View source for Notes:Distribution of the sample median

{{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} } }}
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==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:
* {{MM|\P{\text{Median}(X_1,\ldots,X_{2m+1})\le r} }} - the {{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 {{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+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.
==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.
==Progression: 1==
We are evaluating: {{MM|\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 {{MM|\big((2m+1)!\big)\times}} of this as {{M|t\rightarrow\infty}} ), the full integral follows:
* {{MM|\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})<!--

MARKER: (int^x_m+1 ... \d x_m starts here 
I want to put an underbrace around it.

--><!--\underbrace-->{\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)}<!--

Marker: \d x_m ends here


-->\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:
* {{MM|{\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:
* {{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} }}
==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:
* {{M|m\eq 1}} yields {{MM|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:
*# if {{M|t\le r}} then {{M|x_3\le t\le r}} or {{M|x_3\le r}} over the entire domain of interest, so {{M|\Min{r,x_3}\eq x_3}} over the entire domain, giving:
*#* {{MM|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:
*#*** {{MM|I_1\eq\frac{1}{2!}\int^t_{-\infty}f(x_3)F(x_3)^2\d x_3}}
*#***: {{MM|\eq\frac{1}{3!}F(t)^3}}
*# if {{M|t\ge r}} then we split {{M|(-\infty,t]}} into {{M|(-\infty,r)}} and {{M|[r,t]}}, giving:
*#* {{MM|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]}}
*#*: {{MM|\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:
*#**: {{MM|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]}}
*#**: {{MM|\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: {{MM|\int^t_rf(x)\d x\eq\int_{-\infty}^tf(x)\d x-\int_{-\infty}^rf(x)\d x}} {{MM|\eq F(t)-F(r)}}
*#**: {{MM|\eq\frac{1}{2!}F(r)^2\left[\frac{1}{3}F(r)+\big(F(t)-F(r)\big)\right]}}, note that: {{MM|F(t)-F(r)\eq\frac{3F(t)-3F(r)}{3} }} which we'll use next
*#**: {{MM|\eq\frac{1}{2!}F(r)^2\left[\frac{3F(t)-2F(r)}{3}\right]}}
*#**: {{MM|\eq\frac{1}{3!}F(r)^2\big(3F(t)-2F(r)\big)}}
It is clear that as {{M|t\rightarrow\infty}} that we end up with {{MM|I_1\eq\frac{1}{3!}F(r)^2\big(3-2F(r)\big)}}

Thus: {{MM|\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:
* {{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)}}
===Required corollary===
Recall from [[Probability of i.i.d random variables being in an order and not greater than something]] that:
* {{MM|\frac{1}{k!}\int^r_{-\infty}f(x)F(x)^k\d x\eq \frac{1}{(k+1)!}F(r)^{k+1} }}
So:
* {{MM|\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 {{M|x_2}} integrals):
* {{MM|\int^r_{-\infty}f(x)F(x)^1\d x\eq \frac{1}{2}F(r)^2 }}, we then substitute this for the cases {{M|r:\eq r}} and {{M|r:\eq x_3}}
We'll then apply it to the {{M|x_3}} integrals.
===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)}}
==Progression 3==
Now we look at {{M|m\eq 2}}, or 5 samples.