Dr. Hackney STA Solutions pg 85

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

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Unformatted text preview: Second Edition 5-19 Write P (Zi+1 a) = P (Vi+1 a and Ui+1 i+1 ) + P (Zi a and Ui+1 > i+1 ). Since Zi fY , suppressing the unnecessary subscripts we can write P (Zi+1 a) = P (V a and U (V, Y )) + P (Y a and U > (V, Y )). Add and subtract P (Y a and U (V, Y )) to get P (Zi+1 a) = P (Y a) + P (V a and U (V, Y )) -P (Y a and U (V, Y )). Thus we need to show that P (V a and U (V, Y )) = P (Y a and U (V, Y )). Write out the probability as P (V a and U (V, Y )) a = - a - (v, y)fY (y)fV (v)dydv I(w(v, y) 1) - - a = + - a fY (v)fV (y) fV (v)fY (y) fY (y)fV (v)dydv I(w(v, y) 1)fY (y)fV (v)dydv - = + - I(w(v, y) 1)fY (v)fV (y)dydv - - a I(w(v, y) 1)fY (y)fV (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)fY (v)fV (y)dydv - a - = - I(w(y, v) > 1)fY (v)fV (y)dydv = P (Y a, (V, Y ) = 1, U (V, Y )). The second term is a - I(w(v, y) 1)fY (y)fV (v)dydv - a - a = - I(w(y, v) 1)fY (y)fV (v)dydv I(w(y, v) 1) - a - = = - a - - I(w(y, v) 1) fV (y)fY (v) fV (y)fY (v) fY (y)fV (v) fV (y)fY (v) fY (y)fV (v)dydv fV (y)fY (v)dydv = - I(w(y, v) 1)w(y, v)fV (y)fY (v)dydv = P (Y a, U (V, Y ), (V, Y ) 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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