- #1
FallenApple
- 566
- 61
So ##\vec Y##~MVN(X##\vec\beta##, ##\sigma^2##I)
and
##\hat {\vec Y}##~MVN(X##\vec\beta##, ##\sigma^2##H)
and I want to show
##\hat{\vec e}##~MVN(##\vec 0##, ##\sigma^2##(I-H))
Where ##\hat{\vec e}## is the vector of observed residuals( ##\vec e=\vec Y- \hat{\vec Y}=(I-H) \vec Y ##).
And H=##X(X'X)^{-1}X'## is the projection matrix where the primed means transpose
Since ##\hat{\vec e}=\vec Y- \hat{\vec Y}## its just the distribution of ##\vec Y## minus the distribution of ##\hat{\vec Y}##
So MVN(X##\vec\beta##, ##\sigma^2##I)---MVN(X##\vec\beta##, ##\sigma^2##H) = MVN(##\vec 0##, ##\sigma^2##(I-H))
this uses the fact that X##\vec\beta##-X##\vec\beta##=##\vec 0## and ##\sigma^2##I-##\sigma^2##H=##\sigma^2##(I-H) by simple vector addition.
Can one subtract that simply? I mean, ##\vec Y## and ##\hat {\vec Y}## are not independent, so how can I just total up the expectations and the variances like that?
and
##\hat {\vec Y}##~MVN(X##\vec\beta##, ##\sigma^2##H)
and I want to show
##\hat{\vec e}##~MVN(##\vec 0##, ##\sigma^2##(I-H))
Where ##\hat{\vec e}## is the vector of observed residuals( ##\vec e=\vec Y- \hat{\vec Y}=(I-H) \vec Y ##).
And H=##X(X'X)^{-1}X'## is the projection matrix where the primed means transpose
Since ##\hat{\vec e}=\vec Y- \hat{\vec Y}## its just the distribution of ##\vec Y## minus the distribution of ##\hat{\vec Y}##
So MVN(X##\vec\beta##, ##\sigma^2##I)---MVN(X##\vec\beta##, ##\sigma^2##H) = MVN(##\vec 0##, ##\sigma^2##(I-H))
this uses the fact that X##\vec\beta##-X##\vec\beta##=##\vec 0## and ##\sigma^2##I-##\sigma^2##H=##\sigma^2##(I-H) by simple vector addition.
Can one subtract that simply? I mean, ##\vec Y## and ##\hat {\vec Y}## are not independent, so how can I just total up the expectations and the variances like that?
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