I have the following question at hand.

Suppose $(x_1,y_1),(x_2,y_2),\cdots,(x_{10},y_{10})$ represents a set of bi-variate observations on $(X,Y)$ such that $x_2=x_3=\cdots =x_{10}\ne x_1.$ Under what conditions will the Least Square Regression line of $Y$ on $X$ be identical to the Least Absolute Deviation line?

I know that say we want to find $\hat{\alpha}$ and $\hat\beta$ such that $Y=\hat\alpha+\hat\beta X$; the LSQ method will give $$\hat\beta={\sum\limits_{i=1}^{10} (x_i-\bar x)y_i\over \sum\limits_{i=1}^{10}(x_i-\bar x)x_i}$$ and hence $\hat\alpha$. Can someone help me proceed?

  • 1
    $\begingroup$ Intuitively, there's a trivial case and a non trivial case where pairs of points stabilize a line with equal errors. $\endgroup$
    – Firebug
    Aug 5, 2016 at 18:15

1 Answer 1


Some hints to help you gain some insight

  1. Make up or generate some data consistent with the conditions in the question. Try $x_1=0, y_1=0$ and $x_2..x_{10}=1$ (choosing some values for $y_i$, $i=2,...,10$). Where do the lines pass relative to the first point?

  2. Now start as above but try placing $y_i$, $i=2,...,10$ at say 1,2,3,4,5,6,7,8,9 respectively. Where do the lines go?

plot of (0,0) and nine uniformly-spread points at x=1

  1. Now place $y_i$, $i=2,...,10$ at say 1,2,3,4,5,6,7,8,99 respectively. Where do the lines go?

previous plot but with highest point moved up to y=99

What is special/interesting about the fitted values for the two lines at $x=1$?

(If it's not clear try some other values for $y_{10}$.)

Same data showing fitted lines

  1. Can you prove this is the case more generally?

This ultimately brings us to a slightly simpler question, which relates to when means and medians are equal in the univariate case. (There's a simple, obvious condition that's sufficient, but not necessary.)

There are a number of posts on site that discuss the other case. There are some interesting examples here


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.