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If $E(Ymid X)=a+bX$, show that $b =fracmathrmCov(X,Y)mathrmVar(X)$ without assuming $(X,Y)$ has bivariate normal distribution?



The Next CEO of Stack OverflowProving $a$ (in $Y = aX + b + e$) satisfies $a = Cov(X, Y )/Var(X)$Prove that (X,Y) is bivariate normal if X is normal and Y conditionally on X is normal$P(X>0,Y>0)$ for a bivariate normal distribution with correlation $rho$Symmetric property for bivariate normal distributionBivariate normal distribution questionBivariate distribution with normal conditionsDistribution Theory - bivariate normal distributionQuestion on bivariate normal distributionDo we really need finite variances $mathrmVar(X)$ and $mathrmVar(Y)$ in the definition of covariance $mathrmCov(X,Y)$?Non-normal Bivariate distribution with normal marginsCorrelation coefficient of a bivariate normal distribution










1












$begingroup$



If $E(Ymid X)=a+bX$, show that $b =fracmathrmCov(X,Y)mathrmVar(X)$ where $a$ and $b$ are constants.




This question was asked before:



Proving $a$ (in $Y = aX + b + e$) satisfies $a = Cov(X, Y )/Var(X)$



But can someone suggest an answer without assuming bivariate normal distribution?






probability-theory probability-distributions conditional-probability







share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    It comes down to justifying $operatornamecov(X,,Y)=operatornamecov(X,,a+bX)$.
    $endgroup$
    – J.G.
    Mar 18 at 10:49










  • $begingroup$
    $X$ and $Y$ should be square integrable, this is missing from the statement
    $endgroup$
    – zhoraster
    Mar 18 at 10:55










  • $begingroup$
    Law of total covariance should do then.
    $endgroup$
    – Mann
    Mar 18 at 11:07















1












$begingroup$



If $E(Ymid X)=a+bX$, show that $b =fracmathrmCov(X,Y)mathrmVar(X)$ where $a$ and $b$ are constants.




This question was asked before:



Proving $a$ (in $Y = aX + b + e$) satisfies $a = Cov(X, Y )/Var(X)$



But can someone suggest an answer without assuming bivariate normal distribution?






probability-theory probability-distributions conditional-probability







share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    It comes down to justifying $operatornamecov(X,,Y)=operatornamecov(X,,a+bX)$.
    $endgroup$
    – J.G.
    Mar 18 at 10:49










  • $begingroup$
    $X$ and $Y$ should be square integrable, this is missing from the statement
    $endgroup$
    – zhoraster
    Mar 18 at 10:55










  • $begingroup$
    Law of total covariance should do then.
    $endgroup$
    – Mann
    Mar 18 at 11:07













1












1








1


0



$begingroup$



If $E(Ymid X)=a+bX$, show that $b =fracmathrmCov(X,Y)mathrmVar(X)$ where $a$ and $b$ are constants.




This question was asked before:



Proving $a$ (in $Y = aX + b + e$) satisfies $a = Cov(X, Y )/Var(X)$



But can someone suggest an answer without assuming bivariate normal distribution?






probability-theory probability-distributions conditional-probability







share|cite|improve this question











$endgroup$





If $E(Ymid X)=a+bX$, show that $b =fracmathrmCov(X,Y)mathrmVar(X)$ where $a$ and $b$ are constants.




This question was asked before:



Proving $a$ (in $Y = aX + b + e$) satisfies $a = Cov(X, Y )/Var(X)$



But can someone suggest an answer without assuming bivariate normal distribution?






probability-theory probability-distributions conditional-probability




probability-theory probability-distributions conditional-probability






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 18 at 22:02









Jack

27.6k1782203




27.6k1782203










asked Mar 18 at 10:33









Sri Krishna SahooSri Krishna Sahoo

620217




620217







  • 1




    $begingroup$
    It comes down to justifying $operatornamecov(X,,Y)=operatornamecov(X,,a+bX)$.
    $endgroup$
    – J.G.
    Mar 18 at 10:49










  • $begingroup$
    $X$ and $Y$ should be square integrable, this is missing from the statement
    $endgroup$
    – zhoraster
    Mar 18 at 10:55










  • $begingroup$
    Law of total covariance should do then.
    $endgroup$
    – Mann
    Mar 18 at 11:07












  • 1




    $begingroup$
    It comes down to justifying $operatornamecov(X,,Y)=operatornamecov(X,,a+bX)$.
    $endgroup$
    – J.G.
    Mar 18 at 10:49










  • $begingroup$
    $X$ and $Y$ should be square integrable, this is missing from the statement
    $endgroup$
    – zhoraster
    Mar 18 at 10:55










  • $begingroup$
    Law of total covariance should do then.
    $endgroup$
    – Mann
    Mar 18 at 11:07







