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Math 6010 Solutions to homework 2 4, p. 31. We can write X = µ+AZ and Y = ν +BW , where Z and W are vectors of i.i.d. standard normal random variables. Note that Z and W must be independent. This is intuitively clear. But if you want a completely rigorous proof, then here it is: Indeed, 0 0 0 0 0 0 E et AZ+s BW = e−t µ−s ν E et X+s Y 0 0 0 0 = e−t µ−s ν E et X · E es Y , because X and Y are independent. It follows then that 0 0 0 0 E et AZ+s BW = E et AZ E es BW , for every s and t. Because u := t0 A and v := s0 B are arbitrary vectors [since t and s are], it follows that the MGF of (Z1 , · · · , Zn , W1 , . . . , Wm )0 is the product of the respective MGFs of Z and W . Therefore, W and Z are independent by the independence theorem. Let m := aµ + bν to see that aX + bY = m + AZ + BW := m + CY , where C := A 0 0 B and Y := X W . We are done, because Y is a vector of i.i.d. standard normals. 7, p. 32. It is best to solve this problem directly. First of all, let me emphasize that n 1X Yi . Ȳ := n i=1 Note that the (j , k)th element of vv 0 —for any p-vector v—is vj vk . Apply this with A := Yi − Ȳ to see that the (j , k)th element of (Yi − Ȳ )(Yi − Ȳ )0 is (Yi,j − Ȳ·,j )(Yi,k − Ȳ·,k ) = Yi,j Yi,k − Ȳ·,j Yi,k − Yi,j Ȳ·,k + Ȳ·,j Ȳ·,k , 1 where n Ȳ·,m 1X := Yν,m . n ν=1 Now, E(Yi,j Yi,k ) = Σj,k , E(Yi,j Yν,k ) = 0 for i 6= ν. Therefore, n 1X 1 E(Yi,j Yν,k ) = Σj,k , n ν=1 n E(Yi,j Ȳ·,k ) = and E(Ȳ·,j Ȳ·,k ) = n n 1 1 XX E(Yτ,j Yν,k ) = Σj,k . 2 n ν=1 τ =1 n We can put everything together to find that E h 0 (Yi − Ȳ )(Yi − Ȳ ) i j,k 1 = Σj,k − Σj,k = n n−1 n Σj,k . We can write this, in equivalent form, using linear algebra: n−1 E (Yi − Ȳ )(Yi − Ȳ )0 = Σ. n We add from i = 1, . . . , n and then divide by n − 1 to see that " # n 1 X 0 E (Yi − Ȳ )(Yi − Ȳ ) = Σ, n − 1 i=1 as desired. 15., p. 32. (a) Let us observe that Y1 , Y2 , Y3 , Y4 are i.i.d. standard normals. Therefore, Y1 Y2 and Y3 Y4 are i.i.d. In particular, Q := Y1 Y2 − Y3 Y4 is a symmetric random variable, so that P{Q < 0} = 1/2 6= 0. Therefore, Q cannot have a χ2 distribution of any kind [since the latter are distributions on nonnegative numbers]. (b) We may note that if 0 1/2 A := 0 0 1/2 0 0 0 0 0 0 0 0 −1/2 −1/2 0 then y 0 Ay = y1 y2 − y3 y4 for all four-vectors y = (y1 , y2 , y3 , y4 )0 . [Consider separately the quadratic forms for y1 y2 and y3 y4 , and then subtract the matrices.] In other words, this A is the matrix for the 2 quadratic form Q. The eigenvalues of A are ±1/2 with multiplicity of two each. Therefore, we can apply (2.10) on page 27 in order to see that 1 1 1 1 Q = Y 0 AY = Z12 + Z22 − Z32 − Z42 , 2 2 2 2 where Z1 , . . . , Z4 are i.i.d. standard normals. Consequently, 2 2 2 2 MQ (t) = E etZ1 /2 E etZ2 /2 E e−tZ3 /2 E e−tZ4 /2 = E12 E22 , 2 2 where E1 := E(etZ1 /2 ) and E2 := E(e−tZ1 /2 ). Because Z12 ∼ χ21 , h i2 h i2 1 2 1 2 MQ (t) = Mχ21 (t/2) · Mχ21 (−t/2) = √ · √ , 1−t 1+t if −1 < t < 1 and other wise infinity. In other words, we find once again that MQ (t) = 1/(1 − t2 ) when |t| < 1 and ∞ when |t| ≥ 1. 3