4
$\begingroup$

I'm curious about the responses to some questions I've found on this site -- but I can't post comments yet, so I am starting a new question. At least two people have asked about regression to predict a dependent variable which is a ranking from 1...N where N is the number of datapoints. They have been given suggestions dealing with parts of their problem not related to my concern here -- a general response seems to be using ordinal regression. But in ordinal regression, it's at least theoretically possible that all the rank variables could have the same value.

I am assuming that ranking in the examples below are actually exclusive: only one datum can be rank 1, only one can be rank 2, etc. It seems to me that the dependent (Y) variables, then, are necessarily NOT iid.

Here is the question I am responding to:

Regression with rank order as dependent variable

And for the one below, I notice that there is no response. Is there no real method for this kind of regression?

How can I estimate a "rank" dependent variable in a multivariate dataset?

My question, I suppose, is this: am I right in thinking that rank data are necessarily not IID, and therefore we cannot use standard statistics on them? I know its hard to prove a negative, but I'd appreciate any expertise on this subject. (I don't have a specific test or project in mind: just trying to expand my understanding.)

[edit] Realized I had to add this detail to my question: can ordinal regression really be used in the case where we are ranking data, without violating assumptions?

[edit 2] removed a link .. not sure what the correct one was

I think this is the clearest example: regressing the ranking of countries by GDP on some set of predictors. So the ranks, not the GDP. (I'm not sure why anyone would want to do this, but its a hypothetical example.)

$\endgroup$

1 Answer 1

5
$\begingroup$

You are correct... but we need to be careful.

Ranked data cannot be identically and independently distributed. For example, if you pull $Y_1$, then pull $Y_2$ where $Y_2>Y_1$, then $Y_1$ and $Y_2$ are not independent because the range of possible values for $Y_2$ is necessarily smaller than what it would be beforehand, meaning that the pdf for $Y_2$ must be different from the pdf of $Y_1$.

When I say "we need to be careful," I mean that you need to differentiate between observations that are necessarily ranked and observations that you rank yourselves.

You can of course pull $Y_1$ and $Y_2$ independently and from the same distribution, then rank them. If $Y_{(1)}=\min\{Y_1,Y_2\}$ and $Y_{(2)}=\max\{Y_1,Y_2\}$, the distributions of $Y_{(1)}$ and $Y_{(2)}$ are not independent whereas the distributions of $Y_1$ and $Y_2$ clearly are.

$\endgroup$
3
  • $\begingroup$ Thanks! So I guess I understand the interdependence of ranking in the hypothetical case I'm thinking about. But what would be an example of observations that are necessarily ranked? Ie, in what case could ranking be i.i.d. even when rankings are exclusive? (by exclusive, i mean only one datum per rank) $\endgroup$ Commented Mar 26, 2016 at 3:58
  • $\begingroup$ and -- i had to add a detail to my question: does this non-iid mean we can't use something like ordinal regression on ranked data? $\endgroup$ Commented Mar 26, 2016 at 4:04
  • $\begingroup$ Re-reading your answer, I think perhaps I was a bit unclear: I'm talking specifically about the rankings themselves are the dependent variable. So it's closer to your latter example, comparing Y<sub>(1)</sub> and Y<sub>(2)</sub> but for a case where you're dealing with population level data. Eg, regressing the ranking of countries by GDP on some set of predictors. $\endgroup$ Commented Mar 26, 2016 at 23:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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