Math

Math 362, Problem set 5

Due 3/16/10 1. (3.7.6) Another estimating chi-square: Let the result of a random experiment be classifies as one of the mutually exclusive and exhaustive ways A1 , A2 , A3 and also as one of the mutually exclusive and exhaustive ways B1 , B2 , B3 , B4 . Two hundred independent trials of the experiment result in the following data B1 10 11 6 B2 21 27 19 B3 15 21 27 B4 6 13 24

A1 A2 A3

Test, at the 0.05 significance level the hypothesis of independence of the A and B attribute, namely H0 : P(Ai ∩ Bi ) = P(Ai )P(Bi ), i = 1, 2, 3 and j = 1, 2, 3, 4 against the alternative of dependence. Answer: We estimate P(A1 ) = 18 50 67 P(B2 ) = 200 P(A2 ) =

10+21+15+6 200

=

13 50 .

Likewise we estimate: 27 200 43 P(B4 ) = 200 P(B1 ) =

19 50 63 P(B3 ) = 200 P(A3 ) =

Note that, really, we’ve only estimated 5 probabilities, because P(A3 ) is determined once P(A2 ) and P(A1 ) is estimated, and likewise for P(B4 ). Let nij denote the number of events in Ai ∩Bj (so n23 = 21). We compute (nij − 200P(Ai )P(Bj )2 ≈ 12.941. 200P(Ai )P(Bj ) Since we estimated 5 parameters, we compare to a χ2 (12 − 1 − 5) = χ2 (6) random variable. Since 12.941 > 12.592, we reject H0 . 2. (5.8.5) Determine a method to generate random observations for the following pdf: f (x) = 4x3 for 0 < x < 1, zero elsewhere. 1

3. (5.8.18) For α > 0 and β > 0, consider the following accept/reject algorithm: (1) Generate U1 and U2 iid uniform(0,1) random variables. Set V1 = 1/α 1/β U1 and V2 = U2 . (2) Set W = V1 + V2 . If W ≤ 1, set X = V1 /W , else goto Step(1). (3) Deliver X. Show that X has a beta distribution with parameters α and β. Note/Hint: The analysis is quite similar to the analysis of the algorithm we did in class. That is, look for the cdf P(X ≤ x) and note that it is some conditional probability. Answer: Our goal is to show that fX (x) = Cxα−1 (1 − x)β−1 for some constant C. We note that FX (x) = P(X ≤ x) = P(V1 /W ≤ x|W ≤ 1) P(V1 ≤ xW, W ≤ 1) P(W ≤ 1) = cP(V1...