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
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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
$begingroup$
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
edited Mar 18 at 22:02
Jack
27.6k1782203
asked Mar 18 at 10:33
Sri Krishna SahooSri Krishna Sahoo
620217
$endgroup$
add a comment |
$begingroup$
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
edited Mar 18 at 22:02
Jack
27.6k1782203
asked Mar 18 at 10:33
Sri Krishna SahooSri Krishna Sahoo
620217
$endgroup$
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
add a comment |
$begingroup$
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
edited Mar 18 at 22:02
Jack
27.6k1782203
asked Mar 18 at 10:33
Sri Krishna SahooSri Krishna Sahoo
620217
$endgroup$
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
edited Mar 18 at 22:02
Jack
27.6k1782203
asked Mar 18 at 10:33
Sri Krishna SahooSri Krishna Sahoo
620217
edited Mar 18 at 22:02
Jack
27.6k1782203
asked Mar 18 at 10:33
Sri Krishna SahooSri Krishna Sahoo
620217
edited Mar 18 at 22:02
Jack
27.6k1782203
edited Mar 18 at 22:02
Jack
27.6k1782203
edited Mar 18 at 22:02
Jack
27.6k1782203
27.6k1782203
asked Mar 18 at 10:33
Sri Krishna SahooSri Krishna Sahoo
620217
asked Mar 18 at 10:33
Sri Krishna SahooSri Krishna Sahoo
620217
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
add a comment |
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Observe that: $$mathbbEY=mathbbEleft[mathbbEleft[Ymid Xright]right]=mathbbEleft[a+bXright]=a+bmathbbEX$$
so that:
$$mathsfCovleft(X,Yright)=mathbbEleft[left(X-mathbbEXright)left(Y-mathbbEYright)right]=mathbbEleft[mathbbEleft[left(X-mathbbEXright)left(Y-mathbbEYright)mid Xright]right]=$$$$mathbbEleft[left(X-mathbbEXright)mathbbEleft[left(Y-mathbbEYright)mid Xright]right]=mathbbEleft[left(X-mathbbEXright)left(a+bX-mathbbEYright)right]=bmathbbEleft(X-mathbbEXright)^2=bmathsfVarX$$
answered Mar 18 at 11:05
drhabdrhab
104k545136
$endgroup$
add a comment |
$begingroup$
Least squares method. Consider the squared residual that must be minimized:
$$R^2=sum_i=1^n (y_i-ax_i-b)^2;\
begincases(R^2)_a=-2sum_i=1^n(y_i-ax_i-b)x_i=0\
(R^2)_b=-sum_i=1^n (y_i-ax_i-b)=0 endcases Rightarrow \
begincasescolorredsum_i=1^n x_i^2cdot a+colorbluesum_i=1^n x_icdot b=
colorgreensum_i=1^n x_iy_i\
colorredsum_i=1^n x_icdot a+qquad colorbluencdot b=colorgreensum_i=1^n y_iendcases qquad stackrelCramerRightarrow \
a=fracbeginvmatrixcolorgreensum_i=1^n x_iy_i&colorbluesum_i=1^n x_i\ colorgreensum_i=1^n y_i&colorbluenendvmatrixbeginvmatrixcolorredsum_i=1^n x_i^2&colorbluesum_i=1^n x_i\ colorredsum_i=1^n x_i&colorbluenendvmatrix=
fracnsum_i=1^n x_iy_i-sum_i=1^n x_icdot sum_i=1^n y_insum_i=1^n x_i^2-(sum_i=1^n x_i)^2=\
fracfracsum_i=1^n x_iy_in-fracsum_i=1^n x_incdot fracsum_i=1^n y_infracsum_i=1^n x_i^2n-left(fracsum_i=1^n x_inright)^2=\
fracCov(X,Y)Var(X).$$
answered Mar 18 at 13:04
farruhotafarruhota
21.7k2842
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Observe that: $$mathbbEY=mathbbEleft[mathbbEleft[Ymid Xright]right]=mathbbEleft[a+bXright]=a+bmathbbEX$$
so that:
