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Let $X$ and $Y$ be random variables such that $0<\sigma^2_X<\infty$ and $0<\sigma^2_Y<\infty$. Suppose that $U=aX+b$ and $V=cY+d$, where $a \not= 0$ and $c \not= 0$. Show that $\rho(U,V)=\rho(X,Y)$ if $ac>0$, and $\rho(U,V)=-\rho(X,Y)$ if $ac<0$.

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    $\begingroup$ Have you tried anything? $\endgroup$ – angryavian Feb 16 '17 at 5:36
  • $\begingroup$ Not yet. I'm really lost, so pointers to begin with would help a ton! $\endgroup$ – Amanda R. Feb 16 '17 at 5:51
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Might be best to start with covariance and $a,c > 0$. $$Cov(U,V) = Cov(aX+b, cY+d) = Cov(aX,cY) + Cov(aX,d) + Cov(b,cY) + Cov(b,d)\\ = Cov(aX,cY) = acCov(X,Y).$$ Then $\rho(U,V) = Cor(U,V) = \frac{Cov(U,V)}{SD(U)SD(V)}.$

Finally, finish by finding $\sigma_U = SD(U)$ and $\sigma_V = SD(V).$

I hope my notation is sufficiently similar to notation in your text so that you can follow this.

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  • $\begingroup$ $SD(U)=a\sigma_X$ and $SD(V)=c\sigma_Y$ right? $\endgroup$ – Amanda R. Feb 16 '17 at 7:42
  • $\begingroup$ Right. $Var(aX) = a^2Var(X).$ Then take square root to get SD. $\endgroup$ – BruceET Feb 16 '17 at 7:56
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\begin{eqnarray*} \displaystyle\rho(U, V) & = & Cov\left(\frac{U-\mathbb{E}(U)}{\sigma_U}, \frac{V-\mathbb{E}(V)}{\sigma_V}\right) \\ & = & Cov\left(\frac{aX+b-a\mathbb{E}(X)-b}{|a|\sigma_X}, \frac{cY+d-c\mathbb{E}(Y)-d}{|c|\sigma_Y}\right) \\ & = & \frac{ac}{|a||c|}Cov\left(\frac{X-\mathbb{E}(X)}{\sigma_X}, \frac{Y-\mathbb{E}(Y)}{\sigma_Y}\right) \\ & = & \frac{ac}{|a||c|}\rho(X, Y) \\ & = & \begin{cases} \rho(X, Y) & \text{ if } ac > 0 \\ -\rho(X, Y) & \text{ if } ac < 0 \end{cases} \end{eqnarray*}

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    $\begingroup$ Elegant development (+1) $\endgroup$ – BruceET Feb 16 '17 at 8:02

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