Estimation in multivariate normal

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I need to show $underlinehatmu=underlinebary$:



$$L(y_1, y_2, ...) = prod _i=1^n f(y_i,underlinemu,underlineSigma)$$



$$ = prod _i=1^n frac1^1/2e^-frac12(y_i -mu)'Sigma ^-1(y_i -mu) $$



Taking log:



$$= -np logbiggl(sqrt2pibiggl)-fracn2log(Sigma)-frac12sum_i=1^n(y_i -mu)'Sigma ^-1(y_i -mu)$$



Maximising:



$$ 0 = frac-12 times 2 sum^n_i=1 (y_i -mu)$$



Then, I do not know how to proceed.
I presume



$$ nmu =sum^n_i=1 y_i $$










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  • 1




    After $116$ questions and $65$ answers, I would thought you'd have had enough opportunity to familiarize yourself with how we typeset math on this site. Please see this tutorial and reference. In particular, double dollar signs yield displayed equations, which are much easier to read, especially when they heavily mix fractions, subscripts, superscripts and radicals as in your case; and operator names should be in Roman font, using e.g. log, or operatornamename for operators that don't have a command of their own.
    – joriki
    Sep 3 at 11:47






  • 1




    Advice is well received, and editing done also. Sometimes a bit of sarcasm can change so much. :P
    – Tos Hina
    Sep 3 at 18:34






  • 2




    It looks like the last statement immediately follows from the previous. What are you looking for?
    – herb steinberg
    Sep 3 at 18:37







  • 2




    It would help if you explicitly defined $hatmu$ and $bary$.
    – herb steinberg
    Sep 3 at 19:25














up vote
1
down vote

favorite












I need to show $underlinehatmu=underlinebary$:



$$L(y_1, y_2, ...) = prod _i=1^n f(y_i,underlinemu,underlineSigma)$$



$$ = prod _i=1^n frac1^1/2e^-frac12(y_i -mu)'Sigma ^-1(y_i -mu) $$



Taking log:



$$= -np logbiggl(sqrt2pibiggl)-fracn2log(Sigma)-frac12sum_i=1^n(y_i -mu)'Sigma ^-1(y_i -mu)$$



Maximising:



$$ 0 = frac-12 times 2 sum^n_i=1 (y_i -mu)$$



Then, I do not know how to proceed.
I presume



$$ nmu =sum^n_i=1 y_i $$










share|cite|improve this question



















  • 1




    After $116$ questions and $65$ answers, I would thought you'd have had enough opportunity to familiarize yourself with how we typeset math on this site. Please see this tutorial and reference. In particular, double dollar signs yield displayed equations, which are much easier to read, especially when they heavily mix fractions, subscripts, superscripts and radicals as in your case; and operator names should be in Roman font, using e.g. log, or operatornamename for operators that don't have a command of their own.
    – joriki
    Sep 3 at 11:47






  • 1




    Advice is well received, and editing done also. Sometimes a bit of sarcasm can change so much. :P
    – Tos Hina
    Sep 3 at 18:34






  • 2




    It looks like the last statement immediately follows from the previous. What are you looking for?
    – herb steinberg
    Sep 3 at 18:37







  • 2




    It would help if you explicitly defined $hatmu$ and $bary$.
    – herb steinberg
    Sep 3 at 19:25












up vote
1
down vote

favorite









up vote
1
down vote

favorite











I need to show $underlinehatmu=underlinebary$:



$$L(y_1, y_2, ...) = prod _i=1^n f(y_i,underlinemu,underlineSigma)$$



$$ = prod _i=1^n frac1^1/2e^-frac12(y_i -mu)'Sigma ^-1(y_i -mu) $$



Taking log:



$$= -np logbiggl(sqrt2pibiggl)-fracn2log(Sigma)-frac12sum_i=1^n(y_i -mu)'Sigma ^-1(y_i -mu)$$



