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

Revision as of 07:30, 11 December 2017 (view source) Alec (Talk | contribs) (Saving work)		Revision as of 03:29, 12 December 2017 (view source) Alec (Talk | contribs) (Saving work, added more work!) Newer edit →
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−	{{ProbMacros}}	+	{{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.
		+
		+
		+	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.
		+	* {{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]]
		+	** It's {{MM|\eq\frac{1}{n!}F_X(r)^n}}
		+	===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}}")
		+	* '''Claim:''' {{M|(X\le r\wedge X\le Y)\iff(X\le\Min{r,Y})}}
		+	* '''Proof:'''
		+	** {{M|\implies}}
		+	**# 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
		+	**# 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.
		+	** {{M|\impliedby}}
		+	*** 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
		+	**** 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.
		+	*** So in either case, we have {{M|X\le Y}} and {{M|X\le r}} - as required
		+	==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} } }}
		+	*: {{MM|\eq\frac{\P{\M\ \text{and}\ \O} }{\P{\O} } }}
		+	*: {{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} } }}
		+	** 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.

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Revision as of 03:29, 12 December 2017

$\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:

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

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

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

  • Notice that we use $(X\le\Min{r,Y})\iff(X\le r\wedge X\le Y)$ here. Caveat:But is it enough to get $(X\le r\wedge X\le Y\le Z)\iff(X\le\Min{r,Y}\le Z)$? - we only need an implication.