Math16B - Spring 09 Notes

Math16B - Spring 09 Notes - x dx We now need to integrate...

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MATH 16B - SPRING 2009 - SOLUTION TO VARIANCE EXAMPLE Example 1 . A continuous random variable X has cumulative distribution function F ( x ) = 0 if x 3 π 2 cos( x ) if 3 π 2 x 2 π 1 if x 2 π We determine variance of X . We will compute the variance using the formula (1) var( X ) = Z -∞ x 2 f ( x ) dx - E ( X ) 2 , where f ( x ) is the probability density function of X . To use the formula for variance, we need the probability density function f ( x ), which we find by differentiating the cumulative distribution function: f ( x ) = F 0 ( x ) = 0 if x < 3 π 2 - sin( x ) if 3 π 2 x 2 π 0 if x > 2 π . We also need the expected value E ( X ), which we compute as E ( X ) = Z -∞ xf ( x ) dx = Z 2 π 3 π/ 2 x ( - sin x ) dx. To compute this integral, we use integration by parts, with f 1 ( x ) = x and g 1 ( x ) = - sin( x ). Then f 0 1 ( x ) = 1 and g 1 has antiderivative G 1 ( x ) = cos x , whence E ( X ) = x cos x | 2 π 3 π/ 2 - Z 2 π 3 π/ 2 cos x dx = x cos x | 2 π 3 π/ 2 - sin x | 2 π 3 π/ 2 = 2 π - 1 . We now compute the integral in (1), Z -∞ x 2 f ( x ) dx = Z 2 π 3 π/ 2 x 2 ( - sin x ) dx. We integrate by parts with f 2 ( x ) = x 2 and g 2 ( x ) = - sin x , so that f 0 2 ( x ) = 2 x and G 2 ( x ) = cos x , which gives Z 2 π 3 π/ 2 x 2 ( - sin x ) dx = x 2 cos x ± ± 2 π 3 π/ 2 - Z 2 π 3 π/ 2 2 x cos
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Unformatted text preview: x dx. We now need to integrate by parts again, with f 3 ( x ) = 2 x and g 3 ( x ) = cos x so that f 3 ( x ) = 2 and G 3 ( x ) = sin( x ). This gives us Z 2 π 3 π/ 2 x 2 (-sin x ) dx = x 2 cos x ± ± 2 π 3 π/ 2-2 x sin x | 2 π 3 π/ 2-Z 2 π 3 π/ 2 2 sin( x ) dx ! = x 2 cos x ± ± 2 π 3 π/ 2-[2 x sin x + 2 cos( x )] 2 π 3 π/ 2 = ² x 2 cos x-2 x sin x-2 cos( x ) ³ 2 π 3 π/ 2 = ( (2 π ) 2 cos 2 π-2(2 π ) sin 2 π-2 cos(2 π ) )-´ 3 π 2 µ 2 cos 3 π 2-2 · 3 π 2 sin 3 π 2-2 cos ´ 3 π 2 µ ! = ( 4 π 2 · 1-4 π ·-2 · 1 )-´ 9 π 2 4 ·-2 · 3 π 2 · (-1)-2 · µ = 4 π 2-3 π-2 Therefore, var( X ) = Z 2 π 3 π/ 2 x 2 f ( x ) dx-E ( X ) 2 = (4 π 2-3 π-2)-(2 π-1) 2 = (4 π 2-3 π-2)-(4 π 2-4 π + 1) = π-3 . 1...
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This note was uploaded on 05/12/2010 for the course CHEM 3A taught by Professor Fretchet during the Fall '08 term at Berkeley.

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