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when toggle format what by license comment
Sep 20, 2015 at 16:24 comment added Creosote Let us continue this discussion in chat.
Sep 20, 2015 at 15:25 comment added Creosote See end of updated post.
Sep 20, 2015 at 15:24 history edited Creosote CC BY-SA 3.0
answered a comment with supplementary material
Sep 20, 2015 at 14:04 comment added Dalek Thanks a lot but I can not follow the math. How did you obtain the first equation?
Sep 20, 2015 at 13:54 history undeleted Creosote
Sep 20, 2015 at 13:54 history edited Creosote CC BY-SA 3.0
rewrite
Sep 20, 2015 at 11:42 history deleted Creosote via Vote
Sep 20, 2015 at 10:59 comment added Creosote I could've put more comments in the code ... The black points are the true (but unseen) $x$ and $y$ values, deliberately chosen to be perfectly linear. The red circles are the observed versions of $x$ and $y$, i.e. with added "uncertainty" (Normal errors). The red lines drop down to the most-probable point on the black line. If you changed the code to sd_x=rep(2,n); sd_y=rep(0.2,n) in the relevant place, you'll see that the projections are obviously better than perpendicular would have been.
Sep 20, 2015 at 10:13 comment added Dalek it doesn't seem the black points are the perpendicular projections of the red points on the line.
Sep 20, 2015 at 9:53 comment added Creosote They're projections onto the "best" point on the black line. If the sd_x and sd_y values are equal for a data point, that projection should be perpendicular to the black line, or I messed up my code.
Sep 20, 2015 at 9:47 comment added Dalek are the red lines representing the distances has been computed as perpendicular distances?
Sep 19, 2015 at 21:31 comment added Dalek I appreciate your answer but I am not very familiar with R but I know python. It would be absolutely great if you could post your example by python too, or write just the math if you think it is sufficient.
Sep 19, 2015 at 21:12 history answered Creosote CC BY-SA 3.0