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

Revision as of 17:10, 16 December 2017 (view source) Alec (Talk | contribs) m (Adding conclusion getting ready for next section) ← Older edit		Revision as of 17:50, 16 December 2017 (view source) Alec (Talk | contribs) (Fixing some typos, summing up steps) 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)} \newcommand{\d}[0]{\mathrm{d} } }}		{{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__
		+	==Important results==
		+	# {{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{X_1\le\cdots\le X_{m+1}\le\Min{r,X_{m+2} }\le X_{m+2}\le X_{m+3}\le\cdots\le X_{2m+1} } }{\frac{1}{(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}\le\cdots\le X_{2m+1} } }}
		+	#: {{MM|\eq \lim_{t\rightarrow+\infty}\Bigg(\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}\le\cdots\le X_{2m+1}\le t }\Bigg) }}
		+	#: {{MM|\eq\frac{(2m+1)!}{m!}\lim_{t\rightarrow+\infty}\Bigg[\int^t_{-\infty}f(x_{2m+1})\left(\int^{x_{2m+1} }_{-\infty}f(x_{2m})\left(\cdots\int^{x_{m+3} }_{-\infty}f(x_{m+2})\left(\int^{\Min{r,x_{m+2} } }_{-\infty} f(x_{m+1})F(x_{m+1})^m\d x_{m+1}\right)\d x_{m+2}\cdots\right)\d x_{2m}\right)\d x_{2m+1}\Bigg] }}
		+
	==Problem overview==		==Problem overview==
	Let {{M|X_1,\ldots,X_{2m+1} }} be a sample from a population {{M|X}}, meaning that the {{M|X_i}} are {{iid}} [[random variables]], for some {{M|m\in\mathbb{N}_{0} }}. We wish to find:		Let {{M|X_1,\ldots,X_{2m+1} }} be a sample from a population {{M|X}}, meaning that the {{M|X_i}} are {{iid}} [[random variables]], for some {{M|m\in\mathbb{N}_{0} }}. We wish to find:
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	* {{M|\mathcal{O}:\eq X_1\le\cdots\le X_{2m+1} }} - representing the order part		* {{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{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	+	* {{M|\mathcal{Q}:\eq\P{\text{Median}(X_1,\ldots,X_{2m+1})\le r}\eq\Pcond{\mathcal{M} }{\mathcal{O} } }} - representing the question
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	We substitute this back in to yield:		We substitute this back in to yield:
	* {{MM|\frac{1}{m!}\int^t_{-\infty}f(x_{2m+1})\left(\int^{x_{2m+1} }_{-\infty}f(x_{2m})\left(\cdots\int^{x_{m+3} }_{-\infty}f(x_{m+2})\left(\int^{\Min{r,x_{m+2} } }_{-\infty} f(x_{m+1})F(x_{m+1})^m\d x_{m+1}\right)\d x_{m+2}\cdots\right)\d x_{2m}\right)\d x_{2m+1} }}		* {{MM|\frac{1}{m!}\int^t_{-\infty}f(x_{2m+1})\left(\int^{x_{2m+1} }_{-\infty}f(x_{2m})\left(\cdots\int^{x_{m+3} }_{-\infty}f(x_{m+2})\left(\int^{\Min{r,x_{m+2} } }_{-\infty} f(x_{m+1})F(x_{m+1})^m\d x_{m+1}\right)\d x_{m+2}\cdots\right)\d x_{2m}\right)\d x_{2m+1} }}
		+	===Conclusion of progression 1===
		+	We see here that
	==Progression: 2==		==Progression: 2==
	This'll involve induction and dealing with the {{M|\text{Min}()}} will be "tricky", both for practice and induction we will consider the special cases {{M|m\eq 1}} and {{M|m\eq 2}} by evaluating:		This'll involve induction and dealing with the {{M|\text{Min}()}} will be "tricky", both for practice and induction we will consider the special cases {{M|m\eq 1}} and {{M|m\eq 2}} by evaluating:
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	===Conclusion of progression 2===		===Conclusion of progression 2===
	* {{MM|\Pcond{X_1\le X_2\le r}{X_1\le X_2\le X_3}\eq F(r)^2\big(3-2F(r)\big)}}		* {{MM|\Pcond{X_1\le X_2\le r}{X_1\le X_2\le X_3}\eq F(r)^2\big(3-2F(r)\big)}}
−	==Progression 3==	+	==Progression: 3==
	Now we look at {{M|m\eq 2}}, or 5 samples.		Now we look at {{M|m\eq 2}}, or 5 samples.

