MLE of $lambda$ Given $f(x;lambda)=1-dfrac{2}{3}lambda+lambdasqrt{x}$












1












$begingroup$



$f(x;lambda)=1-dfrac{2}{3}lambda+lambdasqrt{x} ; 0le xle1 $



$0$ otherwise




What is the maximum likelihood estimate of the parameter $lambda$ based on two independent observations $x_1=dfrac{1}{4}$ and $x_2=dfrac{9}{16}$



My input



Liklihood function of the sample is given by



$L(lambda)=prod_{i=1}^{2}f(x_i;lambda)=(1-dfrac{2}{3}lambda+lambdasqrt{x_i})$



=$(1-dfrac{2}{3}lambda+lambdasqrt{x_1})(1-dfrac{2}{3}lambda+lambdasqrt{x_2})$



$log(L(lambda)=log(1-dfrac{2}{3}lambda+lambdasqrt{x_1})(1-dfrac{2}{3}lambda+lambdasqrt{x_2})$



$=log(1-dfrac{2}{3}lambda+lambdasqrt{x_1})+log(1-dfrac{2}{3}lambda+lambdasqrt{x_2})$



Am I following correct path ? Can someone tell me ?






statistical-inference maximum-likelihood







share|cite|improve this question









$endgroup$

















    1












    $begingroup$



    $f(x;lambda)=1-dfrac{2}{3}lambda+lambdasqrt{x} ; 0le xle1 $



    $0$ otherwise




    What is the maximum likelihood estimate of the parameter $lambda$ based on two independent observations $x_1=dfrac{1}{4}$ and $x_2=dfrac{9}{16}$



    My input



    Liklihood function of the sample is given by



    $L(lambda)=prod_{i=1}^{2}f(x_i;lambda)=(1-dfrac{2}{3}lambda+lambdasqrt{x_i})$



    =$(1-dfrac{2}{3}lambda+lambdasqrt{x_1})(1-dfrac{2}{3}lambda+lambdasqrt{x_2})$



    $log(L(lambda)=log(1-dfrac{2}{3}lambda+lambdasqrt{x_1})(1-dfrac{2}{3}lambda+lambdasqrt{x_2})$



    $=log(1-dfrac{2}{3}lambda+lambdasqrt{x_1})+log(1-dfrac{2}{3}lambda+lambdasqrt{x_2})$



    Am I following correct path ? Can someone tell me ?






    statistical-inference maximum-likelihood







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$



      $f(x;lambda)=1-dfrac{2}{3}lambda+lambdasqrt{x} ; 0le xle1 $



      $0$ otherwise




      What is the maximum likelihood estimate of the parameter $lambda$ based on two independent observations $x_1=dfrac{1}{4}$ and $x_2=dfrac{9}{16}$



      My input



      Liklihood function of the sample is given by



      $L(lambda)=prod_{i=1}^{2}f(x_i;lambda)=(1-dfrac{2}{3}lambda+lambdasqrt{x_i})$



      =$(1-dfrac{2}{3}lambda+lambdasqrt{x_1})(1-dfrac{2}{3}lambda+lambdasqrt{x_2})$



      $log(L(lambda)=log(1-dfrac{2}{3}lambda+lambdasqrt{x_1})(1-dfrac{2}{3}lambda+lambdasqrt{x_2})$



      $=log(1-dfrac{2}{3}lambda+lambdasqrt{x_1})+log(1-dfrac{2}{3}lambda+lambdasqrt{x_2})$



      Am I following correct path ? Can someone tell me ?






      statistical-inference maximum-likelihood







      share|cite|improve this question









      $endgroup$





      $f(x;lambda)=1-dfrac{2}{3}lambda+lambdasqrt{x} ; 0le xle1 $



      $0$ otherwise




      What is the maximum likelihood estimate of the parameter $lambda$ based on two independent observations $x_1=dfrac{1}{4}$ and $x_2=dfrac{9}{16}$



      My input



      Liklihood function of the sample is given by



      $L(lambda)=prod_{i=1}^{2}f(x_i;lambda)=(1-dfrac{2}{3}lambda+lambdasqrt{x_i})$



      =$(1-dfrac{2}{3}lambda+lambdasqrt{x_1})(1-dfrac{2}{3}lambda+lambdasqrt{x_2})$



      $log(L(lambda)=log(1-dfrac{2}{3}lambda+lambdasqrt{x_1})(1-dfrac{2}{3}lambda+lambdasqrt{x_2})$



