0
$\begingroup$

I am reading a textbook on statists by Freedman, Pisani, and Purves. In one of the chapters about correlation between two variables, it is given that the vertical distance of a typical point from the standard deviation line (say $s_v$) on a scatter plot is, $s_v=\sqrt{2(1-|r|)} \times \sigma_v$, where $r$ is the correlation coefficient, and $\sigma_v$ is the vertical standard deviation.

How to work out this formula mathematically? I thought about working it out from the slope of the SD line, or formula for $r$ but I can't seem to work it out or find a hint/solution online. It's not a homework problem it's a sort of technical footnote.

It is also mentioned that there are similar formulas for the horizontal direction. Is $s_h=\sqrt{2(1-r)} \times \sigma_h$, the formula for the horizontal distance? If not, what will it be?

$\endgroup$
1
0
$\begingroup$

It is kind of confusing. The vertical distance of a typical point in fact refers to r.m.s(vertical distance of all data). Typical here means average in some sense.

Give $(x_i,y_i)_{i=1}^n$, let $d_i$ be the vertical distance from SD line to $(x_i,y_i)$. Then we have $$d_i = y_i - \left(\frac{\sigma_y}{\sigma_x}(x_i-\bar{x})+\bar{y}\right).$$

Thus r.m.s(vertical distance of all data) is given by \begin{align} \sqrt{\frac{1}{n}\sum d_i^2}=SD[d_i]&=SD[y_i - \left(\frac{\sigma_y}{\sigma_x}(x_i-\bar{x})+\bar{y}\right)]\\ &=SD[y_i - \frac{\sigma_y}{\sigma_x}x_i + const]. \end{align} SD is invariant with constant shift, applying $$SD(X+Y)=\sqrt{Var(X)+Var(Y)+2COV(X,Y)}$$ onto the above: \begin{align} SD[y_i - \frac{\sigma_y}{\sigma_x}x_i]&=\sqrt{\sigma_y^2+\frac{\sigma_y^2}{\sigma_x^2}\sigma_x^2 -2\frac{\sigma_y}{\sigma_x}COV(x,y)}\\ &=\sqrt{2\sigma_y^2 -2\frac{\sigma_y}{\sigma_x}r\sigma_y\sigma_x}\\ &=\sqrt{2(1-r)}\sigma_y \end{align}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.