---

Revision as of 17:50, 16 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} }$

Important results

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

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

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

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

   $$\eq\frac{(2m+1)!}{m!}\lim_{t\rightarrow+\infty}\Bigg[\int^t_{-\infty}f(x_{2m+1})\left(\int^{x_{2m+1} }_{-\infty}f(x_{2m})\left(\cdots\int^{x_{m+3} }_{-\infty}f(x_{m+2})\left(\int^{\Min{r,x_{m+2} } }_{-\infty} f(x_{m+1})F(x_{m+1})^m\d x_{m+1}\right)\d x_{m+2}\cdots\right)\d x_{2m}\right)\d x_{2m+1}\Bigg]$$

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
• $\mathcal{Q}:\eq\P{\text{Median}(X_1,\ldots,X_{2m+1})\le r}\eq\Pcond{\mathcal{M} }{\mathcal{O} }$ - 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.

Progression: 1

We are evaluating:

$$\P{X_1\le\cdots\le X_{m+1}\le\Min{r,X_{m+2} }\le X_{m+2}\le X_{m+3}\cdots\le X_{2m+1}\le t }$$ (our answer is $$\big((2m+1)!\big)\times$$ of this as $t\rightarrow\infty$ ), the full integral follows:

• $$\int^t_{-\infty}f(x_{2m+1})\left(\int^{x_{2m+1} }_{-\infty}f(x_{2m})\left(\cdots\int^{x_{m+3} }_{-\infty}f(x_{m+2})\left(\int^{\Min{r,x_{m+2} } }_{-\infty} f(x_{m+1}){\left(\int^{x_{m+1} }_{-\infty}f(x_{m} )\left(\cdots\int^{x_2}_{-\infty}f(x_1)\d x_1\cdots\right)\d x_m\right)}\d x_{m+1}\right)\d x_{m+2}\cdots\right)\d x_{2m}\right)\d x_{2m+1}$$

We operate on the inner bit:

• $${\int^{x_{m+1} }_{-\infty}f(x_{m} )\left(\cdots\int^{x_2}_{-\infty}f(x_1)\d x_1\cdots\right)\d x_m}\eq \frac{1}{m!}F(x_{m+1})^m$$

We substitute this back in to yield:

• $$\frac{1}{m!}\int^t_{-\infty}f(x_{2m+1})\left(\int^{x_{2m+1} }_{-\infty}f(x_{2m})\left(\cdots\int^{x_{m+3} }_{-\infty}f(x_{m+2})\left(\int^{\Min{r,x_{m+2} } }_{-\infty} f(x_{m+1})F(x_{m+1})^m\d x_{m+1}\right)\d x_{m+2}\cdots\right)\d x_{2m}\right)\d x_{2m+1}$$

Conclusion of progression 1

We see here that

Progression: 2

This'll involve induction and dealing with the $\text{Min}()$ will be "tricky", both for practice and induction we will consider the special cases $m\eq 1$ and $m\eq 2$ by evaluating:

