142 chapter 4 jointly distributed random variables p

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Unformatted text preview: t; 1 the following would have to hold: (u1 + v1 )(u2 + v2 ) = (u1 + v2 )(u2 + v1 ), or equivalently, (u2 − u1 )(v2 − v1 ) = 0, which is a contradiction. 136 CHAPTER 4. JOINTLY DISTRIBUTED RANDOM VARIABLES Example 4.5.1 Suppose X has the binomial distribution with parameters m and p, Y has the binomial distribution with parameters n and p (the same p as for X ), and X and Y are independent. Describe the distribution of S = X + Y. Solution: This problem can be solved with a little thought and no calculation, as follows. Recall that the distribution of X arises as the number of times heads shows if a coin with bias p is flipped m times. Then Y could be thought of as the number of times heads shows in n additional flips of the same coin. Thus, S = X + Y is the number of times heads shows in m + n flips of the coin. So S has the binomial distribution with parameters m + n and p. That was easy, or, at least it didn’t require tedious computation. Let’s try doing it another way, and at the same time get pra...
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