Dr. Hackney STA Solutions pg 63

# Dr. Hackney STA Solutions pg 63 - ) = P ( Q n i =1 X i y )...

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Second Edition 4-19 4.51 a. P ( X/Y t ) = ± 1 2 t t > 1 1 2 + (1 - t ) t 1 P ( XY t ) = t - t log t 0 < t < 1 . b. P ( XY/Z t ) = Z 1 0 P ( XY zt ) dz = ²R 1 0 ³ zt 2 + (1 - zt ) ´ dz if t 1 R 1 t 0 ³ zt 2 + (1 - zt ) ´ dz + R 1 1 t 1 2 zt dz if t 1 = ± 1 - t/ 4 if t 1 t - 1 4 t + 1 2 t log t if t > 1 . 4.53 P (Real Roots) = P ( B 2 > 4 AC ) = P (2 log B > log 4 + log A + log C ) = P ( - 2 log B ≤ - log 4 - log A - log C ) = P ( - 2 log B ≤ - log 4 + ( - log A - log C )) . Let X = - 2 log B , Y = - log A - log C . Then X exponential(2), Y gamma(2 , 1), indepen- dent, and P (Real Roots) = P ( X < - log 4 + Y ) = Z log 4 P ( X < - log 4 + y ) f Y ( y ) dy = Z log 4 Z - log 4+ y 0 1 2 e - x/ 2 dxye - y dy = Z log 4 µ 1 - e - 1 2 log 4 e - y/ 2 ye - y dy. Integration-by-parts will show that R a ye - y/b = b ( a + b ) e - a/b and hence P (Real Roots) = 1 4 (1 + log 4) - 1 24 · 2 3 + log 4 ¸ = . 511 . 4.54 Let Y = Q n i =1 X i . Then P ( Y y
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Unformatted text preview: ) = P ( Q n i =1 X i y ) = P ( n i =1-log X i -log y ). Now,-log X i exponential(1) = gamma(1 , 1). By Example 4.6.8, n i =1-log X i gamma( n, 1). Therefore, P ( Y y ) = Z -log y 1 ( n ) z n-1 e-z dz, and f Y ( y ) = d dy Z -log y 1 ( n ) z n-1 e-z dz =-1 ( n ) (-log y ) n-1 e-(-log y ) d dy (-log y ) = 1 ( n ) (-log y ) n-1 , &lt; y &lt; 1 ....
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## This note was uploaded on 02/03/2012 for the course STA 1014 taught by Professor Dr.hackney during the Spring '12 term at UNF.

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