• $m\eq 1$ yields $$I_1:\eq\frac{1}{1!}\int^t_{-\infty} f(x_3)\left(\int^{\Min{r,x_3} }_{-\infty}f(x_2)F(x_2) \d x_2\right)\d x_3$$ , by case analysis:
  1. if $t\le r$ then $x_3\le t\le r$ or $x_3\le r$ over the entire domain of interest, so $\Min{r,x_3}\eq x_3$ over the entire domain, giving:
     • $$I_1\eq\frac{1}{1!}\int^t_{-\infty}f(x_3)\left(\int^{x_3}_{-\infty}f(x_2)F(x_2)\d x_2\right)\d x_3$$
       • We now use the corollary below to see:
         • $$I_1\eq\frac{1}{2!}\int^t_{-\infty}f(x_3)F(x_3)^2\d x_3$$

           $$\eq\frac{1}{3!}F(t)^3$$

  2. if $t\ge r$ then we split $(-\infty,t]$ into $(-\infty,r)$ and $[r,t]$, giving:
     • $$I_1\eq\frac{1}{1!}\left[\int^r_{-\infty} f(x_3)\left(\int^{\Min{r,x_3} }_{-\infty}f(x_2)F(x_2) \d x_2\right)\d x_3+\int_r^tf(x_3)\left(\int^{\Min{r,x_3} }_{-\infty}f(x_2)F(x_2) \d x_2\right)\d x_3\right]$$

       $$\eq\frac{1}{1!}\left[\int^r_{-\infty}f(x_3)\left(\int^{x_3}_{-\infty}f(x_2)F(x_2) \d x_2\right)\d x_3+\int_r^tf(x_3)\left(\int^r_{-\infty}f(x_2)F(x_2) \d x_2\right)\d x_3\right]$$

       • We now use the required corollary immediately below to yield:

         $$I_1\eq\frac{1}{1!}\left[\int^r_{-\infty}f(x_3)\cdot\frac{1}{2}F(x_3)^2\d x_3+\int_r^tf(x_3)\cdot\frac{1}{2}F(r)^2\d x_3\right]$$

         $$\eq\frac{1}{2!}\left[\frac{1}{3}F(r)^3+F(r)^2\int^t_rf(x_3)\d x_3\right]$$ , note that: $$\int^t_rf(x)\d x\eq\int_{-\infty}^tf(x)\d x-\int_{-\infty}^rf(x)\d x$$ $$\eq F(t)-F(r)$$

         $$\eq\frac{1}{2!}F(r)^2\left[\frac{1}{3}F(r)+\big(F(t)-F(r)\big)\right]$$ , note that: $$F(t)-F(r)\eq\frac{3F(t)-3F(r)}{3}$$ which we'll use next

         $$\eq\frac{1}{2!}F(r)^2\left[\frac{3F(t)-2F(r)}{3}\right]$$

         $$\eq\frac{1}{3!}F(r)^2\big(3F(t)-2F(r)\big)$$

It is clear that as $t\rightarrow\infty$ that we end up with

$$I_1\eq\frac{1}{3!}F(r)^2\big(3-2F(r)\big)$$

Thus:

$$\P{X_1\le X_2\le\Min{r,X_3}\le X_3}\eq\frac{1}{3!}F(r)^2\big(3-2F(r)\big)$$

Finally:

• $$\Pcond{X_1\le X_2\le r}{X_1\le X_2\le X_3}\eq F(r)^2\big(3-2F(r)\big)$$

Required corollary

Recall from Probability of i.i.d random variables being in an order and not greater than something that:

• $$\frac{1}{k!}\int^r_{-\infty}f(x)F(x)^k\d x\eq \frac{1}{(k+1)!}F(r)^{k+1}$$

So:

• $$\int^r_{-\infty}f(x)F(x)^k\d x\eq \frac{1}{k+1}F(r)^{k+1}$$

By applying this to above (with the $x_2$ integrals):

• $$\int^r_{-\infty}f(x)F(x)^1\d x\eq \frac{1}{2}F(r)^2$$ , we then substitute this for the cases $r:\eq r$ and $r:\eq x_3$

We'll then apply it to the $x_3$ integrals.

Conclusion of progression 2

• $$\Pcond{X_1\le X_2\le r}{X_1\le X_2\le X_3}\eq F(r)^2\big(3-2F(r)\big)$$

Progression: 3

Now we look at $m\eq 2$, or 5 samples.