Difference between revisions of "Notes:Distribution of the sample median"
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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. | ||
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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. | ||
| + | * {{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 | ||
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| + | |||
| + | 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. | ||
Revision as of 03:29, 12 December 2017
\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} } \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
- 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
- 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
- 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)
- \implies
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.
