 By Kenneth B Stolarsky

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Extra resources for Algebraic numbers and Diophantine approximation

Example text

Are independent. 5. 2, part (2). 6. 2(a) to find the distribution of (t'1. ) 2 from Exercise 9b in Chapter 2. 7. Consider Exercise 11 of Chapter 2. 2 to show that Y1 + 2Y2 - Y3 is independent of 2Y 2 + y2 + 2y2 _ 2Y1Y2 + 2Y2 I13. 2(a) to nnO a constant c such that c [ 5 Y 2 + 2 Y 2 + 5 Y 2 4Y1Y2 + 2Y1 Y3 + 4Y2 Y3] has a central chi-square distribution. 8. 2. 9. ~ . . . 1. 10. Let the 3 • 1 random vector Y = (Y1, Y2, Y3)' ~ N3(ot13, crzI3) and define Z1 = ( r 2 + r~ - 2 r l r3 and Z2 = r 2 + r 2 + r 2 - rl r2 - rl r3 - r2 r3.

ITt)' =: [(1/s)l' s | I/]Y. Therefore, the sum of squares due to the random factor T is Z(~. j -stf" = j=l i=1 2 .. 1 Two-Way ANOVA Table Source df SS Mean 1 E~=,Etj=l]T2.. t ~'~=, ~-~j=, (]7i. 1 ~-~j=l g/2j s s = Y'A,Y : Y'AzY : Y'A3Y t ~ - ~ j = l (Yij - ]7i. )2 __ Y ' A 4 Y t = Y'AsY 2 Multivariate Normal Distribution _- y , 35 1 (It - ~ J / ) ] Y sJ~| = Y'A3Y 1 where the st x st matrix A3 = [~Js | (It - ~Jt)]. Js| = [Is@ ( I t - ~ J / ) ] : [ ( I s - - 1 j| s ) (I/-~Jt) _ ttjl )] (It--~Jt)] 1.

15. Let the n x 1 random vector Y = (Y1 . . . Yn)'~ Nn(otln, treIn) for n >_ 3. (a) Find the conditional expectation of ~l(Ye + Y3)[ ~" where ~" = ~-~i=1 n Yi/n; that is, find E[ ~l (Ye + I13)1~']. (b) Show that the variance of E[ 1 (Ye + Y3)[~'] is smaller than the variance of 1 ~(Y2 -1- Y3)16. Let F and G be independent normal random variables with 0 means and variances cr 2 and 1 - a 2, respectively, 0 < tr 2 < 1. Find the conditional distribution of F given F + G = c. 17. Let the 3 • 1 random vector Y - (Y1, Y2, Y3)' ~ N3(/z, ~ ) .