Index: The Book of Statistical ProofsGeneral Theorems ▷ Probability theory ▷ Correlation ▷ Range

Theorem: Let $X$ and $Y$ be two random variables. Then, the correlation of $X$ and $Y$ is between and including $-1$ and $+1$:

$\label{eq:corr-range} -1 \leq \mathrm{Corr}(X,Y) \leq +1 \; .$

Proof: Consider the variance of $X$ plus or minus $Y$, each divided by their standard deviations:

$\label{eq:var-XY} \mathrm{Var}\left( \frac{X}{\sigma_X} \pm \frac{Y}{\sigma_Y} \right) \; .$

Because the variance is non-negative, this term is larger than or equal to zero:

$\label{eq:var-XY-0} 0 \leq \mathrm{Var}\left( \frac{X}{\sigma_X} \pm \frac{Y}{\sigma_Y} \right) \; .$

Using the variance of a linear combination, it can also be written as:

$\label{eq:var-XY-s1} \begin{split} \mathrm{Var}\left( \frac{X}{\sigma_X} \pm \frac{Y}{\sigma_Y} \right) &= \mathrm{Var}\left( \frac{X}{\sigma_X} \right) + \mathrm{Var}\left( \frac{Y}{\sigma_Y} \right) \pm 2 \, \mathrm{Cov}\left( \frac{X}{\sigma_X}, \frac{Y}{\sigma_Y} \right) \\ &= \frac{1}{\sigma_X^2} \mathrm{Var}(X) + \frac{1}{\sigma_Y^2} \mathrm{Var}(Y) \pm 2 \, \frac{1}{\sigma_X \sigma_Y} \, \mathrm{Cov}(X,Y) \\ &= \frac{1}{\sigma_X^2} \sigma_X^2 + \frac{1}{\sigma_Y^2} \sigma_Y^2 \pm 2 \, \frac{1}{\sigma_X \sigma_Y} \, \sigma_{XY} \; . \end{split}$

Using the relationship between covariance and correlation, we have:

$\label{eq:var-XY-s2} \mathrm{Var}\left( \frac{X}{\sigma_X} \pm \frac{Y}{\sigma_Y} \right) = 1 + 1 + \pm 2 \, \mathrm{Corr}(X,Y) \; .$

Thus, the combination of \eqref{eq:var-XY-0} with \eqref{eq:var-XY-s2} yields

$\label{eq:var-XY-ineq} 0 \leq 2 \pm 2 \, \mathrm{Corr}(X,Y)$

which is equivalent to

$\label{eq:corr-range-qed} -1 \leq \mathrm{Corr}(X,Y) \leq +1 \; .$
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Metadata: ID: P300 | shortcut: corr-range | author: JoramSoch | date: 2021-12-14, 02:08.