Maximising:



$$ 0 = frac-12 times 2 sum^n_i=1 (y_i -mu)$$



Then, I do not know how to proceed.
I presume



$$ nmu =sum^n_i=1 y_i $$










share|cite|improve this question















I need to show $underlinehatmu=underlinebary$:



$$L(y_1, y_2, ...) = prod _i=1^n f(y_i,underlinemu,underlineSigma)$$



$$ = prod _i=1^n frac1^1/2e^-frac12(y_i -mu)'Sigma ^-1(y_i -mu) $$



Taking log:



$$= -np logbiggl(sqrt2pibiggl)-fracn2log(Sigma)-frac12sum_i=1^n(y_i -mu)'Sigma ^-1(y_i -mu)$$



Maximising:



$$ 0 = frac-12 times 2 sum^n_i=1 (y_i -mu)$$



Then, I do not know how to proceed.
I presume



$$ nmu =sum^n_i=1 y_i $$







probability multivariable-calculus normal-distribution






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edited Sep 3 at 18:32

























asked Sep 3 at 10:45









Tos Hina

1,027518




1,027518







  • 1




    After $116$ questions and $65$ answers, I would thought you'd have had enough opportunity to familiarize yourself with how we typeset math on this site. Please see this tutorial and reference. In particular, double dollar signs yield displayed equations, which are much easier to read, especially when they heavily mix fractions, subscripts, superscripts and radicals as in your case; and operator names should be in Roman font, using e.g. log, or operatornamename for operators that don't have a command of their own.
    – joriki
    Sep 3 at 11:47






  • 1




    Advice is well received, and editing done also. Sometimes a bit of sarcasm can change so much. :P
    – Tos Hina
    Sep 3 at 18:34






  • 2




    It looks like the last statement immediately follows from the previous. What are you looking for?
    – herb steinberg
    Sep 3 at 18:37







  • 2




    It would help if you explicitly defined $hatmu$ and $bary$.
    – herb steinberg
    Sep 3 at 19:25












  • 1




    After $116$ questions and $65$ answers, I would thought you'd have had enough opportunity to familiarize yourself with how we typeset math on this site. Please see this tutorial and reference. In particular, double dollar signs yield displayed equations, which are much easier to read, especially when they heavily mix fractions, subscripts, superscripts and radicals as in your case; and operator names should be in Roman font, using e.g. log, or operatornamename for operators that don't have a command of their own.
    – joriki
    Sep 3 at 11:47






  • 1




    Advice is well received, and editing done also. Sometimes a bit of sarcasm can change so much. :P
    – Tos Hina
    Sep 3 at 18:34






  • 2




    It looks like the last statement immediately follows from the previous. What are you looking for?
    – herb steinberg
    Sep 3 at 18:37







  • 2




    It would help if you explicitly defined $hatmu$ and $bary$.
    – herb steinberg
    Sep 3 at 19:25







1




1




After $116$ questions and $65$ answers, I would thought you'd have had enough opportunity to familiarize yourself with how we typeset math on this site. Please see this tutorial and reference. In particular, double dollar signs yield displayed equations, which are much easier to read, especially when they heavily mix fractions, subscripts, superscripts and radicals as in your case; and operator names should be in Roman font, using e.g. log, or operatornamename for operators that don't have a command of their own.
– joriki
Sep 3 at 11:47




After $116$ questions and $65$ answers, I would thought you'd have had enough opportunity to familiarize yourself with how we typeset math on this site. Please see this tutorial and reference. In particular, double dollar signs yield displayed equations, which are much easier to read, especially when they heavily mix fractions, subscripts, superscripts and radicals as in your case; and operator names should be in Roman font, using e.g. log, or operatornamename for operators that don't have a command of their own.
– joriki
Sep 3 at 11:47




1




1




Advice is well received, and editing done also. Sometimes a bit of sarcasm can change so much. :P
– Tos Hina
Sep 3 at 18:34




Advice is well received, and editing done also. Sometimes a bit of sarcasm can change so much. :P
– Tos Hina
Sep 3 at 18:34




2




2




It looks like the last statement immediately follows from the previous. What are you looking for?
– herb steinberg
Sep 3 at 18:37





It looks like the last statement immediately follows from the previous. What are you looking for?
– herb steinberg
Sep 3 at 18:37





2




2




It would help if you explicitly defined $hatmu$ and $bary$.
– herb steinberg
Sep 3 at 19:25




It would help if you explicitly defined $hatmu$ and $bary$.
– herb steinberg
Sep 3 at 19:25










