By Klotz J.H.

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**Extra resources for A computational approach to statistics**

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N} for finite unions, and ∞ i=1 Ai = {e : e ∈ Ai for some i, i = 1, 2, . . , ∞} for countably infinite unions. To illustrate these definitions, consider the sample space for the roll of two dice . The first die is red and the second die is green and each die has 1,2,3,4,5, 6 on the faces. Then if we use the notation (i, j) where i, j ∈ {1, 2, 3, 4, 5, 6} for an outcome with the first coordinate representing the up face for the red die and the second coordinate the up face for the green die, the sample space is S = { (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) } Let the event A be the 1st (red) die is 3 or less A = { (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)} 40 CHAPTER 2.

X i=k+1 ˜ − X(k+1) ) = 0 we can sum from k + 2. Using For n = 2k + 1, since (X X(k+1) ≤ X(k+2) ≤ · · · ≤ X(r) < C r ˜ +2 (2r − n)(C − X) ˜ − X(i) ) > (2k + 2 − n)(C − X) ˜ >0 (X i=k+2 For n = 2k similarly using X(r) < C replacing X(i) by C, r ˜ +2 (2r − n)(C − X) ˜ − X(i) ) > (2k − n)(C − X) ˜ =0. (X i=k+1 ˜ Case II. C ≤ X. ˜ = X(k+1) gives r < k + 1 or r ≤ k. For n = 2k + 1, X(r) < C ≤ X ˜ = (X(k) + X(k+1) )/2 also gives r ≤ k. For n = 2k, X(r) < C ≤ X Then as in case I and using C ≤ X(i) for i = r + 1, r + 2, .

Where Ai ∩ Aj = φ for i = j}. Note, for finite sample spaces, all but a finite number of the sets are the empty set φ. For these we use the convention that a sum over no outcomes is zero. P( ∞ pj = Ai ) = i=1 Sj : ej ∈ ∞ i=1 ∞ i=1 Ai pj i = ji : eji ∈Ai ∞ P (Ai ) . i=1 Finally, axiom (iii) holds since P (S) = {i : ei ∈S} pi = 1. As an example, consider the sample space S = {H, T H, T T H, T T T H, . } of tossing a coin until it comes up heads. Suppose we assign corresponding probabilities {1/2, 1/4, 1/8, .

### A computational approach to statistics by Klotz J.H.

by George

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