$$mathsfCovleft(X,Yright)=mathbbEleft[left(X-mathbbEXright)left(Y-mathbbEYright)right]=mathbbEleft[mathbbEleft[left(X-mathbbEXright)left(Y-mathbbEYright)mid Xright]right]=$$$$mathbbEleft[left(X-mathbbEXright)mathbbEleft[left(Y-mathbbEYright)mid Xright]right]=mathbbEleft[left(X-mathbbEXright)left(a+bX-mathbbEYright)right]=bmathbbEleft(X-mathbbEXright)^2=bmathsfVarX$$
answered Mar 18 at 11:05
drhabdrhab
104k545136
$endgroup$
add a comment |
$begingroup$
Observe that: $$mathbbEY=mathbbEleft[mathbbEleft[Ymid Xright]right]=mathbbEleft[a+bXright]=a+bmathbbEX$$
so that:
$$mathsfCovleft(X,Yright)=mathbbEleft[left(X-mathbbEXright)left(Y-mathbbEYright)right]=mathbbEleft[mathbbEleft[left(X-mathbbEXright)left(Y-mathbbEYright)mid Xright]right]=$$$$mathbbEleft[left(X-mathbbEXright)mathbbEleft[left(Y-mathbbEYright)mid Xright]right]=mathbbEleft[left(X-mathbbEXright)left(a+bX-mathbbEYright)right]=bmathbbEleft(X-mathbbEXright)^2=bmathsfVarX$$
answered Mar 18 at 11:05
drhabdrhab
104k545136
$endgroup$
add a comment |
$begingroup$
Observe that: $$mathbbEY=mathbbEleft[mathbbEleft[Ymid Xright]right]=mathbbEleft[a+bXright]=a+bmathbbEX$$
so that:
$$mathsfCovleft(X,Yright)=mathbbEleft[left(X-mathbbEXright)left(Y-mathbbEYright)right]=mathbbEleft[mathbbEleft[left(X-mathbbEXright)left(Y-mathbbEYright)mid Xright]right]=$$$$mathbbEleft[left(X-mathbbEXright)mathbbEleft[left(Y-mathbbEYright)mid Xright]right]=mathbbEleft[left(X-mathbbEXright)left(a+bX-mathbbEYright)right]=bmathbbEleft(X-mathbbEXright)^2=bmathsfVarX$$
answered Mar 18 at 11:05
drhabdrhab
104k545136
$endgroup$
Observe that: $$mathbbEY=mathbbEleft[mathbbEleft[Ymid Xright]right]=mathbbEleft[a+bXright]=a+bmathbbEX$$
so that:
$$mathsfCovleft(X,Yright)=mathbbEleft[left(X-mathbbEXright)left(Y-mathbbEYright)right]=mathbbEleft[mathbbEleft[left(X-mathbbEXright)left(Y-mathbbEYright)mid Xright]right]=$$$$mathbbEleft[left(X-mathbbEXright)mathbbEleft[left(Y-mathbbEYright)mid Xright]right]=mathbbEleft[left(X-mathbbEXright)left(a+bX-mathbbEYright)right]=bmathbbEleft(X-mathbbEXright)^2=bmathsfVarX$$
answered Mar 18 at 11:05
drhabdrhab
104k545136
answered Mar 18 at 11:05
drhabdrhab
104k545136
answered Mar 18 at 11:05
drhabdrhab
104k545136
answered Mar 18 at 11:05
drhabdrhab
104k545136
104k545136
add a comment |
add a comment |
$begingroup$
Least squares method. Consider the squared residual that must be minimized:
$$R^2=sum_i=1^n (y_i-ax_i-b)^2;\
begincases(R^2)_a=-2sum_i=1^n(y_i-ax_i-b)x_i=0\
(R^2)_b=-sum_i=1^n (y_i-ax_i-b)=0 endcases Rightarrow \
begincasescolorredsum_i=1^n x_i^2cdot a+colorbluesum_i=1^n x_icdot b=
colorgreensum_i=1^n x_iy_i\
colorredsum_i=1^n x_icdot a+qquad colorbluencdot b=colorgreensum_i=1^n y_iendcases qquad stackrelCramerRightarrow \
a=fracbeginvmatrixcolorgreensum_i=1^n x_iy_i&colorbluesum_i=1^n x_i\ colorgreensum_i=1^n y_i&colorbluenendvmatrixbeginvmatrixcolorredsum_i=1^n x_i^2&colorbluesum_i=1^n x_i\ colorredsum_i=1^n x_i&colorbluenendvmatrix=
fracnsum_i=1^n x_iy_i-sum_i=1^n x_icdot sum_i=1^n y_insum_i=1^n x_i^2-(sum_i=1^n x_i)^2=\
fracfracsum_i=1^n x_iy_in-fracsum_i=1^n x_incdot fracsum_i=1^n y_infracsum_i=1^n x_i^2n-left(fracsum_i=1^n x_inright)^2=\
fracCov(X,Y)Var(X).$$
answered Mar 18 at 13:04
farruhotafarruhota
21.7k2842
$endgroup$
