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problem set

I know that correlation coefficient is scale invariant but my math says otherwise. By calculation, I got that $$\operatorname{std}(x) \operatorname{std}(y) = 5.$$ So by squaring both sides, $$\operatorname{Var}(x)\operatorname{Var}(y) = 25.$$ Then by using $\operatorname{Var}(aX) = a^2\operatorname{Var}(X)$, I got $$\frac{625}{6561} \cdot \operatorname{Var}(t)\operatorname{Var}(s) = 25.$$ Hence $\operatorname{Var}(t)\operatorname{Var}(s) = \frac{6561}{25}$. Then by square rooting both sides, $\operatorname{std}(t)\operatorname{std}(s) = \frac{81}5$. From the first part of the solution, $\operatorname{Cov}(t,s) = \frac{100}{81}$ so $\rho(t,s) = \frac{100}{729}$ which is $\neq \rho(x,y) = 0.8$. But they are supposed to be the same because correlation is scale invariant right? Where did I go wrong?

I found out that my math is in fact not wrong. I used $X = \frac{5}{9} T$ when it should have been $X = \frac{9}{5} T$ . After rectifying that, I got the answer using my method!

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    $\begingroup$ This question would be easier to read using math formatting. math.meta.stackexchange.com/questions/5020/… $\endgroup$
    – Sycorax ♦
    Commented Jan 13, 2020 at 1:54
  • $\begingroup$ You can answer these questions with minimal calculation and no algebra, using only the insights (from simple pictures) derived at the end of stats.stackexchange.com/a/18200/919. They show that $\rho$ will not change and the covariance will be multiplied by $(5/9)^2.$ $\endgroup$
    – whuber ♦
    Commented Jan 13, 2020 at 18:35

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$\newcommand{\Cov}{\operatorname{Cov}}$The multiplication rule for covariances is $\Cov(aX,bY)=ab\Cov(X,Y)$. Note that the constant term ($-\frac{160}{9}$) doesn't affect the standard deviation or covariances of the temperatures, so we can ignore it. Since $T=\frac{5}{9}X$ and $S=\frac{5}{9}Y$, we know that $$\Cov(T,S)=\Cov\left(\frac{5}{9}X,\frac{5}{9}Y\right)=(\frac{5}{9})^2 \Cov(X,Y)=\left(\frac{5}{9}\right)^2 4$$

The multiplication rule for standard deviations comes from the rule for variances: $s_{kX}=ks_X$. So, $$\rho(T,S)=\frac{\Cov(T,S)}{s_T s_S}=\frac{\Cov(\frac{5}{9}X,\frac{5}{9}Y)}{s_{\frac{5}{9}X}s_{\frac{5}{9}Y}}=\frac{(\frac{5}{9})^2 \Cov(X,Y)}{(\frac{5}{9})^2 s_X s_Y}=\frac{\Cov(X,Y)}{s_X s_Y} = \rho(X,Y)=0.8$$

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    $\begingroup$ You probably want $\text{ Cov}(aX+c,bY+d)=ab\text{ Cov}(X,Y)$ and $T=\frac59X-\frac{160}{9}$ etc before you ignore the constant, though this does not affect the second half of what you wrote $\endgroup$
    – Henry
    Commented Jan 13, 2020 at 11:21
  • $\begingroup$ Good point, I was just being lazy. $\endgroup$
    – Noah
    Commented Jan 13, 2020 at 15:27

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