mont4e_sm_ch05_mind

# c 2v x p p a b 0 0 5 96 a f xy x y dydx

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Unformatted text preview: om the definition of the mean. Therefore, each cross-term in the last integral for V(Y) is zero and V (Y ) = [∫ c ( x − μ ) 2 1 1 1 2 [ f X 1 ( x1 )dx1 ... ∫ c 2 ( x p − μ p ) 2 f X p ( x p )dx p p = c12V ( X 1 ) + ... + c 2V ( X p ). p a b 0 0 ∫∫ 5-96 a f XY ( x, y )dydx = ∫ 0 b ∫ cdydx = cab . Therefore, c = 1/ab. Then, 0 b f X ( x) = ∫ cdy = 0 1 a a 1 for 0 < x < a, and fY ( y) = ∫ cdx = b for 0 < y < b. Therefore, 0 f XY (x,y)=f X (x)f Y (y) for all x and y and X and Y are independent. 5-97 The marginal density of X is b b b 0 0 f X ( x) = ∫ g ( x)h(u )du = g ( x) ∫ h(u )du = kg ( x) where k = ∫ h(u )du. Also, 0 a f Y ( y ) = lh( y ) where l = ∫ g (v)dv. Because f XY (x,y) is a probability density function, 0 ⎡ ⎤⎡b ⎤ g ( x)h( y )dydx = ⎢ ∫ g (v)dv ⎥ ⎢ ∫ h(u )du ⎥ = 1. Therefore, kl = 1 and ∫∫ 00 ⎣0 ⎦⎣ 0 ⎦ f XY (x,y)=f X (x)f Y (y) for all x and y. a b a 5-48...
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