1




1




$begingroup$
It comes down to justifying $operatornamecov(X,,Y)=operatornamecov(X,,a+bX)$.
$endgroup$
– J.G.
Mar 18 at 10:49




$begingroup$
It comes down to justifying $operatornamecov(X,,Y)=operatornamecov(X,,a+bX)$.
$endgroup$
– J.G.
Mar 18 at 10:49












$begingroup$
$X$ and $Y$ should be square integrable, this is missing from the statement
$endgroup$
– zhoraster
Mar 18 at 10:55




$begingroup$
$X$ and $Y$ should be square integrable, this is missing from the statement
$endgroup$
– zhoraster
Mar 18 at 10:55












$begingroup$
Law of total covariance should do then.
$endgroup$
– Mann
Mar 18 at 11:07




$begingroup$
Law of total covariance should do then.
$endgroup$
– Mann
Mar 18 at 11:07










2 Answers
2






active

oldest

votes


















6












$begingroup$

Observe that: $$mathbbEY=mathbbEleft[mathbbEleft[Ymid Xright]right]=mathbbEleft[a+bXright]=a+bmathbbEX$$
so that:



$$mathsfCovleft(X,Yright)=mathbbEleft[left(X-mathbbEXright)left(Y-mathbbEYright)right]=mathbbEleft[mathbbEleft[left(X-mathbbEXright)left(Y-mathbbEYright)mid Xright]right]=$$$$mathbbEleft[left(X-mathbbEXright)mathbbEleft[left(Y-mathbbEYright)mid Xright]right]=mathbbEleft[left(X-mathbbEXright)left(a+bX-mathbbEYright)right]=bmathbbEleft(X-mathbbEXright)^2=bmathsfVarX$$






share|cite|improve this answer









$endgroup$




















    0












    $begingroup$

    Least squares method. Consider the squared residual that must be minimized:
    $$R^2=sum_i=1^n (y_i-ax_i-b)^2;\
    begincases(R^2)_a=-2sum_i=1^n(y_i-ax_i-b)x_i=0\
    (R^2)_b=-sum_i=1^n (y_i-ax_i-b)=0 endcases Rightarrow \
    begincasescolorredsum_i=1^n x_i^2cdot a+colorbluesum_i=1^n x_icdot b=
    colorgreensum_i=1^n x_iy_i\
    colorredsum_i=1^n x_icdot a+qquad colorbluencdot b=colorgreensum_i=1^n y_iendcases qquad stackrelCramerRightarrow \
    a=fracbeginvmatrixcolorgreensum_i=1^n x_iy_i&colorbluesum_i=1^n x_i\ colorgreensum_i=1^n y_i&colorbluenendvmatrixbeginvmatrixcolorredsum_i=1^n x_i^2&colorbluesum_i=1^n x_i\ colorredsum_i=1^n x_i&colorbluenendvmatrix=
    fracnsum_i=1^n x_iy_i-sum_i=1^n x_icdot sum_i=1^n y_insum_i=1^n x_i^2-(sum_i=1^n x_i)^2=\
    fracfracsum_i=1^n x_iy_in-fracsum_i=1^n x_incdot fracsum_i=1^n y_infracsum_i=1^n x_i^2n-left(fracsum_i=1^n x_inright)^2=\
    fracCov(X,Y)Var(X).$$






    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









      6












      $begingroup$

      Observe that: $$mathbbEY=mathbbEleft[mathbbEleft[Ymid Xright]right]=mathbbEleft[a+bXright]=a+bmathbbEX$$
      so that:



      $$mathsfCovleft(X,Yright)=mathbbEleft[left(X-mathbbEXright)left(Y-mathbbEYright)right]=mathbbEleft[mathbbEleft[left(X-mathbbEXright)left(Y-mathbbEYright)mid Xright]right]=$$$$mathbbEleft[left(X-mathbbEXright)mathbbEleft[left(Y-mathbbEYright)mid Xright]right]=mathbbEleft[left(X-mathbbEXright)left(a+bX-mathbbEYright)right]=bmathbbEleft(X-mathbbEXright)^2=bmathsfVarX$$






      share|cite|improve this answer









      $endgroup$

















        6












        $begingroup$

        Observe that: $$mathbbEY=mathbbEleft[mathbbEleft[Ymid Xright]right]=mathbbEleft[a+bXright]=a+bmathbbEX$$
        so that:



        $$mathsfCovleft(X,Yright)=mathbbEleft[left(X-mathbbEXright)left(Y-mathbbEYright)right]=mathbbEleft[mathbbEleft[left(X-mathbbEXright)left(Y-mathbbEYright)mid Xright]right]=$$$$mathbbEleft[left(X-mathbbEXright)mathbbEleft[left(Y-mathbbEYright)mid Xright]right]=mathbbEleft[left(X-mathbbEXright)left(a+bX-mathbbEYright)right]=bmathbbEleft(X-mathbbEXright)^2=bmathsfVarX$$






