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# Mathematical Statistics with Applications, Seventh Edition

www.downloadslide.com 5.4 Independent Random Variables 247 *5.42 The number of defects per yard Y for a certain fabric is known to have a Poisson distribution with parameter λ. However, λ itself is a random variable with probability density function given by { e −λ , λ ≥ 0, f (λ) = 0, elsewhere. Find the unconditional probability function for Y . 5.4 Independent Random Variables In Example 5.8 we saw two dependent random variables, for which probabilities associated with Y 1 depended on the observed value of Y 2 . In Exercise 5.24 (and some others), this was not the case: Probabilities associated with Y 1 were the same, regardless of the observed value of Y 2 . We now present a formal definition of independence of random variables. Two events A and B are independent if P(A ∩ B) = P(A) × P(B). When discussing random variables, if a < b and c < d we are often concerned with events of the type (a < Y 1 ≤ b) ∩ (c < Y 2 ≤ d). For consistency with the earlier definition of independent events, if Y 1 and Y 2 are independent, we would like to have P(a < Y 1 ≤ b, c < Y 2 ≤ d) = P(a < Y 1 ≤ b) × P(c < Y 2 ≤ d) for any choice of real numbers a < b and c < d. That is, if Y 1 and Y 2 are independent, the joint probability can be written as the product of the marginal probabilities. This property will be satisfied if Y 1 and Y 2 are independent in the sense detailed in the following definition. DEFINITION 5.8 Let Y 1 have distribution function F 1 (y 1 ), Y 2 have distribution function F 2 (y 2 ), and Y 1 and Y 2 have joint distribution function F(y 1 , y 2 ). Then Y 1 and Y 2 are said to be independent if and only if F(y 1 , y 2 ) = F 1 (y 1 )F 2 (y 2 ) for every pair of real numbers (y 1 , y 2 ). If Y 1 and Y 2 are not independent, they are said to be dependent. It usually is convenient to establish independence, or the lack of it, by using the result contained in the following theorem. The proof is omitted; see “References and Further Readings” at the end of the chapter. THEOREM 5.4 If Y 1 and Y 2 are discrete random variables with joint probability function p(y 1 , y 2 ) and marginal probability functions p 1 (y 1 ) and p 2 (y 2 ), respectively, then Y 1 and Y 2 are independent if and only if p(y 1 , y 2 ) = p 1 (y 1 )p 2 (y 2 ) for all pairs of real numbers (y 1 , y 2 ).

www.downloadslide.com 248 Chapter 5 Multivariate Probability Distributions If Y 1 and Y 2 are continuous random variables with joint density function f (y 1 , y 2 ) and marginal density functions f 1 (y 1 ) and f 2 (y 2 ), respectively, then Y 1 and Y 2 are independent if and only if f (y 1 , y 2 ) = f 1 (y 1 ) f 2 (y 2 ) for all pairs of real numbers (y 1 , y 2 ). We now illustrate the concept of independence with some examples. EXAMPLE 5.9 Solution For the die-tossing problem of Section 5.2, show that Y 1 and Y 2 are independent. In this problem each of the 36 sample points was given probability 1/36. Consider, for example, the point (1, 2). We know that p(1, 2) = 1/36. Also, p 1 (1) = P(Y 1 = 1) = 1/6 and p 2 (2) = P(Y 2 = 2) = 1/6. Hence, p(1, 2) = p 1 (1)p 2 (2). The same is true for all other values for y 1 and y 2 , and it follows that Y 1 and Y 2 are independent. EXAMPLE 5.10 Refer to Example 5.5. Is the number of Republicans in the sample independent of the number of Democrats? (Is Y 1 independent of Y 2 ?) Solution Independence of discrete random variables requires that p(y 1 , y 2 ) = p 1 (y 1 )p 2 (y 2 ) for every choice (y 1 , y 2 ). Thus, if this equality is violated for any pair of values, (y 1 , y 2 ), the random variables are dependent. Looking in the upper left-hand corner of Table 5.2, we see p(0, 0) = 0. But p 1 (0) = 3/15 and p 2 (0) = 6/15. Hence, so Y 1 and Y 2 are dependent. p(0, 0) ≠ p 1 (0)p 2 (0), EXAMPLE 5.11 Let { 6y1 y2 2 f (y 1 , y 2 ) = , 0 ≤ y 1 ≤ 1, 0 ≤ y 2 ≤ 1, 0, elsewhere. Show that Y 1 and Y 2 are independent.

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