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We consider random sampling from a population in which the variable of interest $X$ has some cumulative distribution $F$. Next, we consider a simple random sample of size $n, X_1,\ldots,X_n,$ which are i.i.d. $F$. We then compute a sample statistic $T_n=T(X_1\ldots,X_n)$. Subsequently, we can think about the cumulative distribution of $T_n,$ which I will call $F_{tn}.$

The question is the following: given that we use $T_n$ = "sample median" -- thus, our sample statistic is the sample median -- for which distribution $F$ does the following hold: $F=F_{tn}$?

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  • $\begingroup$ For which $n$ do you require these two distributions to be equal? Just one $n$ (presumably not $n=1,$ though!) or for all $n$? $\endgroup$
    – whuber
    Oct 5 '18 at 13:15
  • $\begingroup$ Thanks for your answer, this is the question that is asked. maybe we can ignore n=1 $\endgroup$
    – Data Lover
    Oct 5 '18 at 19:24
  • $\begingroup$ Right. It may be useful to know $n=1$ and $n=2$ don't impose any constraints. $\endgroup$
    – whuber
    Oct 5 '18 at 20:08
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Let $F$ be the distribution of $X,$ as in the question, and $F_n$ be the distribution for a median $T_n$ of a sample of $n$ iid variables $X_i$ distributed like $X.$ (Since the median is usually not uniquely defined for even $n,$ we have to be a little open-minded about $T_n$ and $F_n$! More about this below.) Then, by the very definition of any median, for all real numbers $x$ the event $$T_n \le x$$ is identical to the event "at least $n/2$ of the $X_i$ are less than or equal to $x.$" For any $i,$ the chance of $X_i \le x$ is given by $F(x)$ and because the $X_i$ are independent, the chance that $T_n \le x$ is given by a tail probability for the Binomial distribution with parameters $n$ and $F(x).$

Suppose $F=F_n$ for some $n$ to be determined as appropriate to produce interesting results. The cases $n=1$ and $n=2$ yield no useful information, but consider the case $n=3.$ Then because "at least $n/2$" means "$2$ or $3$," we obtain

$$F(x) = F_3(x) = \Pr(T_3 \le x) = \binom{3}{2}F(x)^2(1-F(x)) + \binom{3}{3}F(x)^3.$$

This is a cubic equation in $F(x)$ whose only solutions are $\{0,1/2,1\}.$ This limits $X$ to random variables having positive probabilities of attaining at most two values (namely, the point where the jump from $0$ to $1/2$ occurs and the point where the jump from $1/2$ to $1$ occurs). For nonconstant $X$ these are the affine transformations of Bernoulli$(1/2)$ variables. In other words, either $X$ is constant or there exist distinct numbers $x_0, x_1$ and $$\Pr(X=x_0) = \Pr(X=x_1)=\frac{1}{2}.$$

For such variables to have the same distribution as the median $T_n$, you need to define medians in such a way that guarantees they are possible values of $X.$ However you do it, it's clear that for these "affine Bernoulli" variables the median also has equal chances of being either $x_0$ or $x_1$ and so obviously has the same distribution as $X$ itself.

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  • $\begingroup$ Thanks for your answer. I read it several times, but I don't understand the what the answer is. Bernoulli or Binominal? $\endgroup$
    – Data Lover
    Oct 5 '18 at 19:42
  • $\begingroup$ The answer is "the affine transformations of Bernoulli(1/2) variables." The meaning of this is explained by the final equation, beginning with the bold text preceding it. $\endgroup$
    – whuber
    Oct 5 '18 at 20:08
  • $\begingroup$ Many thanks for your answer. I'm sorry but I dont understand why you use Binomial distribution? and why this (This is a cubic equation in F(x) whose only solutions are {0,1/2,1})? $\endgroup$
    – Data Lover
    Oct 5 '18 at 20:12
  • $\begingroup$ Both are immediate applications of the definitions. $\endgroup$
    – whuber
    Oct 5 '18 at 20:15
  • $\begingroup$ I'm sorry but I read several times and It is not understandable for me. However, I will try my best again. Billion Thanks for your reply. $\endgroup$
    – Data Lover
    Oct 5 '18 at 20:23

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