问题及解答

Show that if $Y=aX+b$ ($a\neq 0$), then $\rho=\mathrm{Corr}(X,Y)=+1$ or $-1$. Under what conditions will $\rho=+1$?
 

\[
\begin{split}
\mathrm{Cov}(X,Y)&=E(XY)-\mu_X\cdot\mu_Y\\
&=E\bigl[X(aX+b)\bigr]-E(X)\cdot E(aX+b)\\
&=E(aX^2+bX)-E(X)\cdot\bigl[aE(X)+b\bigr]\\
&=aE(X^2)+bE(X)-a(E(X))^2-bE(X)\\
&=a\Bigl[E(X^2)-(E(X))^2\Bigr]\\
&=aV(X)
\end{split}
\]

\[
V(Y)=V(aX+b)=a^2V(X)
\]

Hence,

\[
\begin{split}
\rho&=\mathrm{Corr}(X,Y)\\
&=\frac{\mathrm{Cov}(X,Y)}{\sigma_X\cdot\sigma_Y}\\
&=\frac{aV(X)}{\sqrt{V(X)}\cdot\sqrt{V(Y)}}\\
&=\frac{aV(X)}{\sqrt{V(X)}\cdot|a|\sqrt{V(X)}}\\
&=\frac{aV(X)}{|a|V(X)}\\
&=\pm 1
\end{split}
\]

Now, we see that $\rho=1$ if and only if $a > 0$.