Notes:Distribution of the sample median

$\newcommand{\P}[2][]{\mathbb{P}#1{\left[{#2}\right]} } \newcommand{\Pcond}[3][]{\mathbb{P}#1{\left[{#2}\!\ \middle\vert\!\ {#3}\right]} } \newcommand{\Plcond}[3][]{\Pcond[#1]{#2}{#3} } \newcommand{\Prcond}[3][]{\Pcond[#1]{#2}{#3} }$

$\newcommand{\E}[1]{ {\mathbb{E}{\left[{#1}\right]} } }$$\newcommand{\Mdm}[1]{\text{Mdm}{\left({#1}\right) } }$$\newcommand{\Var}[1]{\text{Var}{\left({#1}\right) } }$$\newcommand{\ncr}[2]{ \vphantom{C}^{#1}\!C_{#2} }$$\newcommand{\O}[0]{\mathcal{O} } \newcommand{\M}[0]{\mathcal{M} } \newcommand{\Q}[0]{\mathcal{Q} } \newcommand{\Min}[1]{\text{Min}\left({#1}\right)}$

Problem overview

Let $X_1,\ldots,X_{2m+1}$ be a sample from a population $X$, meaning that the $X_i$ are i.i.d random variables, for some $m\in\mathbb{N}_{0}$. We wish to find:

• $$\P{\text{Median}(X_1,\ldots,X_{2m+1})\le r}$$ - the Template:Cdf of the median.

Initial work

Since the variables are independent then any ordering is as likely as any other (which I proved the long way, rather than just jumping to

$$\frac{1}{(2m+1)!}$$ - silly me) however the result, found in Probability of i.i.d random variables being in an order and not greater than something will be useful.

I believe the $\P{\text{Median}(X_1,\ldots,X_{2m+1})\le r}\eq\Pcond{X_1\le\cdots\le X_{m+1}\le r}{X_1\le\cdots\le X_{2m+1} }$. Let us make some definitions to make this shorter.

• $\mathcal{O}:\eq X_1\le\cdots\le X_{2m+1}$ - representing the order part
• $\mathcal{M}:\eq X_1\le\cdots\le X_{m+1}\le r$ - representing the median part
• $\mathcal{Q}:\eq\P{\text{Median}(X_1,\ldots,X_{2m+1})\le r}\eq\Pcond{\mathcal{O} }{\mathcal{O} }$ - representing the question

We should also have some sort of converse, related to $r\le X_{m+2}\le\cdots X_{2m+1}$ or something.

We also have:

• An expression for $\P{X_1\le \cdots\le X_n\le r}$ from Probability of i.i.d random variables being in an order and not greater than something
  • It's $$\eq\frac{1}{n!}F_X(r)^n$$

Analysis

Let us look at $X\le r$ and $X\le Y$ to see what we can say if both are true (the "and")

• Claim: $(X\le r\wedge X\le Y)\iff(X\le\Min{r,Y})$
• Proof:
  • $\implies$
    1. Suppose $r\le Y$, so $\Min{r,Y}\eq r$, obviously $X\le r\ \implies\ X\le r\eq\Min{r,Y}$, so the implication holds in this case
    2. Suppose $Y\le r$, so $\Min{r,Y}\eq Y$, obviously $X\le Y\ \implies\ X\le Y\eq\Min{r,Y}$, so the implication holds in this case too.
  • $\impliedby$
    • We notice either $\Min{r,Y}\eq r$ if $r\le Y$, or $\Min{r,Y}\eq Y$ if $Y\le r$ (slightly modify the language for the equality, it doesn't matter though really)
      • Thus if $r\le Y$ then $X\le r$ and as $r\le Y$ by assumption, we use the transitivity of $\le$ to see $X\le r\le Y$ thus $X\le Y$ too - as required
      • Thus if $Y\le r$ then $X\le Y$ and as $Y\le r$ by assumption, we use the transitivity of $\le$ to see $X\le Y\le r$ and thus $X\le r$ too - as required.
    • So in either case, we have $X\le Y$ and $X\le r$ - as required

Problem statement

Thus we really want to find:

• $\P{\text{Median}(X_1,\ldots,X_{2m+1})\le r}\eq\Pcond{X_1\le\cdots\le X_{m+1}\le r}{X_1\le\cdots\le X_{2m+1} }$

  $$\eq\frac{\P{\M\ \text{and}\ \O} }{\P{\O} }$$

  $$\eq \big((2m+1)!\big)\P{X_1\le\cdots\le X_{m+1}\le\Min{r,X_{m+2} }\le X_{m+2}\le X_{m+3}\cdots\le X_{2m+1} }$$

  • Caveat:We now need: $$\big(X\le r\wedge X\le Y\le Z\big)\implies\big(X\le\Min{r,Y}\le Y\le Z\big)$$ to justify this format. Although that's arguably not that helpful for the integral.