# For two uncorrelated random variables $X,Y$, why does $\rho(X+Y,2X+2Y)=4?$

Given two uncorrelated random variables $X,Y$ with the same variance $\sigma^2$ I need to compute $\rho= \frac{COV(X,Y)}{\sigma(X)\sigma(Y)}$ between $X+Y$ and $2X+2Y$. I know it should be a number between $-1$ and $1$ and I don't understand how come I get $4$.

Here's what I did:

$COV(X+Y,2X+2Y)=COV(X+Y,2X)+COV(X+Y,2Y)=COV(2X,X)+COV(2X,Y)+COV(2Y,Y)+COV(2Y,X)=2COV(X,X)+2COV(Y,Y)+4COV(X,Y)=2\sigma^2+2\sigma^2=4\sigma^2$ so final result is $\rho=4$ since $\sigma(X)=\sqrt{Var(x)}$.

What's wrong with what I did?

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## 2 Answers

The reason things went wrong is probably due to an unfortunate choice of notation, the use of $X$ and $Y$ with two different meanings.

We want the correlation coefficient $\rho(U,V)$, where $U=X+Y$ and $V=2(X+Y)$. So we need to divide $\text{Cov}(U,V)$ by the product of the standard deviations of $U$ and of $V$ (not of $X$ and of $Y$).

You did divide, but by the wrong thing. For the denominator, calculate and use $\sigma_U\sigma_V$.

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@Ben Benli: What I did was to explain what went wrong in your calculation. Please note that the answer by did gives a much smoother path to the answer. –  André Nicolas Oct 25 '12 at 19:48

Show that, for every nondegenerate random variable $Z$ and nonzero real number $a$, $\mathrm{var}(aZ)=a^2\cdot\mathrm{var}(Z)$ and $\mathrm{cov}(Z,aZ)=a\cdot\mathrm{var}(Z)$. Deduce that $\varrho(Z,aZ)=\mathrm{sgn}(a)$.

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