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Unformatted text preview: tion of n. An example is shown in Figure 2.7. It is zero if n ≤ k − 1; it jumps up to k 1/k n
0 k Figure 2.7: The likelihood of {X = k } for k ﬁxed, as a function of n.
at n = k ; as n increases beyond k the function decreases. It is thus maximized at n = k. That is,
nM L (k ) = k. Example 2.8.3 Suppose X has the geometric distribution with some parameter p which is unknown. Suppose a particular value k for X is observed. Find the maximum likelihood estimate,
pM L .
Solution: The pmf is given by pX (k ) = (1 − p)k−1 p for k ≥ 1. If k = 1 we have to maximize
pX (0) = p with respect to p, and p = 1 is clearly the maximizer. So pM L = 1 if k = 1. If k ≥ 2 then
pX (k ) = 0 for p = 0 or p = 1. Since ((1−p)k−1 p) = (−(k −1)p−(1−p))(1−p)k−2 = (1−kp)(1−p)k−2 ,
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we conclude that pX (k ) is increasing in p for 0 ≤ p ≤ k and decreasing in p for k ≤ p ≤ 1. Therefore,
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pM L = k if k ≥ 2. This expression is correct for k = 1 as well, so for any k ≥ 1, pM L = k . 2.9. MARKOV AND CHEBYCHEV INEQUALITIES AND CONFIDENCE INTERV...
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This note was uploaded on 02/09/2014 for the course ISYE 2027 taught by Professor Zahrn during the Spring '08 term at Georgia Institute of Technology.
 Spring '08
 Zahrn
 The Land

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