2 Answers
2






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1
down vote



accepted










You have done everything correct. You've reached
$$0 = frac-12 times 2 sum^n_i=1 (y_i -mu)$$
or more formally (since your maximum is attained at the MLE)
$$0 = frac-12 times 2 sum^n_i=1 (y_i - hatmu)$$
where $hatmu$ is your MLE. Then
$$0 = sum^n_i=1 (y_i - hatmu)$$
i.e.
$$0 = sum^n_i=1 y_i -sum^n_i=1 hatmu$$
But $hatmu$ is a constant being summed up $n$ times, so
$$0 = sum^n_i=1 y_i - n hatmu$$
Finally,
$$hatmu = frac1n sum^n_i=1 y_i = bary$$
is the MLE estimate, i.e. the sample mean of the data is the MLE estimate of $mu$.






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    up vote
    1
    down vote













    Recall that for a vector $x in mathbbR ^ p$, and real symmetric matrix $A$,
    $$
    fracpartialpartial x x ^ T A x = 2 A x,
    $$
    hence,
    $$
    fracpartialpartial mu ln L ( cdot ; mu, Sigma )
    =
    sum frac12fracpartialpartial mu(y_i-mu)^TSigma^-1 (y_i-mu) = - Sigma^-1 sum_i=1^n(y_i-mu) = 0,
    $$
    multiplying by $Sigma$ and re-arranging the equation you get
    $$
    hatmu = frac1nsum_i=1^n y_i.
    $$
    Don't forget that $y_i$ are vectors, hence $hatmu$ is $p times 1$ vector of sample means.






    share|cite|improve this answer




















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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      1
      down vote



      accepted










      You have done everything correct. You've reached
      $$0 = frac-12 times 2 sum^n_i=1 (y_i -mu)$$
      or more formally (since your maximum is attained at the MLE)
      $$0 = frac-12 times 2 sum^n_i=1 (y_i - hatmu)$$
      where $hatmu$ is your MLE. Then
      $$0 = sum^n_i=1 (y_i - hatmu)$$
      i.e.
      $$0 = sum^n_i=1 y_i -sum^n_i=1 hatmu$$
      But $hatmu$ is a constant being summed up $n$ times, so
      $$0 = sum^n_i=1 y_i - n hatmu$$
      Finally,
      $$hatmu = frac1n sum^n_i=1 y_i = bary$$
      is the MLE estimate, i.e. the sample mean of the data is the MLE estimate of $mu$.






      share|cite|improve this answer
























        up vote
        1
        down vote



        accepted










        You have done everything correct. You've reached
        $$0 = frac-12 times 2 sum^n_i=1 (y_i -mu)$$
        or more formally (since your maximum is attained at the MLE)
        $$0 = frac-12 times 2 sum^n_i=1 (y_i - hatmu)$$
        where $hatmu$ is your MLE. Then
        $$0 = sum^n_i=1 (y_i - hatmu)$$
        i.e.
        $$0 = sum^n_i=1 y_i -sum^n_i=1 hatmu$$
        But $hatmu$ is a constant being summed up $n$ times, so
        $$0 = sum^n_i=1 y_i - n hatmu$$
        Finally,
        $$hatmu = frac1n sum^n_i=1 y_i = bary$$
        is the MLE estimate, i.e. the sample mean of the data is the MLE estimate of $mu$.






        share|cite|improve this answer






















          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          You have done everything correct. You've reached
          $$0 = frac-12 times 2 sum^n_i=1 (y_i -mu)$$
          or more formally (since your maximum is attained at the MLE)
          $$0 = frac-12 times 2 sum^n_i=1 (y_i - hatmu)$$
          where $hatmu$ is your MLE. Then
          $$0 = sum^n_i=1 (y_i - hatmu)$$
          i.e.
          $$0 = sum^n_i=1 y_i -sum^n_i=1 hatmu$$
          But $hatmu$ is a constant being summed up $n$ times, so
          $$0 = sum^n_i=1 y_i - n hatmu$$
          Finally,
          $$hatmu = frac1n sum^n_i=1 y_i = bary$$
          is the MLE estimate, i.e. the sample mean of the data is the MLE estimate of $mu$.






