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

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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
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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. Proof. Let us fix a box B j . The random variables I ( X 1 B j ) , . . . , I ( X n B j ) that indicate whether each observation X i is in the box B j or not are i.i.d. with
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