lecture9

# lecture9 - Stat 302 Introduction to Probability Jiahua Chen...

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Unformatted text preview: Stat 302, Introduction to Probability Jiahua Chen January-April 2011 Jiahua Chen () Lecture 9 January-April 2011 1 / 32 Variance of continuous random variables If X has continuous distribution with pdf f ( x ) , then var ( X ) = E ( X − μ ) 2 = integraldisplay ( x − μ ) 2 f ( x ) dx where μ = E ( X ) and the range of integration is over R . We do require the existence of E ( X ) and this integration to claim the variance of X exists. We rarely directly compute E ( X − μ ) 2 but make use of var ( X ) = E ( X 2 ) − { E ( X ) } 2 . Jiahua Chen () Lecture 9 January-April 2011 2 / 32 Three continuous random variables Uniform distribution on [0, 1]: density function f ( x ) = 1 for ≤ x ≤ 1; Exponential distribution with rate parameter λ > 0: density function f ( x ; λ ) = λ exp ( − λ x ) for x > 0. Normal distribution with parameters μ and σ 2 : density function is f ( x ; μ , σ 2 ) = 1 √ 2 πσ exp {− ( x − μ ) 2 2 σ 2 } . Jiahua Chen () Lecture 9 January-April 2011 3 / 32 Variance of Uniform [0, 1] distribution Uniform distribution on [0, 1]: density function f ( x ) = 1 for ≤ x ≤ 1; It is seen that E ( X 2 ) = integraldisplay 1 x 2 · 1 dx = 1 3 . Hence, we have var ( X ) = 1 3 − parenleftbigg 1 2 parenrightbigg 2 = 1 12 . Jiahua Chen () Lecture 9 January-April 2011 4 / 32 Variance of Uniform [0, 1] distribution Uniform distribution on [0, 1]: density function f ( x ) = 1 for ≤ x ≤ 1; It is seen that E ( X 2 ) = integraldisplay 1 x 2 · 1 dx = 1 3 . Hence, we have var ( X ) = 1 3 − parenleftbigg 1 2 parenrightbigg 2 = 1 12 . Remember again that the variance of any random variable is positive (if exists). Jiahua Chen () Lecture 9 January-April 2011 4 / 32 Variance of exponential distribution Exponential distribution with rate parameter λ > 0: density function f ( x ; λ ) = λ exp ( − λ x ) for x > 0. Its mean is given by μ = 1 / λ . It is seen that E ( X 2 ) = integraldisplay ∞ x 2 λ exp ( − λ x ) dx = Γ ( 3 ) λ 2 = 2 λ 2 Hence, we have var ( X ) = 2 λ 2 − parenleftbigg 1 λ parenrightbigg 2 = 1 λ 2 In terms of μ = E ( X ) , we have var ( X ) = μ 2 . Jiahua Chen () Lecture 9 January-April 2011 5 / 32 Variance of standard normal The standard normal distribution has pdf φ ( x ) = 1 √ 2 π exp {− 1 2 x 2 } . Its mean is 0. It is seen that if X has standard normal distribution, E ( X 2 ) = 2 integraldisplay ∞ x 2 φ ( x ) dx = 2 √ π integraldisplay ∞ t 3 / 2 − 1 exp ( − t ) dt = 2 √ π Γ ( 3 / 2 ) = 1 √ π Γ ( 1 / 2 ) = 1. Hence, we have var ( X ) = 1 − 2 = 1. Jiahua Chen () Lecture 9 January-April 2011 6 / 32 Variance of N( μ , σ 2 ) distribution If Y ∼ N ( μ , σ 2 ) , then E ( Y ) = μ shown already....
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lecture9 - Stat 302 Introduction to Probability Jiahua Chen...

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