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Chapter 4 Problems 3. Let X equal the number of dots. Then, f(x) = 16 if x = 1, . . . , 6; else, f(x) = 0. Consequently, 1 7 1 + ··· + 6 × = EX = 1 × 6 6 2 1 1 91 E(X2 ) = 12 × + · · · + 62 × = . 6 6 6 Therefore 91 − Var X = 6 2 35 7 = = 2.916 · · · . 2 12 4. If X is uniform on {1 , . . . , n}, then Var X = (n2 − 1)/12; see “Lecture 12.” P 11. Because E({X − a}2 ) = x (x − a)2 f(x) is assumed to converge nicely, we can differentiate term by term to find that d X 2 d E {X − a}2 = (x − 2xa + a2 )f(x) da da x X = (−2x + 2a)f(x) x = 2a X f(x) − 2 x X xf(x) x = 2a − 2EX. Set this equal to zero to find that a = EX. Also, X d2 E {X − a}2 = 2f(x) = 2, 2 da x which is positive. Positive second derivative means the minimum occurs when the derivative is zero; that is, a = EX. But E({X − EX}2 ) = Var X. 18. By Theorem 4 (p. 121), E X ∞ ∞ ∞ Z 1 k−1 p X 1 k p X q k−1 1 = pq = q = x dx X k q k q k=1 k=1 k=1 0 Z ∞ Z p q X k−1 p q 1 = x dx = dx q 0 q 0 1−x k=1 1 p p 1 = − ln(1 − q) = − ln(p) = ln(p) = ln p p−1 . q 1−p p−1 1 29. (a) “No tail in the first n tosses” means “all heads in the first n tosses.” Therefore, probab. = pn . (b) pn = (1 − p)n−1 p. P P∞ n−1 (c) E(number) = ∞ , which is equal to n=1 npn = p n=1 n(1 − p) 0 ∞ 1 1 d X n = . (1 − p) = −p −p dp n=0 p p 2