Explanation Definition used: If X is a continuous random variable with probability function f on [ a , b ] , then the expected value of X is, E ( X ) = μ = ∫ a b x f ( x ) d x Then the variance of X is V a r ( X ) = ∫ a b x 2 f ( x ) d x − ( μ ) 2 . And the standard deviation value of X is σ ≈ V a r ( X ) . Formula used: Integration by parts If the function u and v differentiable, then ∫ u v d x = u ∫ v d x − ∫ [ d d x ( u ) ∫ v d x ] d x . Verification: The probability density function for the exponential distribution is defined by, f ( x ) = a e − a x for x in [ 0 , ∞ ) Use the above mentioned definition and obtain the expected value as follows, μ = ∫ 0 ∞ x a e − a x d x = [ − x e − a x ] 0 ∞ − ∫ 0 ∞ ( − e − a x ) d x ( by above Formula u = x and d v = a e − a x ) = [ − ∞ e − a × ∞ + 0 × e − a × 0 ] + [ − e − a x − a ] 0 ∞ = 0 − 1 a [ e − a × ∞ − e − a × 0 ] = 1 a ( ∵ e − ∞ = 0 , e 0 = 1 ) Thus, the expected value is 1 a . Use the above mentioned definition and obtain the variance as follows, The variance is var ( X ) = ∫ 0 ∞ x 2 e − a x d x − ( 1 a ) 2

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Chapter 11.3, Problem 20E

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To verify: The expected value and standard deviation of the exponential distribution.

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Chapter 11.3, Problem 19E

Chapter 11.3, Problem 21E

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