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# Note for Probability and Statistics - PS By JNTU Heroes

• Probability and Statistics - PS
• Note
• Jawaharlal Nehru Technological University Anantapur (JNTU) College of Engineering (CEP), Pulivendula, Pulivendula, Andhra Pradesh, India - JNTUACEP
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#### Note for Probability and Statistics - PS By JNTU Heroes

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Random Variable Motivation example In an opinion poll, we might decide to ask 50 people whether they agree or disagree with a certain issue. If we record a “1” for agree and “0” for disagree, the sample space for this experiment has 250 elements. If we define a variable X=number of 1s recorded out of 50, we have captured the essence of the problem. Note that the sample space of X is the set of integers {1, 2, . . . , 50} and is much easier to deal with than the original sample space. In defining the quantity X, we have defined a mapping (a function) from the original sample space to a new sample space, usually a set of real numbers. In general, we have the following definition. Definition of Random Variable A random variable is a function from a sample space S into the real numbers. Example 1.4.2 (Random variables) In some experiments random variables are implicitly used; some examples are these. Experiment Random variable Toss two dice X =sum of the numbers Toss a coin 25 times X =number of heads in 25 tosses Apply different amounts of fertilizer to corn plants X =yield/acre Suppose we have a sample space S = {s1 , . . . , sn } with a probability function P and we define a random variable X with range X = {x1 , . . . , xm }. We can define a probability function PX on X in the following way. We will observe X = xi if and only if the outcome of the random experiment is an sj ∈ S such that X(sj ) = xi . Thus, PX (X = xi ) = P ({sj ∈ S : X(sj ) = xi }). 1 (1)

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Note PX is an induced probability function on X , defined in terms of the original function P . Later, we will simply write PX (X = xi ) = P (X = xi ). Fact The induced probability function defined in (1) defines a legitimate probability function in that it satisfies the Kolmogorov Axioms. Proof: CX is finite. Therefore B is the set of all subsets of X . We must verify each of the three properties of the axioms. (1) If A ∈ B then PX (A) = P (∪xi ∈A {sj ∈ S : X(sj ) = xi }) ≥ 0 since P is a probability function. (2) PX (X ) = P (∪m i=1 {sj ∈ S : X(sj ) = xi }) = P (S) = 1. (3) If A1 , A2 , . . . ∈ B and pairwise disjoint then ∞ PX (∪∞ k=1 Ak ) = P (∪k=1 {∪xi ∈Ak {sj ∈ S : X(sj ) = xi }}) = ∞ X P (∪xi ∈Ak {sj ∈ S : X(sj ) = xi } = ∞ X PX (Ak ), k=1 k=1 where the second inequality follows from the fact P is a probability function. ¤ A note on notation: Random variables will always be denoted with uppercase letters and the realized values of the variable will be denoted by the corresponding lowercase letters. Thus, the random variable X can take the value x. Example 1.4.3 (Three coin tosses-II) Consider again the experiment of tossing a fair coin three times independently. Define the random variable X to be the number of heads obtained in the three tosses. A complete enumeration of the value of X for each point in the sample space is s X(s) HHH HHT 3 HTH THH TTH THT HTT TTT 2 2 1 1 1 0 2 The range for the random variable X is X = {0, 1, 2, 3}. Assuming that all eight points in S have probability 81 , by simply counting in the above display we see that the induced probability function on X is given by 2

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x 0 1 2 3 PX (X = x) 1 8 3 8 3 8 1 8 The previous illustrations had both a finite S and finite X , and the definition of PX was straightforward. Such is also the case if X is countable. If X is uncountable, we define the induced probability function, PX , in a manner similar to (1). For any set A ⊂ X , PX (X ∈ A) = P ({s ∈ S : X(s) ∈ A}). (2) This does define a legitimate probability function for which the Kolmogorov Axioms can be verified. Distribution Functions Definition of Distribution The cumulative distribution function (cdf) of a random variable X, denoted by FX (x), is defined by FX (x) = PX (X ≤ x), for all x. Example 1.5.2 (Tossing three coins) Consider the experiment of tossing three fair coins, and let X =number of heads observed. The    0       1    8   FX (x) = 1 2      7   8      1 cdf of X is if −∞ < x < 0 if 0 ≤ x < 1 if 1 ≤ x < 2 if 2 ≤ x < 3 if 3 ≤ x < ∞. 3

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Remark: 1. FX is defined for all values of x, not just those in X = {0, 1, 2, 3}. Thus, for example, 7 FX (2.5) = P (X ≤ 2.5) = P (X = 0, 1, 2) = . 8 2. FX has jumps at the values of xi ∈ X and the size of the jump at xi is equal to P (X = xi ). 3. FX = 0 for x < 0 since X cannot be negative, and FX (x) = 1 for x ≥ 3 since x is certain to be less than or equal to such a value. FX is right-continuous, namely, the function is continuous when a point is approached from the right. The property of right-continuity is a consequence of the definition of the cdf. In contrast, if we had defined FX (x) = PX (X < x), FX would then be left-continuous. Theorem 1.5.3 The function FX (x) is a cdf if and only of the following three conditions hold: a. limx→−∞ F (x) = 0 and limx→∞ F (x) = 1. b. F (x) is a nondecreasing function of x. c. F (x) is right-continuous; that is, for every number x0 , limx↓x0 F (x) = F (x0 ). Example 1.5.4 (Tossing for a head) Suppose we do an experiment that consists of tossing a coin until a head appears. Let p =probability of a head on any given toss, and define X =number of tosses required to get a head. Then, for any x = 1, 2, . . ., P (X = x) = (1 − p)x−1 p. The cdf is FX (x) = P (X ≤ x) = x X P (X = i) = i=1 x X (1 − p)i−1 p = 1 − (1 − p)x . i=1 It is easy to show that if 0 < p < 1, then FX (x) satisfies the conditions of Theorem 1.5.3. First, lim FX (x) = 0 x→−∞ 4