Asked by ramj
                Argue that the proposed estimators  πΛ  and  πΛ  below are both consistent and asymptotically normal. Then, give their asymptotic variances  π(πΛ)  and  π(πΛ) , and decide if one of them is always bigger than the other.
Let π1,β¦,ππβΌπ.π.π.π―ππππ(π) , for some π>0 . Let πΜ =πβ―β―β―β―β―π and πΜ =βln(πβ―β―β―β―π) , where ππ=1{ππ=0},π=1,β¦,π .
π(πΛ) =? and π(πΛ) =?
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As above, argue that both proposed estimators πΛ and πΛ are consistent and asymptotically normal. Then, give their asymptotic variances π(πΛ) and π(πΛ), and decide if one of them is always bigger than the other.
Let π1,β¦,ππβΌπ.π.π.π€ππ(π) , for some π>0 . Let πΛ=1πβ―β―β―β―β―π and πΜ =βln(πβ―β―β―β―π) , where ππ=1{ππ>1},π=1,β¦,π .
π(πΛ) = ? and π(πΛ)=?
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As above, argue that both proposed estimators πΛ,πΛ, are consistent and asymptotically normal. Then, give their asymptotic variances π(πΛ) and π(πΛ) and decide if one of them is always bigger than the other.
Let π1,β¦,ππβΌπ.π.π.π¦πΎππ(π) , for some πβ(0,1) . That means that
π(π1=π)=π(1βπ)πβ1,for π=1,2,β¦.
 
Let
πΛ = 1πβ―β―β―β―β―π,
and πΛ be the number of ones in the sample divided by π .
π(πΛ)=? and π(πΛ)=?
            
            
        Let π1,β¦,ππβΌπ.π.π.π―ππππ(π) , for some π>0 . Let πΜ =πβ―β―β―β―β―π and πΜ =βln(πβ―β―β―β―π) , where ππ=1{ππ=0},π=1,β¦,π .
π(πΛ) =? and π(πΛ) =?
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As above, argue that both proposed estimators πΛ and πΛ are consistent and asymptotically normal. Then, give their asymptotic variances π(πΛ) and π(πΛ), and decide if one of them is always bigger than the other.
Let π1,β¦,ππβΌπ.π.π.π€ππ(π) , for some π>0 . Let πΛ=1πβ―β―β―β―β―π and πΜ =βln(πβ―β―β―β―π) , where ππ=1{ππ>1},π=1,β¦,π .
π(πΛ) = ? and π(πΛ)=?
----------------------------------------------------------------------------------------------------
As above, argue that both proposed estimators πΛ,πΛ, are consistent and asymptotically normal. Then, give their asymptotic variances π(πΛ) and π(πΛ) and decide if one of them is always bigger than the other.
Let π1,β¦,ππβΌπ.π.π.π¦πΎππ(π) , for some πβ(0,1) . That means that
π(π1=π)=π(1βπ)πβ1,for π=1,2,β¦.
Let
πΛ = 1πβ―β―β―β―β―π,
and πΛ be the number of ones in the sample divided by π .
π(πΛ)=? and π(πΛ)=?
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                    Answered by
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    The proposed estimators  πΛ  and  πΛ  are both consistent and asymptotically normal. Their asymptotic variances are  π(πΛ)=π(1βπ)  and  π(πΛ)=π(1βπ)/π , respectively. The variance of  πΛ  is always bigger than the variance of  πΛ .
    
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