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For a nonnegative continuous random variable X, 5. P(Xu)du = E(X). Note that, X being nonnegative, E(X) is guaranteed to exist, though it may be + ∞. In this exercise, we are going to explore the above equality. (a) For any b>0, show that ["*P(X > u)du = b(1 − F(b)) + [*xƒ(x) dx, where f (respectively, F) is the pdf (respectively, cdf) of X. (b) Use the result in (a) to show that E(X): = + ∞ implies cb 8 "P(X > u du = [1 P(X> u)du = +∞. lim 6+∞ √0 (c) If E(X) < + ∞, show that SOF lim b(1 F(b)) = 0 and b→+∞ P(X> u)du = E(X).

For a nonnegative continuous random variable X, 5. P(Xu)du = E(X). Note that, X being nonnegative, E(X) is guaranteed to exist, though it may be + ∞. In this exercise, we are going to explore the above equality. (a) For any b>0, show that ["*P(X > u)du = b(1 − F(b)) + [*xƒ(x) dx, where f (respectively, F) is the pdf (respectively, cdf) of X. (b) Use the result in (a) to show that E(X): = + ∞ implies cb 8 "P(X > u du = [1 P(X> u)du = +∞. lim 6+∞ √0 (c) If E(X) < + ∞, show that SOF lim b(1 F(b)) = 0 and b→+∞ P(X> u)du = E(X).

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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For a nonnegative continuous random variable X, S Note that, X being nonnegative, E(X) is guaranteed to exist, though it may be + ∞. In this exercise, we are going to explore the above equality. (a) For any b > 0, show that S P(X > u)du = b(1 − F(b)) + where f (respectively, F) is the pdf (respectively, cdf) of X. lim P(X > u)du = E(X). (b) Use the result in (a) to show that E(X) = + ∞ implies b "P(X > u du = [" b- (c) If E(X) < +∞, show that f [xf(x)da, lim b(1 – F(b)) = 0 and b→+∞ P(X> u) du = = + ∞. P(X > u)du = E(X).
Transcribed Image Text:For a nonnegative continuous random variable X, S Note that, X being nonnegative, E(X) is guaranteed to exist, though it may be + ∞. In this exercise, we are going to explore the above equality. (a) For any b > 0, show that S P(X > u)du = b(1 − F(b)) + where f (respectively, F) is the pdf (respectively, cdf) of X. lim P(X > u)du = E(X). (b) Use the result in (a) to show that E(X) = + ∞ implies b "P(X > u du = [" b- (c) If E(X) < +∞, show that f [xf(x)da, lim b(1 – F(b)) = 0 and b→+∞ P(X> u) du = = + ∞. P(X > u)du = E(X).
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