add a comment |
$begingroup$
Least squares method. Consider the squared residual that must be minimized:
$$R^2=sum_i=1^n (y_i-ax_i-b)^2;\
begincases(R^2)_a=-2sum_i=1^n(y_i-ax_i-b)x_i=0\
(R^2)_b=-sum_i=1^n (y_i-ax_i-b)=0 endcases Rightarrow \
begincasescolorredsum_i=1^n x_i^2cdot a+colorbluesum_i=1^n x_icdot b=
colorgreensum_i=1^n x_iy_i\
colorredsum_i=1^n x_icdot a+qquad colorbluencdot b=colorgreensum_i=1^n y_iendcases qquad stackrelCramerRightarrow \
a=fracbeginvmatrixcolorgreensum_i=1^n x_iy_i&colorbluesum_i=1^n x_i\ colorgreensum_i=1^n y_i&colorbluenendvmatrixbeginvmatrixcolorredsum_i=1^n x_i^2&colorbluesum_i=1^n x_i\ colorredsum_i=1^n x_i&colorbluenendvmatrix=
fracnsum_i=1^n x_iy_i-sum_i=1^n x_icdot sum_i=1^n y_insum_i=1^n x_i^2-(sum_i=1^n x_i)^2=\
fracfracsum_i=1^n x_iy_in-fracsum_i=1^n x_incdot fracsum_i=1^n y_infracsum_i=1^n x_i^2n-left(fracsum_i=1^n x_inright)^2=\
fracCov(X,Y)Var(X).$$
answered Mar 18 at 13:04
farruhotafarruhota
21.7k2842
$endgroup$
add a comment |
$begingroup$
Least squares method. Consider the squared residual that must be minimized:
$$R^2=sum_i=1^n (y_i-ax_i-b)^2;\
begincases(R^2)_a=-2sum_i=1^n(y_i-ax_i-b)x_i=0\
(R^2)_b=-sum_i=1^n (y_i-ax_i-b)=0 endcases Rightarrow \
begincasescolorredsum_i=1^n x_i^2cdot a+colorbluesum_i=1^n x_icdot b=
colorgreensum_i=1^n x_iy_i\
colorredsum_i=1^n x_icdot a+qquad colorbluencdot b=colorgreensum_i=1^n y_iendcases qquad stackrelCramerRightarrow \
a=fracbeginvmatrixcolorgreensum_i=1^n x_iy_i&colorbluesum_i=1^n x_i\ colorgreensum_i=1^n y_i&colorbluenendvmatrixbeginvmatrixcolorredsum_i=1^n x_i^2&colorbluesum_i=1^n x_i\ colorredsum_i=1^n x_i&colorbluenendvmatrix=
fracnsum_i=1^n x_iy_i-sum_i=1^n x_icdot sum_i=1^n y_insum_i=1^n x_i^2-(sum_i=1^n x_i)^2=\
fracfracsum_i=1^n x_iy_in-fracsum_i=1^n x_incdot fracsum_i=1^n y_infracsum_i=1^n x_i^2n-left(fracsum_i=1^n x_inright)^2=\
fracCov(X,Y)Var(X).$$
answered Mar 18 at 13:04
farruhotafarruhota
21.7k2842
$endgroup$
Least squares method. Consider the squared residual that must be minimized:
$$R^2=sum_i=1^n (y_i-ax_i-b)^2;\
begincases(R^2)_a=-2sum_i=1^n(y_i-ax_i-b)x_i=0\
(R^2)_b=-sum_i=1^n (y_i-ax_i-b)=0 endcases Rightarrow \
begincasescolorredsum_i=1^n x_i^2cdot a+colorbluesum_i=1^n x_icdot b=
colorgreensum_i=1^n x_iy_i\
colorredsum_i=1^n x_icdot a+qquad colorbluencdot b=colorgreensum_i=1^n y_iendcases qquad stackrelCramerRightarrow \
a=fracbeginvmatrixcolorgreensum_i=1^n x_iy_i&colorbluesum_i=1^n x_i\ colorgreensum_i=1^n y_i&colorbluenendvmatrixbeginvmatrixcolorredsum_i=1^n x_i^2&colorbluesum_i=1^n x_i\ colorredsum_i=1^n x_i&colorbluenendvmatrix=
fracnsum_i=1^n x_iy_i-sum_i=1^n x_icdot sum_i=1^n y_insum_i=1^n x_i^2-(sum_i=1^n x_i)^2=\
fracfracsum_i=1^n x_iy_in-fracsum_i=1^n x_incdot fracsum_i=1^n y_infracsum_i=1^n x_i^2n-left(fracsum_i=1^n x_inright)^2=\
fracCov(X,Y)Var(X).$$
answered Mar 18 at 13:04
farruhotafarruhota
21.7k2842
answered Mar 18 at 13:04
farruhotafarruhota
21.7k2842
answered Mar 18 at 13:04
farruhotafarruhota
21.7k2842
answered Mar 18 at 13:04
farruhotafarruhota
21.7k2842
21.7k2842
add a comment |
add a comment |
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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