This preview shows page 1. Sign up to view the full content.
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...
View Full
Document
 Spring '08
 Zahrn
 The Land

Click to edit the document details