Why is the following true for the covariance $cov(X, Y)$ and variances $Var(X)$, $Var(Y)$ of two rvs $X$ and $Y$: $$ \frac{Cov(c X, Y)}{\sqrt{Var(c X) Var(Y)}} = \frac{c Cov(X, Y)}{\sqrt{c^2 Var(X) Var(Y)}} = \frac{Cov(X, Y)}{\sqrt{Var(X) Var(Y)}} $$ I cannot see how the $c^2$ within the square root can disappear just because we have $c$ as a factor in the numerator.
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Recall that $\sqrt{ab} = \sqrt{a}\sqrt{b}$. Then,
$$\frac{c Cov(X, Y)}{\sqrt{c^2 Var(X) Var(Y)}} = \frac{c Cov(X, Y)}{\sqrt{c^2} \sqrt{ Var(X) Var(Y)}}= \frac{c Cov(X, Y)}{|c|\sqrt{Var(X) Var(Y)}} = \mathrm{sgn}(c)\frac{Cov(X, Y)}{\sqrt{ Var(X) Var(Y)}}$$
Jose Avilez
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But $\sqrt{a b} = \sqrt{a} \sqrt{b}$ is not true? For example with $a=3$ and $b=3$? – That Guy Apr 06 '22 at 03:11
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@sostack Yes it is. $\sqrt{3 \times 3} = \sqrt{9} = 3$. Also, $\sqrt{3} \times \sqrt{3} = \sqrt{3}^2 = 3$. – Jose Avilez Apr 06 '22 at 03:17
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@sostack Please see: https://math.stackexchange.com/questions/1025551/sqrtab-sqrta-sqrtb – Jose Avilez Apr 06 '22 at 03:18
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Ah yea ofc. Sorry. Thank you so much for the answer! I appreciate your time :) – That Guy Apr 06 '22 at 03:19