Let
and
be random variables with finite variance, and let
constant. The covariance of
and
is given through
Show that
Show that
(symmetry)
Show that
Show that
Show that
(linearity)
Does linearity also hold for the second component ?
Show that
Show that for
and
independent, it is
. (hint you can use that for
and
independent it holds
.)
--Friday Sa 30 Jan 2021 17:59:00 CET
![{\displaystyle {\begin{aligned}\mathbb {C} ov(X,Y)&=\mathbb {E} [(X-\mathbb {E} [X])(Y-\mathbb {E} [Y])]\\&=\mathbb {E} [XY]-\mathbb {E} [X\cdot \mathbb {E} [Y]]-\mathbb {E} [Y\cdot \mathbb {E} [X]]+\mathbb {E} [\mathbb {E} [X]\cdot \mathbb {E} [Y]]\\&=\mathbb {E} [XY]-\mathbb {E} [X]\cdot \mathbb {E} [Y]-\mathbb {E} [X]\cdot \mathbb {E} [Y]+\mathbb {E} [X]\cdot \mathbb {E} [Y]\\&=\mathbb {E} [XY]-\mathbb {E} [X]\cdot \mathbb {E} [Y]\quad \Box \end{aligned}}}](/index.php?title=Spezial:MathShowImage&hash=419579a988f0254b6c54190872ed9309&mode=mathml)
1b)[Bearbeiten | Quelltext bearbeiten]
![{\displaystyle \mathbb {C} ov(X,Y)=\mathbb {E} [(X-\mathbb {E} [X])(Y-\mathbb {E} [Y])]=\mathbb {E} [(Y-\mathbb {E} [Y])(X-\mathbb {E} [X])]=\mathbb {C} ov(Y,X)\quad \Box }](/index.php?title=Spezial:MathShowImage&hash=6e10afd670f3ba839c3384e85f6a8380&mode=mathml)
First, we remember the formular for the variance:
![{\displaystyle \mathbb {V} ar(X)=\mathbb {E} (X-\mathbb {E} (X))^{2}}](/index.php?title=Spezial:MathShowImage&hash=8fcac4b9c916904e464c88721bcc6813&mode=mathml)
Now, we can prove the equation:
![{\displaystyle {\begin{aligned}\mathbb {C} ov(X,X)&=\mathbb {E} [(X-\mathbb {E} [X])(X-\mathbb {E} [X])\\&=\mathbb {E} [(X-\mathbb {E} [X]))^{2}\\&=\mathbb {V} ar(X)\quad \Box \end{aligned}}}](/index.php?title=Spezial:MathShowImage&hash=af76360adc3c007ef1caeeab42602edf&mode=mathml)
First, we need to remember that the expection of a constant is itself:
![{\displaystyle \mathbb {E} [a]=a}](/index.php?title=Spezial:MathShowImage&hash=3e19168a9a4e3f76dcc46826c41bf5a9&mode=mathml)
Now we can just solve the equation
![{\displaystyle {\begin{aligned}\mathbb {C} ov(X,a)&=\mathbb {E} [(X-\mathbb {E} [X])(a-\mathbb {E} [a])]\\&=\mathbb {E} [(X-\mathbb {E} [X])(a-a)]\\&=\mathbb {E} [(X-\mathbb {E} [X])\cdot 0]\\&=0\quad \Box \\\end{aligned}}}](/index.php?title=Spezial:MathShowImage&hash=8b7754713e05c074e877772aa70aca7b&mode=mathml)
Of course, from 1b) we know that both components are interchangeable and therefore linearity must also hold for the second component.
![{\displaystyle {\begin{aligned}\mathbb {V} ar(X+Y)&=\mathbb {C} ov(X+Y,X+Y)\\&=\mathbb {C} ov(X+Y,X)+\mathbb {C} ov(X+Y,Y)\\&=\mathbb {C} ov(X,X)+\mathbb {C} ov(X,Y)+\mathbb {C} ov(Y,X)+\mathbb {C} ov(Y,Y)\\&=\mathbb {V} ar(X)+\mathbb {V} ar(Y)+2\cdot \mathbb {C} ov(X,Y)\quad \Box \\\end{aligned}}}](/index.php?title=Spezial:MathShowImage&hash=5fa209db9bdece8e358b9d07d3762d54&mode=mathml)
![{\displaystyle {\begin{aligned}\mathbb {C} ov(X,Y)&=\mathbb {E} [XY]-\mathbb {E} [X]\cdot \mathbb {E} [Y]\\&=\mathbb {E} [X]\cdot \mathbb {E} [Y]-\mathbb {E} [X]\cdot \mathbb {E} [Y]\\&=0\quad \Box \end{aligned}}}](/index.php?title=Spezial:MathShowImage&hash=c42468f27ab4a0e752dbb001d9d94eee&mode=mathml)