# Which cumulative distribution of F(X) is equal to the cumulative distribution of its sample median (as sample statistics)

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: $$F=F_{tn}$$?

• For which $n$ do you require these two distributions to be equal? Just one $n$ (presumably not $n=1,$ though!) or for all $n$?
– whuber
Oct 5 '18 at 13:15
• Thanks for your answer, this is the question that is asked. maybe we can ignore n=1 Oct 5 '18 at 19:24
• Right. It may be useful to know $n=1$ and $n=2$ don't impose any constraints.
– whuber
Oct 5 '18 at 20:08

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.

• Thanks for your answer. I read it several times, but I don't understand the what the answer is. Bernoulli or Binominal? Oct 5 '18 at 19:42
• 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.
– whuber
Oct 5 '18 at 20:08
• 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})? Oct 5 '18 at 20:12
• Both are immediate applications of the definitions.
– whuber
Oct 5 '18 at 20:15
• 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. Oct 5 '18 at 20:23