Is the correlation coefficient just the standard deviations of x and y multiplied together?

Question

Is $r = \frac{\sum{(x_i-\bar{x})(y_i-\bar{y})}}{\sqrt{\sum(x_i-\bar{x})^2}\sqrt{\sum(y_i-\bar{y})^2}} $ the same as $\sum{\frac{(x_i-\bar{x})}{\sigma_x}\frac{(y_i-\bar{y})}{\sigma_y}}$ which is just multiplication of the standard deviation of $x_i$ multiplied by the standard deviation $y_i$ for every $i$ summed up?

Answer 1

No it is not. The mistake in the equations you posted is that the product of the standard deviations is not

$$\sum_i{\frac{(x_i-\bar{x})}{\sigma_x}\frac{(y_i-\bar{y})}{\sigma_y}}$$

but rather

$$\left(\sum_i \frac{x_i-\bar{x}}{\sigma_x}\right) \left(\sum_i \frac{y_i-\bar{y}}{\sigma_y}\right)$$

In the correlation coefficient you sum only once, thus "connecting" the two quantities $x$ and $y$, whereas the standard deviation is something you can calculate for each quantity on its own.