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Random Variables AND DISTRIBUTION FUNCTION Consider the experiment of tossing a coin twice. If we are interested in the number of heads that show on the top face, describe the sample space. Solution: S={ HH , HT , TH , TT } 2 1 1 0 Definition (1): A random variable is a function that associates a real number with each element in the sample space. Remark: We shall use a capital letter, say X, to denote a random variable and its corresponding small letter, x in this case, for one of its values Definition 3.3 If the space of random variable X is countable, then X is called a discrete random variable. Definition 3.4 If the space of random variable X is uncountable, then X is called a continuous random variable. 3.2. Distribution Functions of Discrete Random Variables Definition 3.5. Let ๐ ๐ฅ be the space of the random variable X. The function f : ๐ ๐ฅ IR defined by f(x) = P(X = x) is called the probability mass function (p m f) of X. Example 3.5 A pair of dice consisting of a six-sided die and a four-sided die is rolled and the sum is determined. Let the random variable X denote this sum. โข Find the sample space, โข the space of the random variable, โข And probability mass function of X. Answer: The sample space of this random experiment is given by {(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6) S= (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)} The space of the random variable X is given by ๐ ๐ = {2, 3, 4, 5, 6, 7, 8, 9, 10}. Therefore, the probability mass function of X is given by f(2) = P(X = 2) = ๐ ๐๐ f(5) = P(X = 5) = ๐ ๐๐ f(8) = P(X = 8) = ๐ ๐๐ , f(3) = P(X = 3) = ๐ ๐๐ , f(6) = P(X = 6) = , f(9) = P(X = 9) = , f(4) = P(X = 4) = ๐ ๐๐ ๐ ๐๐ ๐ ๐๐ , f(7) = P(X = 7) = ๐ ๐๐ , f(10) = P(X = 10) = ๐ ๐๐ Example 3.6. A fair coin is tossed 3 times. Let the random variable X denote the number of heads in 3 tosses of the coin. Find the sample space, the space of the random variable, and the probability density function of X. Answer: The sample space S of this experiment consists of all binary sequences of length 3, that is S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}. The space of this random variable is given by ๐ ๐ = {0, 1, 2, 3}. Therefore, the probability density function of X is given by f(0) = P(X = 0) = ๐ 8 3 8 f(1) = P(X = 1) = f(2) = P(X = 2) = 3 8 f(3) = P(X = 3) = ๐ 8 Theorem 3.1 If X is a discrete random variable with space ๐ ๐ and probability density function f(x), then (a). f(x) โฅ 0 for all x in, and (b). f(x) = 1. Example 3.7 If the probability of a random variable X with space ๐ ๐ = {1, 2, 3, ..., 12} is given by f(x) = k (2x โ 1), then, what is the value of the constant k? Answer: 1= f(x) 1 = 12 1 k 1=k (2x โ 1) 12 1 (2x โ 1) 1= k 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 1 = k 144. 1 144 K= Definition 3.6. The cumulative distribution function F(x) of a random variable X is defined as F(x) = P(X=x) for all real numbers x. Theorem 3.2. If X is a random variable with the space๐ ๐ฅ , then F(X) = ๐กโค๐ฅ ๐(๐ก) Example 3.8 If the probability density function of the random variable X is given by 1 (2x 144 โ 1) for x = 1, 2, 3, ..., 12 then find the cumulative distribution function of X. Answer: The space of the random variable X is given by ๐ ๐ = {1, 2, 3, ..., 12}. Then ๐ 144 F(1) = ๐กโค1 ๐(๐ก)= f(1) = F(2) = ๐กโค2 ๐(๐ก)= f(1) + f(2) = F(3) = .. ........ .. ........ ๐กโค3 ๐(๐ก) = f(1) + f(2) + f(3) = F(12) = ๐กโค12 ๐(๐ก) 4 ๐ 3 + = 144 14๐ 144 ๐ 3 5 9 + + = 144 144 144 144 = f(1) + f(2) + · · · + f(12) = 1. This part is in the book from pg 45 best of luck