View source for 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]].
{{ProbMacros}}
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==Set up==
Let {{M|m\in\mathbb{N}_0}} describe the size of the sample, {{M|n}}, by the relation: {{M|n\eq 2m+1}} - forcing {{M|n}} to be [[odd number|odd]].

Let {{M|X_1,\ldots,X_{2m+1} }} be {{iid}} samples from a population distribution {{M|X}}; let {{M|M}} denote the {{link|median|sample}} of the sample, {{M|X_1,\ldots,X_n}}, and let {{M|F(r):\eq \P{X_i\le r}\eq\P{X\le r} }} for any {{M|i}}<ref group="Note">As the {{M|X_i}} are {{iid}} this will be the same for all {{M|i}}</ref> then:
* For {{M|m\eq 1}} / {{M|n\eq 3}} we have: {{M|\P{M\le r}\eq F(r)^2\big(-2F(r)+3\big)}}
* For {{M|m\eq 2}} / {{M|n\eq 5}} we have: {{M|\P{M\le r}\eq F(r)^3\big(6F(r)^2-15F(r)+10\big)}}
* For {{M|m\eq 3}} / {{M|n\eq 7}} we have: {{M|\P{M\le r}\eq F(r)^4\big(-20F(r)^3+70F(r)^2-84F(r)+35\big)}}
* For {{M|m\eq 4}} / {{M|n\eq 9}} we have: {{M|\P{M\le r}\eq F(r)^5\big(70F(r)^4-315F(r)^3+540F(r)^2-420F(r)+126\big)}}<sup>{{Caveat|}}</sup><ref group="Note">This one is predicted from a formula I found however this and further predictions seem to work. They have not been experimentally confirmed to the same high standard as previous results though</ref>
===In general===
: {{Warning|This is written from memory, not from my notes! - Alec check the notes! [[User:Alec|Alec]] ([[User talk:Alec|talk]]) 21:57, 19 December 2017 (UTC)}}
In general I believe for {{M|m\in\mathbb{N}_0}} given and {{M|n:\eq 2m+1}} that:
* {{MM|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 {{M|x^k}} factor removed for our {{M|\P{M\le r} }} equations above. We must then multiply this reversed polynomial by {{M|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 {{M|X\sim\text{Exp}(\lambda)}} for any {{M|\lambda\in\mathbb{R} }} and {{M|\lambda>0}} 
Then {{M|\E{X} }} is above the true median of {{M|X}}, which is {{MM|\frac{\ln(2)}{\lambda} }} but appears to show convergence

Continuing the exponential example, the following hold for all real {{M|\lambda>0}}, here {{M|M}} is the true median of the distribution, {{M|M_i}} is the median random variable for {{M|n\eq i}} samples and the constants are accurate to [[significant figures|2 s.f]]
* {{M|0.83\E{M_3}\approx M}}
* {{M|0.88\E{M_5}\approx M}}
* {{M|0.91\E{M_7}\approx M}}
* {{M|0.93\E{M_9}\approx M}}
* {{M|0.94\E{M_{11} } \approx M}}
* {{M|0.95\E{M_{13} } \approx M}}
* {{M|0.96\E{M_{15} } \approx M}}



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

==Ideas==
# 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:
#* {{MM|\E{X}\eq\int_0^\infty \P{X\ge x}\mathrm{d} x}} - but the requirement for {{M|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.

==Notes==
<references group="Note"/>

=OLD WORK - unsaved addendum==
==Reversing polynomial order==
Let {{MM|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}}<ref group="Note">Usually we use indexes starting at one, so for a polynomial of order n we would write {{M|\sum^{n+1}_{i\eq 1}a_ix^{i-1} }} but the convenience outweighs the "minor gains" (if any) we'd make here</ref> be a [[polynomial]] in {{M|x}}

Our goal is to "reverse the [[coefficients]]" of {{M|f}}, to obtain a polynomial {{M|f'}} such that:
* {{MM|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: {{MM|f'(x)\eq a_n+a_{n-1}x+a_{n-2}x^2+\cdots+a_1x^{n-1}+a_0x^n}}

==Process==
Let {{M|f(x)}} be a {{M|n^\text{th} }}-[[order (polynomial)|order]] polynomial over some field {{XXX|TODO - just assume {{M|\mathbb{R} }} for now}}, then:
* Define {{M|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 {{M|x\neq 0}} }}
** Define {{M|f_2(x):\eq f_1\left(\frac{1}{x}\right)}} so {{M|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}}
**: {{M|\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==
<references group="Note"/>

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