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STAT 330 (53)
Lecture

# Tutorial2_soln.pdf

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School
University of Waterloo
Department
Statistics
Course
STAT 330
Professor
Christine Dupont
Semester
Fall

Description
Stat 330 - Tutorial 2 In this tutorial, we will use the Binomial series quite often. That is, n n X n x n−x (a + b) = a b . x=0 x 1. Sol: (a) By deﬁnition, X n E(X (k) = x(x − 1)··· (x − k + 1)p (1 − p)x x x=0 (Note that x(x − 1)··· (x − k + 1) = 0 for x = 0,... ,k − 1.) n X n! x n−x = x(x − 1)··· (x − k + 1) p (1 − p) . x=k x!(n − x)! We simplify the above form by noting that n! = n(n − 1)··· (n − k + 1)(n − k)!, x! = x(x − 1)··· (x − k + 1)(x − k)!. Therefore n (k) X n(n − 1)··· (n − k + 1)(n −xk)! n−x E(X ) = p (1 − p) . x=k (x − k)!(n − k)! Changing the index x − k = t, then t = 0,... ,n − k and X−k X−k (k) n − k k+t n−k−t (k) k n − k t n−k−t E(X ) = n(n−1)···(n−k+1) t p (1−p) = n p t p (1−p) . t=0 t=0 Using the Binomial series, we have E(X (k)) = nkp . From this result, we get 2 µ = E(X) = np, E{X(X − 1)} = n(n − 1)p . Therefore V ar(X) = E(X )−µ = E{X(X−1)}+E(X)−(np) = n(n−1)p +np−(np) = np(1−p). (b) By deﬁnition, n n tX X txn x n−x X n t x n−x M(t) = E(e ) = e p (1 − p) = (pe ) (1 − p) . x=0 x x=0 x Using Binomial series, we get M(t) = (pe + 1 − p) with |t| < ∞. 1 Note that ′ t n−1 t ′′ t n−1 t t n−2 t 2 M (t) = n(pe +1−p) pe and M (t) = n(pe +1−p) pe +n(n−1)(pe +1−p) (pe ) . Therefore E(X) = M (0) = np, E(X ) = M (0) = np + n(n − 1)p = (np) + np(1 − p), which imply that V ar(X) = (np) + np(1 − p) − (np) = np(1 − p). 2 2 2. Sol: Let 1 = θ ,2p = 2θ(1 − θ), and3p = (1 − θ) . Note th1t p 2 p 3 p = 1. (a) The joint pf of X and Y is given n! x y n−x−y f(x,y) = P(X = x,Y = y) = p1p2 3 , x!y!(n − x − y)! where x = 0,... ,n, y = 0,... ,n, and x + y ≤ n. (b) For x = 0,... ,n, the mar
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