How to work out the variance of this estimator?












1












$begingroup$


enter image description here



We want the variance of this biased estimator.



The answer is: $$σ^2/4n$$



Why is it not this? $$σ^2/2n$$



I.e. wouldn't it be $$1/2n^2 * n * σ^2$$



Any clarification appreciated.



Thanks!






statistics variance parameter-estimation







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    enter image description here



    We want the variance of this biased estimator.



    The answer is: $$σ^2/4n$$



    Why is it not this? $$σ^2/2n$$



    I.e. wouldn't it be $$1/2n^2 * n * σ^2$$



    Any clarification appreciated.



    Thanks!






    statistics variance parameter-estimation







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      enter image description here



      We want the variance of this biased estimator.



      The answer is: $$σ^2/4n$$



      Why is it not this? $$σ^2/2n$$



      I.e. wouldn't it be $$1/2n^2 * n * σ^2$$



      Any clarification appreciated.



      Thanks!






      statistics variance parameter-estimation







      share|cite|improve this question









      $endgroup$




      enter image description here



      We want the variance of this biased estimator.



      The answer is: $$σ^2/4n$$



      Why is it not this? $$σ^2/2n$$



      I.e. wouldn't it be $$1/2n^2 * n * σ^2$$



      Any clarification appreciated.



      Thanks!






      statistics variance parameter-estimation




      statistics variance parameter-estimation






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      asked May 20 '18 at 17:09









      MathsHelpMathsHelp

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      16810






















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          Note that, if $lambdain Bbb R$, then $text{Var}(lambda X) = lambda^{bf{2}}text{Var}(X)$, not $lambdatext{Var}(X)$.



          Thus



          $$text{Var}(hat{theta}) = left(frac{1}{2n}right)^2.text{Var}left(sum_{i=1}^n X_iright) = frac{1}{4n^2}.n.sigma^2 = frac{sigma^2}{4n}.$$






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            $begingroup$

            Note that, if $lambdain Bbb R$, then $text{Var}(lambda X) = lambda^{bf{2}}text{Var}(X)$, not $lambdatext{Var}(X)$.



            Thus



            $$text{Var}(hat{theta}) = left(frac{1}{2n}right)^2.text{Var}left(sum_{i=1}^n X_iright) = frac{1}{4n^2}.n.sigma^2 = frac{sigma^2}{4n}.$$






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              Note that, if $lambdain Bbb R$, then $text{Var}(lambda X) = lambda^{bf{2}}text{Var}(X)$, not $lambdatext{Var}(X)$.



              Thus



              $$text{Var}(hat{theta}) = left(frac{1}{2n}right)^2.text{Var}left(sum_{i=1}^n X_iright) = frac{1}{4n^2}.n.sigma^2 = frac{sigma^2}{4n}.$$






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                Note that, if $lambdain Bbb R$, then $text{Var}(lambda X) = lambda^{bf{2}}text{Var}(X)$, not $lambdatext{Var}(X)$.



                Thus



                $$text{Var}(hat{theta}) = left(frac{1}{2n}right)^2.text{Var}left(sum_{i=1}^n X_iright) = frac{1}{4n^2}.n.sigma^2 = frac{sigma^2}{4n}.$$






                share|cite|improve this answer











                $endgroup$



                Note that, if $lambdain Bbb R$, then $text{Var}(lambda X) = lambda^{bf{2}}text{Var}(X)$, not $lambdatext{Var}(X)$.



                Thus



                $$text{Var}(hat{theta}) = left(frac{1}{2n}right)^2.text{Var}left(sum_{i=1}^n X_iright) = frac{1}{4n^2}.n.sigma^2 = frac{sigma^2}{4n}.$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 9 at 6:01

























                answered May 20 '18 at 17:22









                pafpaf

                4,0281824




                4,0281824






























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