        share|cite|improve this answer









        $endgroup$















          6












          6








          6





          $begingroup$

          Observe that: $$mathbbEY=mathbbEleft[mathbbEleft[Ymid Xright]right]=mathbbEleft[a+bXright]=a+bmathbbEX$$
          so that:



          $$mathsfCovleft(X,Yright)=mathbbEleft[left(X-mathbbEXright)left(Y-mathbbEYright)right]=mathbbEleft[mathbbEleft[left(X-mathbbEXright)left(Y-mathbbEYright)mid Xright]right]=$$$$mathbbEleft[left(X-mathbbEXright)mathbbEleft[left(Y-mathbbEYright)mid Xright]right]=mathbbEleft[left(X-mathbbEXright)left(a+bX-mathbbEYright)right]=bmathbbEleft(X-mathbbEXright)^2=bmathsfVarX$$






          share|cite|improve this answer









          $endgroup$



          Observe that: $$mathbbEY=mathbbEleft[mathbbEleft[Ymid Xright]right]=mathbbEleft[a+bXright]=a+bmathbbEX$$
          so that:



          $$mathsfCovleft(X,Yright)=mathbbEleft[left(X-mathbbEXright)left(Y-mathbbEYright)right]=mathbbEleft[mathbbEleft[left(X-mathbbEXright)left(Y-mathbbEYright)mid Xright]right]=$$$$mathbbEleft[left(X-mathbbEXright)mathbbEleft[left(Y-mathbbEYright)mid Xright]right]=mathbbEleft[left(X-mathbbEXright)left(a+bX-mathbbEYright)right]=bmathbbEleft(X-mathbbEXright)^2=bmathsfVarX$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Mar 18 at 11:05









          drhabdrhab

          104k545136




          104k545136





















              0












              $begingroup$

              Least squares method. Consider the squared residual that must be minimized:
              $$R^2=sum_i=1^n (y_i-ax_i-b)^2;\
              begincases(R^2)_a=-2sum_i=1^n(y_i-ax_i-b)x_i=0\
              (R^2)_b=-sum_i=1^n (y_i-ax_i-b)=0 endcases Rightarrow \
              begincasescolorredsum_i=1^n x_i^2cdot a+colorbluesum_i=1^n x_icdot b=
              colorgreensum_i=1^n x_iy_i\
              colorredsum_i=1^n x_icdot a+qquad colorbluencdot b=colorgreensum_i=1^n y_iendcases qquad stackrelCramerRightarrow \
              a=fracbeginvmatrixcolorgreensum_i=1^n x_iy_i&colorbluesum_i=1^n x_i\ colorgreensum_i=1^n y_i&colorbluenendvmatrixbeginvmatrixcolorredsum_i=1^n x_i^2&colorbluesum_i=1^n x_i\ colorredsum_i=1^n x_i&colorbluenendvmatrix=
              fracnsum_i=1^n x_iy_i-sum_i=1^n x_icdot sum_i=1^n y_insum_i=1^n x_i^2-(sum_i=1^n x_i)^2=\
              fracfracsum_i=1^n x_iy_in-fracsum_i=1^n x_incdot fracsum_i=1^n y_infracsum_i=1^n x_i^2n-left(fracsum_i=1^n x_inright)^2=\
              fracCov(X,Y)Var(X).$$






              share|cite|improve this answer









              $endgroup$

















                0












                $begingroup$

                Least squares method. Consider the squared residual that must be minimized:
                $$R^2=sum_i=1^n (y_i-ax_i-b)^2;\
                begincases(R^2)_a=-2sum_i=1^n(y_i-ax_i-b)x_i=0\
                (R^2)_b=-sum_i=1^n (y_i-ax_i-b)=0 endcases Rightarrow \
                begincasescolorredsum_i=1^n x_i^2cdot a+colorbluesum_i=1^n x_icdot b=
                colorgreensum_i=1^n x_iy_i\
                colorredsum_i=1^n x_icdot a+qquad colorbluencdot b=colorgreensum_i=1^n y_iendcases qquad stackrelCramerRightarrow \
                a=fracbeginvmatrixcolorgreensum_i=1^n x_iy_i&colorbluesum_i=1^n x_i\ colorgreensum_i=1^n y_i&colorbluenendvmatrixbeginvmatrixcolorredsum_i=1^n x_i^2&colorbluesum_i=1^n x_i\ colorredsum_i=1^n x_i&colorbluenendvmatrix=
                fracnsum_i=1^n x_iy_i-sum_i=1^n x_icdot sum_i=1^n y_insum_i=1^n x_i^2-(sum_i=1^n x_i)^2=\
                fracfracsum_i=1^n x_iy_in-fracsum_i=1^n x_incdot fracsum_i=1^n y_infracsum_i=1^n x_i^2n-left(fracsum_i=1^n x_inright)^2=\
                fracCov(X,Y)Var(X).$$






