Chapter7

Chapter7 - Chapter 7 Sums of Independent Random Variables...

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Chapter 7 Sums of Independent Random Variables 7.1 Sums of Discrete Random Variables In this chapter we turn to the important question of determining the distribution of a sum of independent random variables in terms of the distributions of the individual constituents. In this section we consider only sums of discrete random variables, reserving the case of continuous random variables for the next section. We consider here only random variables whose values are integers. Their distri- bution functions are then defined on these integers. We shall find it convenient to assume here that these distribution functions are defined for all integers, by defining them to be 0 where they are not otherwise defined. Convolutions Suppose X and Y are two independent discrete random variables with distribution functions m 1 ( x ) and m 2 ( x ). Let Z = X + Y . We would like to determine the dis- tribution function m 3 ( x )of Z . To do this, it is enough to determine the probability that Z takes on the value z , where z is an arbitrary integer. Suppose that X = k , where k is some integer. Then Z = z if and only if Y = z - k . So the event Z = z is the union of the pairwise disjoint events ( X = k ) and ( Y = z - k ) , where k runs over the integers. Since these events are pairwise disjoint, we have P ( Z = z )= X k = -∞ P ( X = k ) · P ( Y = z - k ) . Thus, we have found the distribution function of the random variable Z . This leads to the following definition. 285
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286 CHAPTER 7. SUMS OF RANDOM VARIABLES Definition 7.1 Let X and Y be two independent integer-valued random variables, with distribution functions m 1 ( x ) and m 2 ( x ) respectively. Then the convolution of m 1 ( x ) and m 2 ( x ) is the distribution function m 3 = m 1 * m 2 given by m 3 ( j )= X k m 1 ( k ) · m 2 ( j - k ) , for j = ..., - 2 , - 1 , 0 , 1 , 2 , ... . The function m 3 ( x ) is the distribution function of the random variable Z = X + Y . 2 It is easy to see that the convolution operation is commutative, and it is straight- forward to show that it is also associative. Now let S n = X 1 + X 2 + ··· + X n be the sum of n independent random variables of an independent trials process with common distribution function m defined on the integers. Then the distribution function of S 1 is m . We can write S n = S n - 1 + X n . Thus, since we know the distribution function of X n is m , we can find the distribu- tion function of S n by induction. Example 7.1 A die is rolled twice. Let X 1 and X 2 be the outcomes, and let S 2 = X 1 + X 2 be the sum of these outcomes. Then X 1 and X 2 have the common distribution function: m = ± 123456 1 / 61 / / / / / 6 . The distribution function of S 2 is then the convolution of this distribution with itself. Thus, P ( S 2 =2) = m (1) m (1) = 1 6 · 1 6 = 1 36 , P ( S 2 =3) = m (1) m (2) + m (2) m (1) = 1 6 · 1 6 + 1 6 · 1 6 = 2 36 , P ( S 2 =4) = m (1) m (3) + m (2) m (2) + m (3) m (1) = 1 6 · 1 6 + 1 6 · 1 6 + 1 6 · 1 6 = 3 36 . Continuing in this way we would find P ( S 2 =5 )=4 / 36, P ( S 2 =6 )=5 / 36, P ( S 2 =7 )=6 / 36, P ( S 2 =8 / 36, P ( S 2 =9 / 36, P ( S 2 =10 )=3 / 36, P ( S 2 =11)=2 / 36, and P ( S 2 =12)=1 / 36.
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Chapter7 - Chapter 7 Sums of Independent Random Variables...

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