Distribution of the sample median

Warning:This page is currently in the "notes" stage, and is a staging area from conclusions drawn from Notes:Distribution of the sample median

• This page is not "formal" yet. However the contents are accurate (to whatever they apply to) - I hope to distil the essence of an "ordered" unit from Alec's taxonomy of units and describe the median's distribution purely on that foundation, then anything which is itself ordinal (in theory both additive and real units) will be a corollary.

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

$\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} }$

Set up

Let $m\in\mathbb{N}_0$ describe the size of the sample, $n$, by the relation: $n\eq 2m+1$ - forcing $n$ to be odd.

Let $X_1,\ldots,X_{2m+1}$ be i.i.d samples from a population distribution $X$; let $M$ denote the median of the sample, $X_1,\ldots,X_n$, and let $F(r):\eq \P{X_i\le r}\eq\P{X\le r}$ for any $i$ then:

• For $m\eq 1$ / $n\eq 3$ we have: $\P{M\le r}\eq F(r)^2\big(-2F(r)+3\big)$
• For $m\eq 2$ / $n\eq 5$ we have: $\P{M\le r}\eq F(r)^3\big(6F(r)^2-15F(r)+10\big)$
• For $m\eq 3$ / $n\eq 7$ we have: $\P{M\le r}\eq F(r)^4\big(-20F(r)^3+70F(r)^2-84F(r)+35\big)$
• For $m\eq 4$ / $n\eq 9$ we have: $\P{M\le r}\eq F(r)^5\big(70F(r)^4-315F(r)^3+540F(r)^2-420F(r)+126\big)$^Caveat:

In general

Warning:This is written from memory, not from my notes! - Alec check the notes! Alec (talk) 21:57, 19 December 2017 (UTC)

In general I believe for $m\in\mathbb{N}_0$ given and $n:\eq 2m+1$ that:

• $$g(x):\eq \sum_{i\eq 0}^m{}^{(2m+1)}C_i * (x-1)^i$$ - which we then expand and reverse the coefficients of to obtain the polynomials in the brackets with the $x^k$ factor removed for our $\P{M\le r}$ equations above. We must then multiply this reversed polynomial by $x^{m+1}$ I believe and job done!

Findings

The median seems to be a rather crappy estimator for the median of a distribution, for example with the Exponential distribution, so $X\sim\text{Exp}(\lambda)$ for any $\lambda\in\mathbb{R}$ and $\lambda>0$ Then $\E{X}$ is above the true median of $X$, which is

$$\frac{\ln(2)}{\lambda}$$ but appears to show convergence

Continuing the exponential example, the following hold for all real $\lambda>0$, here $M$ is the true median of the distribution, $M_i$ is the median random variable for $n\eq i$ samples and the constants are accurate to 2 s.f

• $0.83\E{M_3}\approx M$
• $0.88\E{M_5}\approx M$
• $0.91\E{M_7}\approx M$
• $0.93\E{M_9}\approx M$
• $0.94\E{M_{11} } \approx M$
• $0.95\E{M_{13} } \approx M$
• $0.96\E{M_{15} } \approx M$

However $X$ were normally distributed then the $M_{2m+1}$s would not be biased and multiplying by these constants would be bad.

Ideas

1. Use Alec's alternate expectation formula to try and find the expected value - again this is written from memory not notes but it states something like:
   • $$\E{X}\eq\int_0^\infty \P{X\ge x}\mathrm{d} x$$ - but the requirement for $X\ge 0$ may not be needed (it could work for ALL RVs maybe - can't see my notes right now)... I'm not sure.

OLD WORK - unsaved addendum=

Reversing polynomial order

Let

$$f(x):\eq\sum^n_{i\eq 0}a_ix^i\eq a_0+a_1x+a_2x^2+\cdots+a_{n-1}x^{n-1}+a_nx^n$$ be a polynomial in $x$

Our goal is to "reverse the coefficients" of $f$, to obtain a polynomial $f'$ such that:

• $$f'(x)\eq \sum^n_{i\eq 0}a_ix^{n-i} \eq\sum^n_{i\eq 0}a_{n-i}x^i\eq a_0x^n+a_1x^{n-1}+a_2x^{n-2}+\cdots+a_{n-1}x+a_n$$ , perhaps better written as: $$f'(x)\eq a_n+a_{n-1}x+a_{n-2}x^2+\cdots+a_1x^{n-1}+a_0x^n$$

Process

Let $f(x)$ be a $n^\text{th}$-order polynomial over some field

TODO: TODO - just assume $\mathbb{R}$ for now

, then:

• Define $f_1(x):\eq \frac{1}{x^n}f(x)\eq x^{-n}f(x)\eq\sum^n_{i\eq 0}a_ix^{i-n}\eq a_0x^{-n}+a_1x^{1-n}+a_2x^{2-n}+\cdots+ a_{n-1}x^{-1}+a_n$ - Caveat:Provided $x\neq 0$
  • Define $f_2(x):\eq f_1\left(\frac{1}{x}\right)$ so $f_2(x)\eq \sum^n_{i\eq 0}a_ix^{-(i-n)}\eq a_0x^n+a_1x^{-(1-n)}+a_2x^{-(2-n)}+\cdots+ a_{n-1}x^{-(-1)}+a_n$

    $\eq a_0x^n+a_1x^{n-1}+a_2x^{n-2}+\cdots+a_{n-1}x^1+a_n$ - which is what we want!

Notes

Found this in an old tab, I then made the above content, so adding it here