lec23 - Lecture 23 23.1 Pearson’s theorem. Today we will...

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Unformatted text preview: Lecture 23 23.1 Pearson’s theorem. Today we will prove one result from probability that will be useful in several statistical tests. Let us consider r boxes B 1 , . . . , B r as in figure 23.1 ... B1 B2 Br Figure 23.1: Assume that we throw n balls X 1 , . . . , X n into these boxes randomly independently of each other with probabilities ( X i ∈ B 1 ) = p 1 , . . . , ( X i ∈ B r ) = p r , where probabilities add up to one p 1 + . . . + p r = 1 . Let ν j be a number of balls in the j th box: ν j = # { balls X 1 , . . . , X n in the box B j } = n X l =1 I ( X l ∈ B j ) . On average, the number of balls in the j th box will be np j , so random variable ν j should be close to np j . One can also use Central Limit Theorem to describe how close ν j is to np j . The next result tells us how we can describe in some sense the closeness of ν j to np j simultaneously for all j ≤ r. The main difficulty in this Thorem comes from the fact that random variables ν j for j ≤ r are not independent, for example, because the total number of balls is equal to n, ν 1 + . . . + ν r = n, 89 LECTURE 23. 90 i.e. if we know these numbers in n- 1 boxes we will automatically know their number in the last box. Theorem. We have that the random variable r X j =1 ( ν j- np j ) 2 np j → χ 2 r- 1 converges in distribution to χ 2 r- 1 distribution with ( r- 1) degrees of freedom....
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This note was uploaded on 10/11/2009 for the course STATISTICS 18.443 taught by Professor Dmitrypanchenko during the Spring '09 term at MIT.

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lec23 - Lecture 23 23.1 Pearson’s theorem. Today we will...

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