TU Wien:Statistik und Wahrscheinlichkeitstheorie UE (Bura)/Übungen 2020W/HW11.1

Covariance

Let ${\textstyle X,Y}$ and ${\textstyle Z}$ be random variables with finite variance, and let ${\textstyle a,b\in \mathbb {R} }$ constant. The covariance of ${\textstyle X}$ and ${\textstyle Y}$ is given through ${\textstyle \operatorname {\mathbb {C} ov} (X,Y):=\mathbb {E} [(X-\mathbb {E} [X])(Y-\mathbb {E} [Y])].}$

1a)

Show that ${\textstyle \mathbb {C} ov(X,Y)=\mathbb {E} [XY]-\mathbb {E} [X]\mathbb {E} [Y]}$

1b)

Show that ${\textstyle \mathbb {C} ov(X,Y)=\mathbb {C} ov(Y,X)}$ (symmetry)

1c)

Show that ${\textstyle \mathbb {C} ov(X,X)=\mathbb {V} ar(X)}$

1d)

Show that ${\textstyle \mathbb {C} ov(X,a)=0}$

1e)

Show that ${\textstyle \mathbb {C} ov(aX+bZ,Y)=a\mathbb {C} ov(X,Y)+b\mathbb {C} ov(Z,Y)}$ (linearity)

1f)

Does linearity also hold for the second component ?

1g)

Show that ${\textstyle \mathbb {V} ar(X+Y)=\mathbb {V} ar(X)+\mathbb {V} ar(Y)+2\mathbb {C} ov(X,Y)}$

1h)

Show that for ${\textstyle X}$ and ${\textstyle Y}$ independent, it is ${\textstyle \mathbb {C} ov(X,Y)=0}$. (hint you can use that for ${\textstyle X}$ and ${\textstyle Y}$ independent it holds ${\textstyle \mathbb {E} [XY]=\mathbb {E} [X]\cdot \mathbb {E} [Y]}$.)

Lösungsvorschlag von Friday

--Friday Sa 30 Jan 2021 17:59:00 CET

$${\begin{aligned}\mathbb {C} ov(X,Y)&=\mathbb {E} [(X-\mathbb {E} [X])(Y-\mathbb {E} [Y])]\\&=\mathbb {E} [XY]-\mathbb {E} [X\cdot \mathbb {E} [Y]]-\mathbb {E} [Y\cdot \mathbb {E} [X]]+\mathbb {E} [\mathbb {E} [X]\cdot \mathbb {E} [Y]]\\&=\mathbb {E} [XY]-\mathbb {E} [X]\cdot \mathbb {E} [Y]-\mathbb {E} [X]\cdot \mathbb {E} [Y]+\mathbb {E} [X]\cdot \mathbb {E} [Y]\\&=\mathbb {E} [XY]-\mathbb {E} [X]\cdot \mathbb {E} [Y]\quad \Box \end{aligned}}$$

1b)

$$\mathbb {C} ov(X,Y)=\mathbb {E} [(X-\mathbb {E} [X])(Y-\mathbb {E} [Y])]=\mathbb {E} [(Y-\mathbb {E} [Y])(X-\mathbb {E} [X])]=\mathbb {C} ov(Y,X)\quad \Box$$

1c)

First, we remember the formular for the variance:

$$\mathbb {V} ar(X)=\mathbb {E} (X-\mathbb {E} (X))^{2}$$

Now, we can prove the equation:

$${\begin{aligned}\mathbb {C} ov(X,X)&=\mathbb {E} [(X-\mathbb {E} [X])(X-\mathbb {E} [X])\\&=\mathbb {E} [(X-\mathbb {E} [X]))^{2}\\&=\mathbb {V} ar(X)\quad \Box \end{aligned}}$$

1d)

First, we need to remember that the expection of a constant is itself:

$$\mathbb {E} [a]=a$$

Now we can just solve the equation

$${\begin{aligned}\mathbb {C} ov(X,a)&=\mathbb {E} [(X-\mathbb {E} [X])(a-\mathbb {E} [a])]\\&=\mathbb {E} [(X-\mathbb {E} [X])(a-a)]\\&=\mathbb {E} [(X-\mathbb {E} [X])\cdot 0]\\&=0\quad \Box \\\end{aligned}}$$

1e)

$${\begin{aligned}\mathbb {V} ar(X+Y)&=\mathbb {C} ov(X+Y,X+Y)\\&=\mathbb {C} ov(X+Y,X)+\mathbb {C} ov(X+Y,Y)\\&=\mathbb {C} ov(X,X)+\mathbb {C} ov(X,Y)+\mathbb {C} ov(Y,X)+\mathbb {C} ov(Y,Y)\\&=\mathbb {V} ar(X)+\mathbb {V} ar(Y)+2\cdot \mathbb {C} ov(X,Y)\quad \Box \\\end{aligned}}$$