Question

1. (a). Consider a random sample of size n from the Normal(0,1) distribution. For kan define X, EX and X3 X. Find the distribution of the following statistics: (). X.-X. nk (). *X-in-k)X? (II). X-X, (w). (1x. -8,7 (m). Ś(x-X.)

Answer 1

k (i) x 14 N(0,1); i = 1(1)n and X Xi is linear function of X's so k i=1 n 1 XK No. (1. A). Similarly, řnk= xin N (0 1 ) ) n - n - i=k+1 1 n Again XX and Xn-k are independent, so X1-Xn-k~ N (0, (0, + K n k kan – k) (ii) Y1 = VkXx - Vn – kXn-k~ N(0,2); Y2 = VkXx + Vn – kXn-k~ N(0,2). The joint distribution of (Y1, Y2) is 1 frix.(. ya) = a exp (_+ - < y1, y2 < 0 Transform (Y1,Y2) + (Z,Y) 1 where, 2 = y1y2y1 dy1 dz Jacobianſ = Y2 Y2 42

The joint distribution of (Z, Y2) is 1 1 fz.ya(z,y2) = exp + y2 ; - <2,42 < 0. 47|y2 4 It is very difficult to marginal distribution of 2 (the distribution may not normal) Note: kX*~ xí and (n-k)X-k ~ xỉ and VkXk and Vn – kXn-k are independent hence kX+(n-k)X-k~ xã (Using additive property of x' distribution). (ii) X* ~NO, and Xn ~ N(0,1). Moreover Xx and Xn are independent 1 then XK – XnN0, +1 k k+ k *). k k (Xn - XR) ~ N(0,1) k +1 (Xn – Xx)~ x1 k +1 K (u) (X; - Xx)?~XX-1 (using orthogonal transformation on Xi; i = 1(1)k). i=1