Dr. Hackney STA Solutions pg 23

# Dr. Hackney STA Solutions pg 23 - 2-8 Solutions Manual for...

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Unformatted text preview: 2-8 Solutions Manual for Statistical Inference Since a f (x)dx + - a a f (x)dx = - f (x)dx = 1, it must be that f (x) dx = - a f (x) dx = 1/2. Therefore, a is a median. c. EX - a = = E(X - a) = - a (x - a)f (x)dx (x - a)f (x)dx + - a (x - a)f (x)dx = 0 (- )f (a - )d + 0 f (a + )d With a change of variable, = a - x in the first integral, and = x - a in the second integral we obtain that EX - a = E(X - a) = - 0 f (a - )d + 0 f (a - )d (f (a + ) = f (a - ) for all > 0) = 0. (two integrals are same) Therefore, EX = a. d. If a > > 0, f (a - ) = e-(a- ) > e-(a+ ) = f (a + ). Therefore, f (x) is not symmetric about a > 0. If - < a 0, f (a - ) = 0 < e-(a+ ) = f (a + ). Therefore, f (x) is not symmetric about a 0, either. e. The median of X = log 2 < 1 = EX. 2.27 a. The standard normal pdf. b. The uniform on the interval (0, 1). c. For the case when the mode is unique. Let a be the point of symmetry and b be the mode. Let assume that a is not the mode and without loss of generality that a = b+ > b for > 0. Since b is the mode then f (b) > f (b + ) f (b + 2 ) which implies that f (a - ) > f (a) f (a + ) which contradict the fact the f (x) is symmetric. Thus a is the mode. For the case when the mode is not unique, there must exist an interval (x1 , x2 ) such that f (x) has the same value in the whole interval, i.e, f (x) is flat in this interval and for all b (x1 , x2 ), b is a mode. Let assume that a (x1 , x2 ), thus a is not a mode. Let also assume without loss of generality that a = (b + ) > b. Since b is a mode and a = (b + ) (x1 , x2 ) then f (b) > f (b + ) f (b + 2 ) which contradict the fact the f (x) is symmetric. Thus a (x1 , x2 ) and is a mode. d. f (x) is decreasing for x 0, with f (0) > f (x) > f (y) for all 0 < x < y. Thus f (x) is unimodal and 0 is the mode. ...
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