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Suppose $X_1,\ldots,X_n$ are continuous random variables and IID with sample median $m$ for some $n\geq 2$. Is it then true that $1(X_1\geq m),\ldots,1(X_n\geq m)$ are IID?

Here the definition of $1(X_i\geq m)$ is that it is 1 if $X_i\geq m$ and 0 otherwise.

I ask because I often see researchers create these dummy variables (i.e. indicator variables) in regression analyses and I am worried that there is dependence between them by design (i.e. by construction). Mathematically I feel that since the dummy variables are functions of independent random variables, they ought to be independent. On the other hand, by the law of large numbers, for large $n$ we intuitively have that $(1/n)\sum_i1(X_i\geq m)$ is approximately equal to the expectation of $1(X_i\geq m)$ with $m$ being close to the population median, i.e. 0.5, with probability 1, suggesting that there is a linear dependence between the dummy variables for large $n$. (EDIT: Since $m$ is the sample median, we also know that $(1/n)\sum_i1(X_i\geq m)$ is at least 0.5.) I am unsure what I should conclude from this simple line of reasoning and unable to proceed in a more mathematically rigorous manner.

EDIT: I guess the answer to my question is simple. The indicator variables are not independent because all of them cannot be equal to 1 by the definition of the sample median.

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  • $\begingroup$ This seems to take care of the "independent" part. $\endgroup$
    – Dave
    Jul 21 at 20:29
  • $\begingroup$ I don't see how the law of large numbers "... imply[s] that there is a linear dependence between the dummy variables...". This seems to me to be related to the classical fallacy that, e.g., a run of heads in a coin-flip experiment must make tails more likely so that the fraction of heads can get closer to 0.5. Am I misunderstanding something? $\endgroup$
    – jbowman
    Jul 21 at 20:43
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    $\begingroup$ The ambiguity of the question is whether or not the OP means the population median (as opposed to the sample median). $\endgroup$
    – Xi'an
    Jul 21 at 21:44
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    $\begingroup$ @Xi'an Excellent point. The question is a little more interesting when the sample median is intended, for then these indicators are exchangeable but not independent. $\endgroup$
    – whuber
    Jul 22 at 16:30
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    $\begingroup$ @Xi'an Thanks for emphasizing this. Here I meant $m$ to be equal to the sample median of $(X_1,\ldots,X_n)$. My question has been updated. $\endgroup$
    – Elias
    Jul 23 at 5:32

2 Answers 2

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A fairly rigorous argument is that $$\mathbb P(X_1\ge m(X_1,\ldots,X_n),\ldots,X_n\ge m(X_1,\ldots,X_n))=0\ne \mathbb P(X_1\ge m(X_1,\ldots,X_n))^n$$

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    $\begingroup$ What is interesting about the question is that although it specifies a continuous distribution, I believe the result is true for all distributions (except for the obvious exceptions $n=1$ and when the $X_i$ are constant a.s.) But for discrete distributions your final inequality does not necessarily hold. $\endgroup$
    – whuber
    Jul 23 at 23:26
  • $\begingroup$ Right, all $X_i$'s could be identical. $\endgroup$
    – Xi'an
    Jul 24 at 8:52
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Another argument is that:

  1. If $n$ is even, then $\sum_{i=1}^n1(x_i \geq m) = n/2$, therefore the $1(x_i \geq m)$ cannot be independent,

  2. If $n$ is odd, then $\sum_{i=1}^n1(x_i \geq m) = (n+1)/2$, therefore the $1(x_i \geq m)$ cannot be independent.

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    $\begingroup$ The two "therefores" still need justification. Admittedly it is easy for continuous variables--but it still needs to be there. $\endgroup$
    – whuber
    Jul 23 at 23:24

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