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If $u=cx+dy$ and $v=cx-dy$ and $R$ is the co-efficient of correlation between variables $x$ and $y$ and variables $u$ and $v$ have 0 correlation, then how can I prove that $$s_us_v=2cds_xs_y\sqrt{1-R^2}$$.I do not think I made any progress at all and I request a solution so that I can study it.

I have a test tomorrow at school so I really need it.

Can anyone please post a complete solution?Sorry for appearing lazy but I would appreciate some help here. Thanks.

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3 Answers 3

To start you off: If you add constants to $x$ and $y$ it doesn't change any correlations or standard deviations, so we may assume for simplicity that $x$ and $y$ have mean $0$, and then so do $u$ and $v$.

The covariance of $u$ and $v$ is $0$, which says $0 = E[uv] = E(c^2 x^2 - d^2 y^2] = c^2 s_x^2 - d^2 s_y^2$.
Next look at $E[u^2]$ and $E[v^2]$.

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I am not sure I am technically sound.Can you please post a full solution.I shall study it in detail. –  user34522 Sep 2 '12 at 9:15
    
Your answer has been cited in each of the other answers so far, so it definitely deserves at least an upvote (I should say even more) :-) –  robjohn Sep 2 '12 at 20:10

I'll leave this here in case anyone can find my error.

Assume the same setup as in @RobertIsrael's response. As he showed, we have that $\textrm{Cov}\left[u,v\right]=\textrm{E}\left[uv\right]=c^2s_x^2-d^2s_y^2=0$. Now you know something about $s_x$ and $s_y$, but you still can't say anything about $s_u$ and $s_v$, so let's try to find out more about them. If $u$ and $v$ have mean zero, then their variance can be written as $\textrm{Var}\left[u\right]=\textrm{E}\left[u^2\right]=s_u^2$ and $\textrm{Var}\left[v\right]=\textrm{E}\left[v^2\right]=s_v^2$.

$\begin{align} s_u^2 &= \textrm{E}\left[\left(cx+dy\right)^2\right]\\ &= \textrm{E}\left[c^2x^2+2cdxy+d^2y^2\right]\\ &= c^2s_x^2+2cd\textrm{Cov}\left[x,y\right]+d^2s_y^2\\ &= c^2s_x^2+2cds_xs_yR+d^2s_y^2\\ \end{align}$

$\begin{align} s_v^2 &= \textrm{E}\left[\left(cx-dy\right)^2\right]\\ &= \textrm{E}\left[c^2x^2-2cdxy+d^2y^2\right]\\ &= c^2s_x^2-2cd\textrm{Cov}\left[x,y\right]+d^2s_y^2\\ &= c^2s_x^2-2cds_xs_yR+d^2s_y^2\\ \end{align}$

$\begin{align} s_us_v &= \sqrt{\left(c^2s_x^2+2cds_xs_yR+d^2s_y^2\right)\left(c^2s_x^2-2cds_xs_yR+d^2s_y^2\right)}\\ &= \sqrt{c^4s_x^4-4c^2d^2s_x^2s_y^2R^2+2c^2d^2s_x^2s_y^2+d^4s_y^4}\\ &= \sqrt{\left(c^2s_x^2-d^2s_y^2\right)^2+4c^2d^2s_x^2s_y^2\left(1-R^2\right)}\\ &= \sqrt{4c^2d^2s_x^2s_y^2\left(1-R^2\right)}\\ &= 2cds_xs_y\sqrt{1-R^2}\\ \end{align}$

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The third step in your last seqence of equations is in error. The sign of $d^4s_y^4$ seems to have changed. –  robjohn Sep 2 '12 at 19:25
1  
The third step should be $\sqrt{(c^2s_x^2-d^2s_y^2)^2+4c^2d^2s_x^2s_y^2(1-R^2)}$ –  robjohn Sep 2 '12 at 19:46
    
That's better! (+1) –  robjohn Sep 2 '12 at 20:19
    
@robjohn Thanks –  Max Sep 2 '12 at 20:19

After translation so that $\bar{x}=\bar{y}=0$ (and thus $\bar{u}=\bar{v}=0$) as Robert Israel suggests, we get $$ \begin{align} s_u^2 &=\mathrm{E}\left[(cx+dy)^2\right]\\ &=c^2s_x^2+2cdR\,s_xs_y+d^2s_y^2\tag{1} \end{align} $$ and $$ \begin{align} s_v^2 &=\mathrm{E}\left[(cx-dy)^2\right]\\ &=c^2s_x^2-2cdR\,s_xs_y+d^2s_y^2\tag{2} \end{align} $$ Combining $(1)$ and $(2)$ yields $$ s_u^2-s_v^2=4cdR\,s_xs_y\tag{3} $$ and $$ s_u^2+s_v^2=2c^2s_x^2+2d^2s_y^2\tag{4} $$ Subtracting the square of $(3)$ from the square of $(4)$ gives us $$ \begin{align}4s_u^2s_v^2 &=(2c^2s_x^2+2d^2s_y^2)^2-(4cd\,s_xs_y\,R)^2\\ &=(2c^2s_x^2-2d^2s_y^2)^2+(4cd\,s_xs_y)^2(1-R^2)\tag{5} \end{align} $$ Given that $u$ and $v$ have no correlation, we see that $$ \begin{align} 0 &=\mathrm{E}[uv]\\ &=c^2s_x^2-d^2s_y^2\tag{6} \end{align} $$ Using $(6)$ in $(5)$, canceling and taking square roots yields $$ s_us_v=2cd\,s_xs_y\sqrt{1-R^2}\tag{7} $$

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