Difference between revisions of "Notes:Distribution of the sample median"

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(Saving work, added more work!)
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{{ProbMacros}}
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{{ProbMacros}}{{M|\newcommand{\O}[0]{\mathcal{O} } \newcommand{\M}[0]{\mathcal{M} } \newcommand{\Q}[0]{\mathcal{Q} } \newcommand{\Min}[1]{\text{Min}\left({#1}\right)} }}
 
__TOC__
 
__TOC__
 
==Problem overview==
 
==Problem overview==
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==Initial work==
 
==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 {{MM|\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.
 
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 {{MM|\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.
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 +
 +
I believe the {{M|\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.
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* {{M|\mathcal{O}:\eq X_1\le\cdots\le X_{2m+1} }} - representing the order part
 +
* {{M|\mathcal{M}:\eq X_1\le\cdots\le X_{m+1}\le r}} - representing the median part
 +
* {{M|\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 {{M|r\le X_{m+2}\le\cdots X_{2m+1} }} or something.
 +
 +
 +
We also have:
 +
* An expression for {{M|\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]]
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** It's {{MM|\eq\frac{1}{n!}F_X(r)^n}}
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===Analysis===
 +
Let us look at {{M|X\le r}} and {{M|X\le Y}} to see what we can say if both are true (the "{{link|and|logic}}")
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* '''Claim:''' {{M|(X\le r\wedge X\le Y)\iff(X\le\Min{r,Y})}}
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* '''Proof:'''
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** {{M|\implies}}
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**# Suppose {{M|r\le Y}}, so {{M|\Min{r,Y}\eq r}}, obviously {{M|X\le r\ \implies\ X\le r\eq\Min{r,Y} }}, so the implication holds in this case
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**# Suppose {{M|Y\le r}}, so {{M|\Min{r,Y}\eq Y}}, obviously {{M|X\le Y\ \implies\ X\le Y\eq\Min{r,Y} }}, so the implication holds in this case too.
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** {{M|\impliedby}}
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*** We notice either {{M|\Min{r,Y}\eq r}} if {{M|r\le Y}}, or {{M|\Min{r,Y}\eq Y}} if {{M|Y\le r}} (slightly modify the language for the equality, it doesn't matter though really)
 +
**** Thus if {{M|r\le Y}} then {{M|X\le r}} and as {{M|r\le Y}} by assumption, we use the {{link|transitivity|relation}} of {{M|\le}} to see {{M|X\le r\le Y}} thus {{M|X\le Y}} too - as required
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**** Thus if {{M|Y\le r}} then {{M|X\le Y}} and as {{M|Y\le r}} by assumption, we use the transitivity of {{M|\le}} to see {{M|X\le Y\le r}} and thus {{M|X\le r}} too - as required.
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*** So in either case, we have {{M|X\le Y}} and {{M|X\le r}} - as required
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==Problem statement==
 +
Thus we really want to find:
 +
* {{M|\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} } }}
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*: {{MM|\eq\frac{\P{\M\ \text{and}\ \O} }{\P{\O} } }}
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*: {{MM|\eq \big((2m+1)!\big)\P{X_1\le\cdots\le X_{m+1}\le\Min{r,X_{m+2} }\le X_{m+3}\cdots\le X_{2m+1} } }}
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** Notice that we use {{M|(X\le\Min{r,Y})\iff(X\le r\wedge X\le Y)}} here. {{Caveat|But is it enough to get {{M|(X\le r\wedge X\le Y\le Z)\iff(X\le\Min{r,Y}\le Z)}}?}} - we only need an implication.

Revision as of 03:29, 12 December 2017

[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:

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+3}\cdots\le X_{2m+1} } [/math]
    • Notice that we use [ilmath](X\le\Min{r,Y})\iff(X\le r\wedge X\le Y)[/ilmath] here. Caveat:But is it enough to get [ilmath](X\le r\wedge X\le Y\le Z)\iff(X\le\Min{r,Y}\le Z)[/ilmath]? - we only need an implication.
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