2, 3, 4, 5, 6}. If we believe the die to be fair then we would also assign \u00b5({1}) = \u00b5({2}) = \u00b7 \u00b7 \u00b7 = \u00b5({6}) = 1\/6. The laws of probability then imply various other values such as<\/p>\n

    \u00b5({1, 2}) = 1\/3
    \n\u00b5({2, 4, 6}) = 1\/2
    \netc.<\/p>\n

    Often we omit the braces and write \u00b5(2), \u00b5(5), etc. Setting \u00b5(i) = 1\/6 is not automatic simply because a die has six faces. We set \u00b5(i) = 1\/6 because we believe the die to be fair.<\/p>\n

    We usually use the word \u201cprobability<\/a>\u201d or the symbol P in place of \u00b5. For example, we would use the following phrases interchangeably:<\/p>\n

      \n
    • The probability that the die lands 1<\/li>\n
    • P(1)<\/li>\n
    • P[the die lands 1]<\/li>\n
    • \u00b5({1})<\/li>\n<\/ul>\n

      We also use the word distribution in place of probability measure. The next example illustrates how probabilities of complicated events can be calculated<\/p>\n

      from probabilities of simple events.<\/p>\n

      Example 1.1<\/strong> (The Game of Craps) Craps<\/a> is a gambling game played with two dice. Here are the rules, as explained on the website www.online-craps-gambling.com\/craps-rules.html<\/a>.<\/p>\n

      For the dice thrower (shooter) the object of the game is to throw a 7 or an 11 on the first roll (a win) and avoid throwing a 2, 3 or 12 (a loss). If none of these numbers (2, 3, 7, 11 or 12) is thrown on the first throw (the Come-out roll) then a Point is established (the point is the number rolled) against which the shooter plays. The shooter continues to throw until one of two numbers is thrown, the Point number or a Seven. If the shooter rolls the Point before rolling a Seven he\/she wins, however if the shooter throws a Seven before rolling the Point he\/she loses.<\/p>\n

      Ultimately we would like to calculate P(shooter wins). But for now, let\u2019s just calculate<\/p>\n

      P(shooter wins on Come-out roll) = P(7 or 11) = P(7) + P(11).<\/p>\n

      Using the language of page 1, what is X in this case? Let d1 denote the number showing on the first die and d2 denote the number showing on the second die. d1 and d2 are integers from 1 to 6. So X is the set of ordered pairs (d1, d2) or<\/p>\n

      \"\"<\/p>\n

      If the dice are fair, then the pairs are all equally likely. Since there are 36 of them, we assign P(d1, d2) = 1\/36 for any combination (d1, d2). Finally, we can calculate<\/p>\n

      P(7 or 11) = P(6, 5) + P(5, 6) + P(6, 1) + P(5, 2) + P(4, 3) + P(3, 4) + P(2, 5) + P(1, 6) = 8\/36 = 2\/9.<\/p>\n

      The previous calculation uses desideratum 3 for probability measures. The different pairs (6, 5), (5, 6), . . . , (1, 6) are disjoint, so the probability of their union is the sum of their probabilities.<\/p>\n

      Example 1.1 illustrates a common situation. We know the probabilities of some simple events like the rolls of individual dice, and want to calculate the probabilities of more complicated events like the success of a Come-out roll. Sometimes those probabilities can be calculated mathematically as in the example. Other times it is more convenient to calculate them by computer simulation. We frequently use R to calculate probabilities. To illustrate, Example 1.2 uses R to calculate by simulation the same probability we found directly in Example 1.1.<\/p>\n","protected":false},"excerpt":{"rendered":"