      $=log(1-dfrac{2}{3}lambda+lambdasqrt{x_1})+log(1-dfrac{2}{3}lambda+lambdasqrt{x_2})$



      Am I following correct path ? Can someone tell me ?






      statistical-inference maximum-likelihood




      statistical-inference maximum-likelihood






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 10 at 12:04









      Daman deepDaman deep

      751318




      751318






















          1 Answer
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          $begingroup$

          You are correct, but in this case it is easier to optimize the likelihood instead of the log-likelihood as it is a polynomial. So begin{align}L(lambda)&=left(1+left(sqrt{x_1}-frac23right)lambdaright)left(1+left(sqrt{x_2}-frac23right)lambdaright)\&=left(sqrt{x_1}-frac23right)left(sqrt{x_2}-frac23right)lambda^2+left(sqrt{x_1}+sqrt{x_2}-frac43right)lambda+1\ implies L'left(hatlambdaright)&=2left(sqrt{x_1}-frac23right)left(sqrt{x_2}-frac23right)hatlambda+sqrt{x_1}+sqrt{x_2}-frac43=0end{align} There you can simply solve for $hatlambda$ and check whether it is a maximum or minimum.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            $hat lambda =8 $ But my lambda coming out to be $3 $ can you check if its coming out $8$ or not ?
            $endgroup$
            – Daman deep
            Jan 10 at 13:22












          • $begingroup$
            Even to optimize a polynomial, especially expressed as a product, using logarithms is often the way to go.
            $endgroup$
            – Did
            Jan 10 at 13:32










          • $begingroup$
            @Did I suppose for quadratics and cubics then; once you differentiate them they are easy to factor
            $endgroup$
            – TheSimpliFire
            Jan 10 at 14:16











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          1 Answer
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          0












          $begingroup$

          You are correct, but in this case it is easier to optimize the likelihood instead of the log-likelihood as it is a polynomial. So begin{align}L(lambda)&=left(1+left(sqrt{x_1}-frac23right)lambdaright)left(1+left(sqrt{x_2}-frac23right)lambdaright)\&=left(sqrt{x_1}-frac23right)left(sqrt{x_2}-frac23right)lambda^2+left(sqrt{x_1}+sqrt{x_2}-frac43right)lambda+1\ implies L'left(hatlambdaright)&=2left(sqrt{x_1}-frac23right)left(sqrt{x_2}-frac23right)hatlambda+sqrt{x_1}+sqrt{x_2}-frac43=0end{align} There you can simply solve for $hatlambda$ and check whether it is a maximum or minimum.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            $hat lambda =8 $ But my lambda coming out to be $3 $ can you check if its coming out $8$ or not ?
            $endgroup$
            – Daman deep
            Jan 10 at 13:22












          • $begingroup$
            Even to optimize a polynomial, especially expressed as a product, using logarithms is often the way to go.
            $endgroup$
            – Did
            Jan 10 at 13:32










          • $begingroup$
            @Did I suppose for quadratics and cubics then; once you differentiate them they are easy to factor
            $endgroup$
            – TheSimpliFire
            Jan 10 at 14:16
















          0












          $begingroup$

          You are correct, but in this case it is easier to optimize the likelihood instead of the log-likelihood as it is a polynomial. So begin{align}L(lambda)&=left(1+left(sqrt{x_1}-frac23right)lambdaright)left(1+left(sqrt{x_2}-frac23right)lambdaright)\&=left(sqrt{x_1}-frac23right)left(sqrt{x_2}-frac23right)lambda^2+left(sqrt{x_1}+sqrt{x_2}-frac43right)lambda+1\ implies L'left(hatlambdaright)&=2left(sqrt{x_1}-frac23right)left(sqrt{x_2}-frac23right)hatlambda+sqrt{x_1}+sqrt{x_2}-frac43=0end{align} There you can simply solve for $hatlambda$ and check whether it is a maximum or minimum.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            $hat lambda =8 $ But my lambda coming out to be $3 $ can you check if its coming out $8$ or not ?
            $endgroup$
            – Daman deep
            Jan 10 at 13:22












