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Dr. Hackney STA Solutions pg 85

# Dr. Hackney STA Solutions pg 85 - Second Edition 5-19 Write...

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Second Edition 5-19 Write P ( Z i +1 a ) = P ( V i +1 a and U i +1 ρ i +1 ) + P ( Z i a and U i +1 > ρ i +1 ) . Since Z i f Y , suppressing the unnecessary subscripts we can write P ( Z i +1 a ) = P ( V a and U ρ ( V, Y )) + P ( Y a and U > ρ ( V, Y )) . Add and subtract P ( Y a and U ρ ( V, Y )) to get P ( Z i +1 a ) = P ( Y a ) + P ( V a and U ρ ( V, Y )) - P ( Y a and U ρ ( V, Y )) . Thus we need to show that P ( V a and U ρ ( V, Y )) = P ( Y a and U ρ ( V, Y )) . Write out the probability as P ( V a and U ρ ( V, Y )) = a -∞ -∞ ρ ( v, y ) f Y ( y ) f V ( v ) dydv = a -∞ -∞ I ( w ( v, y ) 1) f Y ( v ) f V ( y ) f V ( v ) f Y ( y ) f Y ( y ) f V ( v ) dydv + a -∞ -∞ I ( w ( v, y ) 1) f Y ( y ) f V ( v ) dydv = a -∞ -∞ I ( w ( v, y ) 1) f Y ( v ) f V ( y ) dydv + a -∞ -∞ I ( w ( v, y ) 1) f Y ( y ) f V ( v ) dydv. Now, notice that w ( v, y ) = 1 /w ( y, v ), and thus first term above can be written a -∞ -∞ I ( w ( v, y ) 1) f Y ( v ) f V ( y ) dydv = a -∞ -∞ I ( w ( y, v ) > 1) f Y ( v ) f V ( y ) dydv = P ( Y a, ρ ( V, Y ) = 1 , U
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