Explanation Given information: f ( x ) = 1 / ( 2 k 3 ) x 2 e − x / k , x ≥ 0 . Explanation: The given function is probability density function if it satisfies the condition below. (1) f ( x ) ≥ 0 for x ≥ 0 For any positive number k , that is, k > 0 , the function 1 / ( 2 k 3 ) x 2 e − x / k ≥ 0 Hence, f ( x ) = 1 / ( 2 k 3 ) x 2 e − x / k ≥ 0 for x ≥ 0 (2) Next, check that ∫ A B f ( x ) d x = 1 Substitute given function f ( x ) = 1 / ( 2 k 3 ) x 2 e − x / k , x ≥ 0 ∫ A B f ( x ) d x = ∫ 0 ∞ 1 2 k 3 x 2 e − x / k d x Take out the constant 1 2 k 3 . = 1 2 k 3 ∫ 0 ∞ x 2 e − x / k d x Use integration by parts rule to solve ∫ 0 ∞ x 2 e − x / k d x . Integration by parts rule is ∫ f ( x ) g ( x ) d x = f ( x ) G ( x ) − ∫ f ' ( x ) G ( x ) d x . Here, f ( x ) = x 2 and g ( x ) = e − x / k . By using integration by parts rule, ⇒ ∫ x 2 e − x / k d x = x 2 . ( − k e − x / k ) − ∫ 2 x . ( − k e − x / k ) d x ⇒ ∫ x 2 e − x / k d x = − k x 2 e − x / 10 + 2 ∫ x e − x / k d x … (1) The integral ∫ x e − x / k needs to be integrated using integration by parts rule. Let f ( x ) = x , g ( x ) = e − x / k . ∫ x e − x / k d x = − k x e − x / k − ∫ ( − k ) ( − e − x / k ) d x = − k x e − x / k − k 2 e − x / k Substitute the value of ∫ x e − x / k in (1)

12th Edition

ISBN: 9780137590810

Author: Larry J. Goldstein, David C. Lay, David I. Schneider, Nakhle H. Asmar, William Edward Tavernetti

Publisher: PEARSON

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Chapter 12, Problem 6RE

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Whether the function f(x)=1/(2k3)x2e−x/k,x≥0 is a probability density function for any positive integer k.

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Whether the function f ( x ) = 1 / ( 2 k 3 ) x 2 e − x / k , x ≥ 0 is a probability density function for any positive integer k .

Solution Summary: The author explains that the given function is a probability density function for any positive integer k if it satisfies the condition below.

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Whether the function f(x)=1/(2k3)x2e−x/k,x≥0 is a probability density function for any positive integer k.

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