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

Revision as of 06:10, 12 December 2017 (view source) Alec (Talk | contribs) m (→‎Problem statement) ← Older edit		Revision as of 07:15, 12 December 2017 (view source) Alec (Talk | contribs) m (Saving work) Newer edit →
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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)} }}	+	{{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)} \newcommand{\d}[0]{\mathrm{d} } }}
	__TOC__		__TOC__
	==Problem overview==		==Problem overview==
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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+2}\le X_{m+3}\cdots\le X_{2m+1} } }}		*: {{MM|\eq \big((2m+1)!\big)\P{X_1\le\cdots\le X_{m+1}\le\Min{r,X_{m+2} }\le X_{m+2}\le X_{m+3}\cdots\le X_{2m+1} } }}
	** {{Caveat|We now need:}} {{MM|\big(X\le r\wedge X\le Y\le Z\big)\implies\big(X\le\Min{r,Y}\le Y\le Z\big)}} to justify this format. Although that's arguably not that helpful for the integral.		** {{Caveat|We now need:}} {{MM|\big(X\le r\wedge X\le Y\le Z\big)\implies\big(X\le\Min{r,Y}\le Y\le Z\big)}} to justify this format. Although that's arguably not that helpful for the integral.
		+	==Initial integral==
		+	: This isn't about the median specifically, this is just looking at the specific integral.
		+	Suppose we have a sample of length 3, {{M|X,Y,Z}} then we are looking at:
		+	* {{M|\P{X\le\Min{r,Y}\le Y\le Z\le t} }} (where {{M|t}} will be used for a limit towards {{m|\infty}} to get {{M|\P{X\le \Min{r,Y}\le Y\le Z} }} in the end), or as an integral:
		+	** {{MM|\int^t_{-\infty}f(z)\left(\int^z_{-\infty}f(y)\left(\int^{\Min{r,y} }_{-\infty} f(x)\d x\right)\d y\right)\d z}}
		+	*** if {{M|t>r}} then the minimum will get involved (for some {{M|z}}s anyway) and limit it to {{M|r}}, otherwise it'll always stay under {{M|r}} - of course in practice (as we'll take {{M|t\rightarrow\infty}}) this will certainly happen.

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Revision as of 07:15, 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)} \newcommand{\d}[0]{\mathrm{d} }$

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:

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

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

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

  • Caveat:We now need: $$\big(X\le r\wedge X\le Y\le Z\big)\implies\big(X\le\Min{r,Y}\le Y\le Z\big)$$ to justify this format. Although that's arguably not that helpful for the integral.

Initial integral

This isn't about the median specifically, this is just looking at the specific integral.

Suppose we have a sample of length 3, $X,Y,Z$ then we are looking at:

• $\P{X\le\Min{r,Y}\le Y\le Z\le t}$ (where $t$ will be used for a limit towards $\infty$ to get $\P{X\le \Min{r,Y}\le Y\le Z}$ in the end), or as an integral:
  • $$\int^t_{-\infty}f(z)\left(\int^z_{-\infty}f(y)\left(\int^{\Min{r,y} }_{-\infty} f(x)\d x\right)\d y\right)\d z$$
    • if $t>r$ then the minimum will get involved (for some $z$s anyway) and limit it to $r$, otherwise it'll always stay under $r$ - of course in practice (as we'll take $t\rightarrow\infty$) this will certainly happen.