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Given that correlation coefficient between X and Y is 𝜌(𝑋, 𝑌) and a, b, c, d are constants, prove that the correlation coefficient between U = 𝑎𝑋 + 𝑏 and 𝑉 = 𝑐𝑌 + 𝑑 is equal to 𝜌(𝑈, 𝑉) = 𝜌(𝑋, 𝑌)

My take on the problem:

Prove that $p(U, V)=p(X, Y)$ $$ \begin{aligned} &\operatorname{Cov}(X, Y)=E[X Y]-E[X] \cdot E[Y] \\ &\rho(U, V)=p(X, Y) \\ &\frac{\operatorname{Cov}(U, V)}{\sqrt{\operatorname{Var}(U) \operatorname{Var}(V)}}=\frac{\operatorname{Cov}(X, Y)}{\sqrt{\operatorname{Var}(X) \operatorname{Var}(Y)}} \\ &\frac{E[U V]-E[U] E[V]}{\sqrt{\operatorname{Var}(U) \operatorname{Var}(V)}}=\frac{E[X Y]-E[X] E[Y]}{\sqrt{\operatorname{Var}(X) \operatorname{Var}(U)}} \end{aligned} $$ $$ \begin{aligned} &\frac{a c \cdot E[X Y]+a d E[X]+b c E[Y]-E[X] \cdot a E[Y] \cdot c}{a c \sqrt{\operatorname{Var}(X) \operatorname{Var}(Y)}} \\ &=\frac{E[X Y]-E[X] E[Y]}{\sqrt{\operatorname{Var}(X) \operatorname{Var}(Y)}} \\ &\frac{E[X Y]+\frac{a d}{a c} E[X]+\frac{b c}{a c} E[Y]-E[Y] E[Y]}{a c / a c} \\ &=E[X Y]-E[X] E[Y] \\ &\frac{d}{c} E[X]+\frac{b}{a} E[Y]=0 \\ &\frac{d}{c} E[X]=-\frac{b}{a} E[Y] \end{aligned} $$

How to get to a conclusion from here?

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2 Answers 2

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A huge problem with what you're doing is you're assuming what you want to show to start with, which is a bad practice. You should also make use of covariance properties to make your life easier.

Let $U = aX + b$ and $V = cY + d$. Then \begin{align} \text{Cov}(U, V) &= \text{Cov}(aX + b, cY + d) \\ &= \text{Cov}(aX, cY) \\ &= ac\text{Cov}(X, Y)\text{.} \end{align} Furthermore, we have that $$\text{Var}(U) = \text{Var}(aX + b) = \text{Var}(aX) = a^2\text{Var}(X)$$ and similarly, $\text{Var}(V) = c^2\text{Var}(Y)$.

Do not forget that $\sqrt{x^2} = |x|$, hence

\begin{align} &\sqrt{\text{Var}(U)} = \sqrt{a^2\text{Var}(X)} = |a|\sqrt{\text{Var}(X)} \\ &\sqrt{\text{Var}(V)} = \sqrt{c^2\text{Var}(Y)} = |c|\sqrt{\text{Var}(Y)} \end{align}

Hence the correlation coefficient \begin{align} \rho(U, V) &= \dfrac{\text{Cov}(U, V)}{\sqrt{\text{Var}(U)}\sqrt{\text{Var}(V)}} \\ &= \dfrac{ac\text{Cov}(X, Y)}{|a|\sqrt{\text{Var}(X)} \cdot |c|\sqrt{\text{Var}(Y)}} \\ &= \dfrac{ac}{|a||c|} \cdot \dfrac{\text{Cov}(X, Y)}{\sqrt{\text{Var}(X)}\sqrt{\text{Var}(Y)}} \\ &= \dfrac{ac}{|a||c|} \cdot \rho(X, Y)\text{.} \end{align}

We now have to consider $\dfrac{ac}{|a||c|}$.

Obviously if both $a, c$ are positive, we have $|a||c| = ac$, so $\dfrac{ac}{|a||c|} = 1$.

Suppose, without loss of generality, that $a$ is negative and $c$ is positive. Then $|a| = -a$ by definition of the absolute value and $|c| = c$. Hence $$\dfrac{ac}{|a||c|} = \dfrac{ac}{(-a)c} = -1\text{.}$$

If both $a$ and $c$ are negative, then $$\dfrac{ac}{|a||c|} = \dfrac{ac}{(-a)(-c)} = \dfrac{ac}{ac} = 1\text{.}$$

Thus, we conclude:

$\fbox{If both $a$ and $c$ are of the same sign, then $\rho(U, V) = \rho(X, Y)$. Otherwise, $\rho(U, V) = -\rho(X, Y)$.}$

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Hint: use $\rho(X,Y)=Cov(\frac{X-\mathbb{E}X}{\sigma(X)},\frac{Y-\mathbb{E}Y}{\sigma(Y)})$.

Should be a one line exercise. Here $\sigma$ is the standard deviation. Explore the properties of variance and mean value.

$\rho(aX+b,cY+d)=Cov(\frac{aX+b-a\mathbb{E}X-b}{\sigma(aX+b)},\frac{cY+d-c\mathbb{E}Y-d}{\sigma(cY+d)})=Cov(\frac{a(X-\mathbb{E}X)}{|a|\sigma(X)},\frac{c(Y-\mathbb{E}Y)}{|c|\sigma(Y)})=\rho(X,Y)$

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