问题及解答

• Use the general formula for the variance of a linear combination to write an expression for $V(aX+Y)$. Then let $a=\frac{\sigma_Y}{\sigma_X}$ and show that $\rho\geqslant -1$. [{\it Hint:} Variance is always $\geqslant 0$, and $\mathrm{Cov}(X,Y)=\sigma_X\cdot\sigma_Y\cdot\rho$.]
• By considering $V(aX-Y)$, conclude that $\rho\leqslant 1$.
• Use the fact that $V(W)=0$ only if $W$ is a constant to show that $\rho=1$ only if $Y=aX+b$.
   

(1)

The general formula for the variance of a linear combination is

\[
V(a_1X_1+a_2X_2+\cdots+a_nX_n)=\sum_{i=1}^{n}\sum_{j=1}^{n}a_i a_j\mathrm{Cov}(X_i,X_j)
\]

Then, let $a_1=a$, $X_1=X$, $a_2=1$, $X_2=Y$, we have

\[
\begin{split}
V(aX+Y)&=V(a_1X_1+a_2X_2)\\
&=\sum_{i=1}^{2}\sum_{j=1}^{n}a_i a_j\mathrm{Cov}(X_i,X_j)\\
&=a_1^2\mathrm{Cov}(X_1,X_1)+a_1 a_2\mathrm{Cov}(X_1,X_2)+a_2 a_1\mathrm{Cov}(X_2,X_1)+a_2^2\mathrm{Cov}(X_2,X_2)\\
&=a^2\mathrm{Cov}(X,X)+2a\mathrm{Cov}(X,Y)+\mathrm{Cov}(Y,Y)\\
&=a^2V(X)+2a\rho\sigma_X\cdot\sigma_Y+V(Y)
\end{split}
\]

Now, if let $a=\frac{\sigma_Y}{\sigma_X}$, then we have

\[
\begin{split}
V(aX+Y)&=\frac{\sigma_Y^2}{\sigma_X^2}V(X)+2\frac{\sigma_Y}{\sigma_X}\cdot\rho\sigma_X\cdot\sigma_Y+V(Y)\\
&=\sigma_Y^2+2\rho\sigma_Y^2+V(Y)\\
&=\sigma_Y^2(1+2\rho+1)
\end{split}
\]

Not that $V(aX+Y)$ is always nonnegative, so $2\rho+2\geqslant 0$. It infers that $\rho\geqslant -1$.

(2)

By considering $V(aX-Y)$, i.e., let $a_1=a$, $a_2=-1$, then we have

\[
\begin{split}
V(aX-Y)&=V(a_1X_1+a_2X_2)\\
&=\sum_{i=1}^{2}\sum_{j=1}^{n}a_i a_j\mathrm{Cov}(X_i,X_j)\\
&=a_1^2\mathrm{Cov}(X_1,X_1)+a_1 a_2\mathrm{Cov}(X_1,X_2)+a_2 a_1\mathrm{Cov}(X_2,X_1)+a_2^2\mathrm{Cov}(X_2,X_2)\\
&=a^2\mathrm{Cov}(X,X)-2a\mathrm{Cov}(X,Y)+\mathrm{Cov}(Y,Y)\\
&=a^2V(X)-2a\rho\sigma_X\cdot\sigma_Y+V(Y)
\end{split}
\]

令 $a=\frac{\sigma_Y}{\sigma_X}$, 并注意到 $V(X)=\sigma_X^2$, $V(Y)=\sigma_Y^2$, 以及方差总是非负, 即 $V(aX-Y)\geqslant 0$, 我们得

\[
\begin{split}
&\frac{\sigma_Y^2}{\sigma_X^2}\cdot\sigma_X^2-2\cdot\frac{\sigma_Y}{\sigma_X}\cdot\rho\sigma_X\sigma_Y+\sigma_Y^2\geqslant 0\\
\Rightarrow\ &\sigma_Y^2-2\rho\sigma_Y^2+\sigma_Y^2\geqslant 0\\
\Rightarrow\ &2\rho\sigma_Y^2\leqslant 2\sigma_Y^2\\
\Rightarrow\ &\rho\leqslant 1.
\end{split}
\]

注: 我们这里假设 $\sigma_Y\neq 0$. 这是合理的, 因为相关系数 $\rho_{XY}=\rho$ 定义为 $\dfrac{\mathrm{Cov}(X,Y)}{\sigma_X\sigma_Y}$. 

如果 $\sigma_Y=0$, 也就是说 $V(Y)=0$, 则 $Y$ 一定是常数. 不妨设 $Y=W$, 这里 $W=\mathrm{const.}$ 于是

\[
\mathrm{Cov}(X,Y)=E(XY)-\mu_X\mu_Y=E(XW)-\mu_X\cdot W=W\cdot E(X)-W\cdot\mu_X=W(E(X)-\mu_X)=0.
\] 

于是 $\rho\sigma_X\sigma_Y=\mathrm{Cov}(X,Y)$ 也成立.

(3)

If $\rho=1$, let $a=\frac{\sigma_Y}{\sigma_X}$, then by the computation above, we have

\[
\begin{split}
V(aX-Y)&=a^2V(X)-2a\rho\sigma_X\sigma_Y+V(Y)\\
&=\frac{\sigma_Y^2}{\sigma_X^2}\cdot\sigma_X^2-2\frac{\sigma_Y}{\sigma_X}\cdot 1\cdot\sigma_X\sigma_Y+\sigma_Y^2\\
&=\sigma_X^2-2\sigma_Y^2+\sigma_Y^2\\
&=0
\end{split}
\]

于是 $aX-Y$ 必为常数, 比如 $aX-Y=b$, 即有 $Y=aX+b$.