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 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!

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

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?