Notes:Distribution of the sample median

[ilmath]\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} }[/ilmath]

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

Problem overview

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

• [math]\P{\text{Median}(X_1,\ldots,X_{2m+1})\le r} [/math] - 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 [math]\frac{1}{(2m+1)!} [/math] - 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 [ilmath]\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} } [/ilmath]. Let us make some definitions to make this shorter.

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

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

We also have:

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

Analysis

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

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

Problem statement

Thus we really want to find:

• [ilmath]\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} } [/ilmath]

  [math]\eq\frac{\P{\M\ \text{and}\ \O} }{\P{\O} } [/math]

  [math]\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} } [/math]

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