Global Journal of Science Frontier Research, F: Mathematics and Decision Science, Volume 22 Issue 4

1 Year 2022 39 © 2022 Global Journals Global Journal of Science Frontier Research Volume XXII Issue ersion I V IV ( F ) On Baysian Estimation of Loss of Estimators of Unknown Parameter of Binomial Distribution E θ L ( θ, δ 1 B , γ 1 B ) = E θ [ h ( θ )( θ − ( X + α − 1)( A − 2) − 1 ) 2 ]( A − 2) 1 / 2 + ( A − 2) − 1 / 2 (2.11) Or, E θ L ( θ, δ 1 B , γ 1 B ) = [ n + h ( θ )(1 − α + θ ( α + β − 2)) 2 ]( A − 2) − 3 / 2 +( A − 2) − 1 / 2 < ∞ (2.12) In this case, R ( θ, δ 1 B ) = E θ { h ( θ )( θ − δ 1 B ) } 2 (2.13) Or, R ( θ, δ 1 B ) = [ n + h ( θ ) { 1 − α + θ ( α + β − 2) } 2 ]( A − 2) − 2 (2.14) As mentioned by Keifer (1977),an estimator γ ( X )is said to be conservatively biased if, E θ { γ ( X ) } ≥ R ( θ, δ ) = E θ { w ( θ, δ ) } (2.15) In the light of this condition, γ 0 ( X ) as given by Rukhin (1988) is not conservatively biased. In this case, E θ { γ 1 B ( X ) } = 1 A − 2 (2.16) Let δ 0 B ( X ) and γ 0 B ( X )be values of δ 1 B ( X ) and γ 1 B ( X ) ,respectively when, α = β = 0.If possible let , E θ { γ 0 B ( X ) } ≥ R ( θ, δ 0 B (2.17) which holds if, − 2 θ 2 + 2 θ − 1 ≥ 0 (2.18) which is a contradiction,since 0 < θ < 1 and maximum value of − 2 θ 2 +2 θ − 1 is − 1 2 which corresponds to θ = 1 2 .Moreover, − 2 θ 2 + 2 θ − 1 = − 1 for θ = 1 and θ = 0 Thus, γ 0 B ( X ) is not conservatively biased. When α = β = 1,we have, E θ { γ 1 B ( X ) } = R ( θ, δ 1 B ) = 1 n (2.19) ) Or, γ 1 B ( X ) = 1 A − 2 (2.10) on simplification,provided, A = n + α + β > 2. We,see that, in this case γ 1 B ( X ) does not depend upon X and is function of n, α and β N otes