          share|cite|improve this answer












          You have done everything correct. You've reached
          $$0 = frac-12 times 2 sum^n_i=1 (y_i -mu)$$
          or more formally (since your maximum is attained at the MLE)
          $$0 = frac-12 times 2 sum^n_i=1 (y_i - hatmu)$$
          where $hatmu$ is your MLE. Then
          $$0 = sum^n_i=1 (y_i - hatmu)$$
          i.e.
          $$0 = sum^n_i=1 y_i -sum^n_i=1 hatmu$$
          But $hatmu$ is a constant being summed up $n$ times, so
          $$0 = sum^n_i=1 y_i - n hatmu$$
          Finally,
          $$hatmu = frac1n sum^n_i=1 y_i = bary$$
          is the MLE estimate, i.e. the sample mean of the data is the MLE estimate of $mu$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Sep 3 at 19:52









          Ahmad Bazzi

          5,7791623




          5,7791623




















              up vote
              1
              down vote













              Recall that for a vector $x in mathbbR ^ p$, and real symmetric matrix $A$,
              $$
              fracpartialpartial x x ^ T A x = 2 A x,
              $$
              hence,
              $$
              fracpartialpartial mu ln L ( cdot ; mu, Sigma )
              =
              sum frac12fracpartialpartial mu(y_i-mu)^TSigma^-1 (y_i-mu) = - Sigma^-1 sum_i=1^n(y_i-mu) = 0,
              $$
              multiplying by $Sigma$ and re-arranging the equation you get
              $$
              hatmu = frac1nsum_i=1^n y_i.
              $$
              Don't forget that $y_i$ are vectors, hence $hatmu$ is $p times 1$ vector of sample means.






              share|cite|improve this answer
























                up vote
                1
                down vote













                Recall that for a vector $x in mathbbR ^ p$, and real symmetric matrix $A$,
                $$
                fracpartialpartial x x ^ T A x = 2 A x,
                $$
                hence,
                $$
                fracpartialpartial mu ln L ( cdot ; mu, Sigma )
                =
                sum frac12fracpartialpartial mu(y_i-mu)^TSigma^-1 (y_i-mu) = - Sigma^-1 sum_i=1^n(y_i-mu) = 0,
                $$
                multiplying by $Sigma$ and re-arranging the equation you get
                $$
                hatmu = frac1nsum_i=1^n y_i.
                $$
                Don't forget that $y_i$ are vectors, hence $hatmu$ is $p times 1$ vector of sample means.






                share|cite|improve this answer






















                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  Recall that for a vector $x in mathbbR ^ p$, and real symmetric matrix $A$,
                  $$
                  fracpartialpartial x x ^ T A x = 2 A x,
                  $$
                  hence,
                  $$
                  fracpartialpartial mu ln L ( cdot ; mu, Sigma )
                  =
                  sum frac12fracpartialpartial mu(y_i-mu)^TSigma^-1 (y_i-mu) = - Sigma^-1 sum_i=1^n(y_i-mu) = 0,
                  $$
                  multiplying by $Sigma$ and re-arranging the equation you get
                  $$
                  hatmu = frac1nsum_i=1^n y_i.
                  $$
                  Don't forget that $y_i$ are vectors, hence $hatmu$ is $p times 1$ vector of sample means.






                  share|cite|improve this answer












                  Recall that for a vector $x in mathbbR ^ p$, and real symmetric matrix $A$,
                  $$
                  fracpartialpartial x x ^ T A x = 2 A x,
                  $$
                  hence,
                  $$
                  fracpartialpartial mu ln L ( cdot ; mu, Sigma )
                  =
                  sum frac12fracpartialpartial mu(y_i-mu)^TSigma^-1 (y_i-mu) = - Sigma^-1 sum_i=1^n(y_i-mu) = 0,
                  $$
                  multiplying by $Sigma$ and re-arranging the equation you get
                  $$
                  hatmu = frac1nsum_i=1^n y_i.
                  $$
                  Don't forget that $y_i$ are vectors, hence $hatmu$ is $p times 1$ vector of sample means.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Sep 3 at 19:45









                  V. Vancak

                  10.1k2926




                  10.1k2926



























                       

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