Correlation

Study guides: Exam MAS-I, Exam 9, Exam INV-201

Correlation $\rho(X, Y)$ measures the strength and direction of the linear relationship between two random variables $X$ and $Y$ . The Pearson correlation coefficient standardizes the Covariance by the product of the standard deviations, forcing it into the range $[-1, 1]$ .

$\rho(X, Y) = \frac{\text{Cov}(X, Y)}{\sigma_X \, \sigma_Y}$

$\rho = 1$ is a perfect positive linear relationship, $\rho = -1$ a perfect negative one, and $\rho = 0$ no linear association.
Zero correlation does not in general imply independence: with $X$ uniform on $[-1, 1]$ , $Y = X^2$ is uncorrelated with $X$ yet a function of it. It does for jointly normal $X$ and $Y$ , and for two indicator variables — events $A$ and $B$ are independent exactly when their indicators are uncorrelated.
Being unit-free, $\rho$ is comparable across variables on different scales, unlike the raw covariance.

From data, the sample correlation $r$ replaces each moment with its sample counterpart — this is the $r$ read off a Scatter Plot:

$r = \frac{\sum_{i=1}^n (x_i - \bar{x})(y_i - \bar{y})}{\sqrt{\sum_{i=1}^n (x_i - \bar{x})^2}\,\sqrt{\sum_{i=1}^n (y_i - \bar{y})^2}}$

In simple linear regression the R-Squared of the fitted line equals $r^2$ . With several predictors the idea does not carry over — correlation describes one pair of variables at a time — and $R^2$ fills that role.

Correlation Between Insurance Loss and Expense Two random variables $X$ and $Y$ have $\text{Var}(X) = 25$ , $\text{Var}(Y) = 36$ , and $\text{Cov}(X, Y) = -18$ . Find the correlation coefficient and interpret it.

Answer The standard deviations are $\sigma_X = \sqrt{25} = 5$ and $\sigma_Y = \sqrt{36} = 6$ : $\rho(X, Y) = \frac{-18}{5 \times 6} = \frac{-18}{30} = -0.6$ A correlation of $-0.6$ indicates a moderately strong negative linear relationship: as $X$ increases, $Y$ tends to decrease.