TU Wien:Statistik und Wahrscheinlichkeitstheorie UE (Levajkovic)/Übungen 2023W/HW12.2

Correlation 2

Let $X$ and $Z$ be random variables such that $\mathbb {E} (X)=\mu x,\mathbb {V} ar(X)=\sigma _{x}^{2},\mathbb {E} (Z)=0{\text{ and }}\mathbb {V} ar(Z)=1$. Let $Y$ be a random variable with known expectation $\mathbb {E} (Y)=\mu y$ and variance $\mathbb {V} ar(Y)=\sigma _{y}^{2}>0$ such that the correlation $\rho =\mathbb {C} orr(X,Y)\in [-1,1]$ is also given. Reperesent $Y$ in the form $Y=\beta _{0}+\beta _{1}X+\sigma Z$

and determine the unknown coefficients b0, b1 and σ.

Dieses Beispiel ist als solved markiert. Ist dies falsch oder ungenau? Aktualisiere den Lösungsstatus (Details: Vorlage:Beispiel)

Lösungsvorschlag von Lessi

--Lessi 2024-02-07T13:04:11Z

By expanding $\rho =\mathbb {C} orr(X,Y)$ we get:

${\begin{aligned}\rho &={\mathbb {C} ov(X,\beta _{0}+\beta _{1}\cdot X+\sigma \cdot Z) \over \sigma _{X}\sigma _{Y}}\\&={\mathbb {C} ov(X,\beta _{1}\cdot X)+\mathbb {C} ov(X,\sigma \cdot Z) \over \sigma _{X}\sigma _{Y}}\\&={\beta _{1}\mathbb {C} ov(X,X)+\sigma \mathbb {C} ov(X,Z) \over \sigma _{X}\sigma _{Y}}\\&={\beta _{1}\sigma _{X}^{2} \over \sigma _{X}\sigma _{Y}}\\&={\beta _{1}\sigma _{X} \over \sigma _{Y}}\\\\&\implies \beta _{1}=\rho {\frac {\sigma _{Y}}{\sigma _{X}}}\end{aligned}}$

We can determine $\sigma$ by the fact that $X$ and $Z$ are uncorrelated:

${\begin{aligned}\mathbb {V} ar(Y)&=\mathbb {V} ar(\beta _{0}+\beta _{1}X+\sigma Z)\\&=\beta _{1}^{2}\mathbb {V} ar(X)+\sigma ^{2}\mathbb {V} ar(Z)\\&=\beta _{1}^{2}\sigma _{X}+\sigma ^{2}\\\\&\implies \sigma _{Y}^{2}=\beta _{1}^{2}*\sigma _{X}+\sigma ^{2}\\&\implies \sigma =\sigma _{Y}{\sqrt {1-{\frac {\rho ^{2}}{\sigma _{X}}}}}\end{aligned}}$

Lastly we can get $\beta _{0}$ by simply inserting all known coefficients to the definition of $Y$:

${\begin{aligned}\mu _{Y}&=\beta _{0}+\beta _{1}\mu _{X}+\sigma \mu _{Z}\\&\implies \beta _{0}=\mu _{Y}-\beta _{1}\mu _{X}\end{aligned}}$