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# Chapter 2 - Probability

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Chapter 2 of Essential Math and Statistics for Finance and Risk Management
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## Probability Density Function

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ESSENTIAL MATH AND STATISTICS FORFINANCE AND RISK MANAGEMENT   Alan Anderson, Ph.D.Western Connecticut State UniversityECI Risk Traininghttp://www.ecirisktraining.com   (c) ECI Risk Training 2009 2 CHAPTER 2: PROBABILITY THEORY Probability theory is based on the notion of a random experiment , which is a  process  that generates outcomes in such a way that: ã   all possible outcomes are known in advance ã   the actual outcome is not known in advanceSome examples of a random experiment are: ã   spinning a roulette wheel ã   choosing a card from a deck  ã   rolling a pair of dice EXAMPLE Suppose that a die is rolled twice; assume that the variable of interest is whether thenumber that turns up is odd or even. There are four possible outcomes of this randomexperiment: {even, even}, {even, odd}, {odd, even}, {odd, odd}.The set of all possible outcomes of a random experiment is known as the sample space .The sample space for this experiment consists four  sample points : S = {EE, EO, OE,OO}. A subset of the sample space is known as an event . EXAMPLE Based on the die-rolling experiment, suppose that the event F is defined as follows: “atmost one even number turns up”. The event F is a set containing the following sample points: F = {EO, OE, OO}. COMPUTING PROBABILITIES OF EVENTS For a random experiment in which there are n equally likely outcomes, each outcome hasa probability of occurring equal to 1/n. In the die-rolling example, each of the four sample points is equally likely, giving it a probability of 1/4. The probability of event F istherefore:P(F) = P{EO} + P{OE} + P{OO} = 1/4 + 1/4 + 1/4 = 3/4Alternatively, the probability of an event E can be computed as P(E) = #E/#S  (c) ECI Risk Training 2009 3where: #E = number of sample points in event E#S = number of sample points in the sample spaceIn this example, the probability of event F would be calculated as P(F) = #F/#S = 3/4. AXIOMS OF PROBABILITY An axiom is a logical statement that is assumed to be true without formal proof; allfurther results are derived using axioms as a starting point. In probability theory, thethree fundamental axioms are:Axiom 1: The probability of any event A is non-negative; P(A) ≥ 0Axiom 2: The probability of the entire sample space is one; P(S) = 1Axiom 3: For any collection of disjoint (non-overlapping) events: RULES OF PROBABILITY The rules of probability are derived from these axioms. Three of the most important rulesare the Addition Rule, Multiplication Rule and the Complement Rule. ADDITION RULE For two events A and B, the addition rule is:P(A ∪ B) = P(A) + P(B) – P(A ∩ B) EXAMPLE Referring to the die-rolling experiment, suppose that the following events are defined:A = “the first roll is even”B = “exactly one roll is oddWhat is the probability that the first roll is even or at exactly one roll is odd?Events A, B and A ∩ B contain the following sample points: P (  A 1 ∪  A 2 ∪ ... ∪  A n ) = P (  A i ) i = 1 n ∑  (c) ECI Risk Training 2009 4A = {EE, EO}B = {EO, OE}A ∩ B = {EO}Since the sample space consists of four equally likely sample points,P(A) = 1/2P(B) = 1/2P(A ∩ B) = 1/4P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = 1/2 + 1/2 – 1/4 = 3/4 ADDITION RULE FOR MUTUALLY EXCLUSIVE (DISJOINT) EVENTS If events A and B cannot both occur at the same time, they are said to be mutuallyexclusive or  disjoint ; in this case, P(A ∩ B) = 0. Therefore, the addition rule formutually exclusive events is: P(A ∪ B) = P(A) + P(B). CONDITIONAL PROBABILITY The conditional probability of an event is the probability that it occurs  given that   another event occurs. For two events A and B, the probability that B occurs  given that  Aoccurs is written as: P(B|A).This can be computed as: P(B|A) = P(A ∩ B) / P(A) = P(B ∩ A) / P(A)Equivalently, the probability that A occurs  given that  B occurs is computed as:P(A|B) = P(A ∩ B) / P(B) = P(B ∩ A) / P(B) EXAMPLE For the die-rolling experiment, where:A = “the first roll is even” = {EE, EO}B = “exactly one roll is odd” = {EO, OE}Since (B ∩ A) = {EO}, P(B ∩ A) = 1/4Since A = {EE, EO}, P(A) = 1/2Since B = {EO, OE}, P(B) = 1/2
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