                share|cite|improve this answer









                $endgroup$















                  0












                  0








                  0





                  $begingroup$

                  Least squares method. Consider the squared residual that must be minimized:
                  $$R^2=sum_i=1^n (y_i-ax_i-b)^2;\
                  begincases(R^2)_a=-2sum_i=1^n(y_i-ax_i-b)x_i=0\
                  (R^2)_b=-sum_i=1^n (y_i-ax_i-b)=0 endcases Rightarrow \
                  begincasescolorredsum_i=1^n x_i^2cdot a+colorbluesum_i=1^n x_icdot b=
                  colorgreensum_i=1^n x_iy_i\
                  colorredsum_i=1^n x_icdot a+qquad colorbluencdot b=colorgreensum_i=1^n y_iendcases qquad stackrelCramerRightarrow \
                  a=fracbeginvmatrixcolorgreensum_i=1^n x_iy_i&colorbluesum_i=1^n x_i\ colorgreensum_i=1^n y_i&colorbluenendvmatrixbeginvmatrixcolorredsum_i=1^n x_i^2&colorbluesum_i=1^n x_i\ colorredsum_i=1^n x_i&colorbluenendvmatrix=
                  fracnsum_i=1^n x_iy_i-sum_i=1^n x_icdot sum_i=1^n y_insum_i=1^n x_i^2-(sum_i=1^n x_i)^2=\
                  fracfracsum_i=1^n x_iy_in-fracsum_i=1^n x_incdot fracsum_i=1^n y_infracsum_i=1^n x_i^2n-left(fracsum_i=1^n x_inright)^2=\
                  fracCov(X,Y)Var(X).$$






                  share|cite|improve this answer









                  $endgroup$



                  Least squares method. Consider the squared residual that must be minimized:
                  $$R^2=sum_i=1^n (y_i-ax_i-b)^2;\
                  begincases(R^2)_a=-2sum_i=1^n(y_i-ax_i-b)x_i=0\
                  (R^2)_b=-sum_i=1^n (y_i-ax_i-b)=0 endcases Rightarrow \
                  begincasescolorredsum_i=1^n x_i^2cdot a+colorbluesum_i=1^n x_icdot b=
                  colorgreensum_i=1^n x_iy_i\
                  colorredsum_i=1^n x_icdot a+qquad colorbluencdot b=colorgreensum_i=1^n y_iendcases qquad stackrelCramerRightarrow \
                  a=fracbeginvmatrixcolorgreensum_i=1^n x_iy_i&colorbluesum_i=1^n x_i\ colorgreensum_i=1^n y_i&colorbluenendvmatrixbeginvmatrixcolorredsum_i=1^n x_i^2&colorbluesum_i=1^n x_i\ colorredsum_i=1^n x_i&colorbluenendvmatrix=
                  fracnsum_i=1^n x_iy_i-sum_i=1^n x_icdot sum_i=1^n y_insum_i=1^n x_i^2-(sum_i=1^n x_i)^2=\
                  fracfracsum_i=1^n x_iy_in-fracsum_i=1^n x_incdot fracsum_i=1^n y_infracsum_i=1^n x_i^2n-left(fracsum_i=1^n x_inright)^2=\
                  fracCov(X,Y)Var(X).$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Mar 18 at 13:04









                  farruhotafarruhota

                  21.6k2842




                  21.6k2842



























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                      FrenchArabicFrenchAir TamajeqChadian ArabicDazaKaadoTamahaqStandardArabicEnglishDazaDinkaDoniFurKadugliKangaKeigaKrongoTemeinTeseTulishiTumtumZaghawaAcheronAmiraDengebuEbangJomangKatlaKoKwalibLaroLogolLumunMoroNdingNgileShirumbaTagoiTegaliTegemTimaTiroTochoToronaUtoroWarnangBejaDomariSudanese ArabicSudanese Sign LanguagesBertiZaghawa Tebu languagesSaharan languagesLanguages of ChadLanguages of NigerToubou people Nilo-Saharan languageDaza peopleChadGouranDjurab DesertBorkouFaya-LargeauTibesti MountainsChadNigerN'guigmiLibyaSudanOmdurmanJeddahSaudi ArabiaTedaga languageTedaToubou peopleSabhaSaharanKanuri languageKanembu languageTebu languagesZaghawa languageBerti languageFrenchArabicliterary languagebilingualHausaZarmaTuaregromanisation Not to be confused with Dazawa language. "Goran language" redirects here. It is not to be confused with Gorani language (Zaza-Gorani) or Gora dialect. Daza Dazaga Native to Chad, Niger, Sudan, Libya Region Chad, BET Region, bahr el Gazel R...
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