          • $begingroup$
            Even to optimize a polynomial, especially expressed as a product, using logarithms is often the way to go.
            $endgroup$
            – Did
            Jan 10 at 13:32










          • $begingroup$
            @Did I suppose for quadratics and cubics then; once you differentiate them they are easy to factor
            $endgroup$
            – TheSimpliFire
            Jan 10 at 14:16














          0












          0








          0





          $begingroup$

          You are correct, but in this case it is easier to optimize the likelihood instead of the log-likelihood as it is a polynomial. So begin{align}L(lambda)&=left(1+left(sqrt{x_1}-frac23right)lambdaright)left(1+left(sqrt{x_2}-frac23right)lambdaright)\&=left(sqrt{x_1}-frac23right)left(sqrt{x_2}-frac23right)lambda^2+left(sqrt{x_1}+sqrt{x_2}-frac43right)lambda+1\ implies L'left(hatlambdaright)&=2left(sqrt{x_1}-frac23right)left(sqrt{x_2}-frac23right)hatlambda+sqrt{x_1}+sqrt{x_2}-frac43=0end{align} There you can simply solve for $hatlambda$ and check whether it is a maximum or minimum.






          share|cite|improve this answer









          $endgroup$



          You are correct, but in this case it is easier to optimize the likelihood instead of the log-likelihood as it is a polynomial. So begin{align}L(lambda)&=left(1+left(sqrt{x_1}-frac23right)lambdaright)left(1+left(sqrt{x_2}-frac23right)lambdaright)\&=left(sqrt{x_1}-frac23right)left(sqrt{x_2}-frac23right)lambda^2+left(sqrt{x_1}+sqrt{x_2}-frac43right)lambda+1\ implies L'left(hatlambdaright)&=2left(sqrt{x_1}-frac23right)left(sqrt{x_2}-frac23right)hatlambda+sqrt{x_1}+sqrt{x_2}-frac43=0end{align} There you can simply solve for $hatlambda$ and check whether it is a maximum or minimum.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 10 at 13:15









          TheSimpliFireTheSimpliFire

          12.4k62460




          12.4k62460












          • $begingroup$
            $hat lambda =8 $ But my lambda coming out to be $3 $ can you check if its coming out $8$ or not ?
            $endgroup$
            – Daman deep
            Jan 10 at 13:22












          • $begingroup$
            Even to optimize a polynomial, especially expressed as a product, using logarithms is often the way to go.
            $endgroup$
            – Did
            Jan 10 at 13:32










          • $begingroup$
            @Did I suppose for quadratics and cubics then; once you differentiate them they are easy to factor
            $endgroup$
            – TheSimpliFire
            Jan 10 at 14:16


















          • $begingroup$
            $hat lambda =8 $ But my lambda coming out to be $3 $ can you check if its coming out $8$ or not ?
            $endgroup$
            – Daman deep
            Jan 10 at 13:22












          • $begingroup$
            Even to optimize a polynomial, especially expressed as a product, using logarithms is often the way to go.
            $endgroup$
            – Did
            Jan 10 at 13:32










          • $begingroup$
            @Did I suppose for quadratics and cubics then; once you differentiate them they are easy to factor
            $endgroup$
            – TheSimpliFire
            Jan 10 at 14:16
















          $begingroup$
          $hat lambda =8 $ But my lambda coming out to be $3 $ can you check if its coming out $8$ or not ?
          $endgroup$
          – Daman deep
          Jan 10 at 13:22






          $begingroup$
          $hat lambda =8 $ But my lambda coming out to be $3 $ can you check if its coming out $8$ or not ?
          $endgroup$
          – Daman deep
          Jan 10 at 13:22














          $begingroup$
          Even to optimize a polynomial, especially expressed as a product, using logarithms is often the way to go.
          $endgroup$
          – Did
          Jan 10 at 13:32




          $begingroup$
          Even to optimize a polynomial, especially expressed as a product, using logarithms is often the way to go.
          $endgroup$
          – Did
          Jan 10 at 13:32












          $begingroup$
          @Did I suppose for quadratics and cubics then; once you differentiate them they are easy to factor
          $endgroup$
          – TheSimpliFire
          Jan 10 at 14:16




          $begingroup$
          @Did I suppose for quadratics and cubics then; once you differentiate them they are easy to factor
          $endgroup$
          – TheSimpliFire
          Jan 10 at 14:16


















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