问题及解答

• (a) Use the rules of expected value to show that $\mathrm{Cov}(aX+b,cY+d)=ac\mathrm{Cov}(X,Y)$.
• (b) Use part (a) along with the rules of variance and standard deviation to show that $\mathrm{Corr}(aX+b,cY+d)=\mathrm{Corr}(X,Y)$ when $a$ and $c$ have the same sign.
• (c) What happens if $a$ and $c$ have opposite signs?
   

(a)

By the property of covariance, 

\[
\begin{split}
\mathrm{Cov}(aX+b,\ cY+d)&=E\bigl[(aX+b)(cY+d)\bigr]-\mu_{aX+b}\cdot\mu_{cY+d}\\
&=E[acXY+adX+bcY+bd]-E(aX+b)\cdot E(cY+d)\\
&=acE(XY)+adE(X)+bcE(Y)+bd-\Bigl[(aE(X)+b)(cE(Y)+d)\Bigr]\\
&=acE(XY)+adE(X)+bcE(Y)+bd-\Bigl[acE(X)E(Y)+adE(X)+bcE(Y)+bd\Bigr]\\
&=acE(XY)-acE(X)E(Y)\\
&=ac\Bigl[E(XY)-\mu_X\cdot\mu_Y\Bigr]\\
&=ac\mathrm{Cov}(X,Y).
\end{split}
\]

(b)

By definition of the corelation coeficient,

\[
\mathrm{Corr}(aX+b,\ cY+d)=\frac{\mathrm{Cov}(aX+b,\ cY+d)}{\sigma_{aX+b}\cdot\sigma_{cY+d}}=\frac{\mathrm{Cov}(aX+b,\ cY+d)}{\sqrt{V(aX+b)}\cdot\sqrt{V(cY+d)}}
\]

Note that $V(aX+b)=a^2V(X)$, hence

\[\sigma_{aX+b}=\sqrt{V(aX+b)}=\sqrt{a^2V(X)}=|a|\sigma_X\]

Therefore, if $a$ and $c$ have the same sign,

\[
\mathrm{Corr}(aX+b,\ cY+d)=\frac{ac\mathrm{Cov}(X,Y)}{|a|\sigma_X\cdot|c|\sigma_Y}=\frac{\mathrm{Cov}(X,Y)}{\sigma_X\cdot\sigma_Y}=\mathrm{Corr}(X,Y).
\]

(c)

If $a$ and $c$ have opposite signs, 

\[
\mathrm{Corr}(aX+b,\ cY+d)=\frac{ac\mathrm{Cov}(X,Y)}{|a|\sigma_X\cdot|c|\sigma_Y}=-\frac{\mathrm{Cov}(X,Y)}{\sigma_X\cdot\sigma_Y}=-\mathrm{Corr}(X,Y).
\]