User:HKIDz404520(4)

I am a non-degree sci-crank. I have a trivial discover on college-level Statistics.

It is about finite population correction FPC($H_{n}^{N}$) factor.

I hope this little discover could help me to pursuit a degree and change my life. Thank thank everyone.

FPC($C_{n}^{N}$)	FPC($H_{n}^{N}$)
very popular and well-known	missing from many texts. I am not sure whether FPC($H_{n}^{N}$) is already well-known or worthy to be published.
1. sample drawn with no replacement	1. sample drawn with replacement 2. samples contain same elements but in different orders are considered as same case
total number of possible samples = $C_{n}^{N}$	total number of possible samples = $H_{n}^{N}$
$FPC(C_{n}^{N})={\sqrt {\frac {N-n}{N-1}}}$	$FPC(H_{n}^{N})={\sqrt {\frac {N+n}{N+1}}}$

I am not tertiary educated. My background is worse.

I came from a grass-root family and my education history is full of grievance.

I faultily be picked in to art subjects and lost the chance to study science and math during my secondary education.

I was not eligible to the vocational training of any engineering field. Finally, I self-studied all the topics till the year-1 level.

name	Alfred Wong
ID card no.	z404520(4)
email	hardislave@gmail.com
mobile	002-852-66000934
country	Hong Kong (SARC)

Summary

Population size = $n$ Sample size = $r$	With replacement	Without replacement
Same elements in different orders are considered as different cases	$n^{r}$	$P_{r}^{n}$
Same elements in different orders are considered as same case	$H_{r}^{n}$	$C_{r}^{n}$

Biased population variance $V_{r}$ , sample variance $V_{n}$ and population variance$V[{\bar {n}}]$ have following relations:

sample drawn with replacement number of possible samples $=n^{r}$	combinations without repetition number of possible samples $=C_{r}^{n}$
$[r-{\bar {r}}]=[{\bar {n}}-{\bar {r}}]+[r-{\bar {n}}]$	$[r-{\bar {r}}]=[{\bar {n}}-{\bar {r}}]+[r-{\bar {n}}]$
${\frac {V_{r}}{r-1}}={\frac {V_{\bar {r}}}{1}}={\frac {V_{n}}{r}}$	${\frac {V_{r}}{[r-1]n}}={\frac {V_{\bar {r}}}{[n-r]}}={\frac {V_{n}}{r[n-1]}}$
${\frac {V_{r}+V_{\bar {r}}}{[r-1]+1}}={\frac {V_{n}}{r}}$	${\frac {V_{r}+V_{\bar {r}}}{[r-1]n+[n-r]}}={\frac {V_{n}}{r[n-1]}}$
combinations with repetition number of possible samples $=H_{r}^{n}$	permutations number of possible samples $=P_{r}^{n}$
$[r-{\bar {r}}]=[{\bar {n}}-{\bar {r}}]+[r-{\bar {n}}]$	$[r-{\bar {r}}]=[{\bar {n}}-{\bar {r}}]+[r-{\bar {n}}]$
${\frac {V_{r}}{[r-1]n}}={\frac {V_{\bar {r}}}{[n+r]}}={\frac {V_{n}}{r[n+1]}}$	${\frac {V_{r}}{[r-1]n}}={\frac {V_{\bar {r}}}{[n-r]}}={\frac {V_{n}}{r[n-1]}}$
${\frac {V_{r}+V_{\bar {r}}}{[r-1]n+[n+r]}}={\frac {V_{n}}{r[n+1]}}$	${\frac {V_{r}+V_{\bar {r}}}{[r-1]n+[n-r]}}={\frac {V_{n}}{r[n-1]}}$

How to proof

Summation notations	Sample size
$\sum _{j=1}^{n}\sum _{i=1}^{n}$	$n^{2}$
$\sum _{k=j+1}^{n}\sum _{j=i+1}^{n-1}\sum _{i=1}^{n-2}$	${\frac {n(n-1)(n-2)}{3!}}=C_{3}^{n}$
$\sum _{k=j}^{n}\sum _{j=i}^{n}\sum _{i=1}^{n}$	${\frac {n(n+1)(n+2)}{3!}}=H_{3}^{n}$
$\sum _{r=q}^{n}\sum _{q=p}^{n}...\sum _{k=j}^{n}\sum _{j=i}^{n}\sum _{i=1}^{n}$	${\frac {n(n+1)(n+2)(n+3)...(n+r-1)}{r!}}=H_{r}^{n}$
$\sum _{j=i}^{n}\sum _{i=1}^{n}+\sum _{j=i+1}^{n}\sum _{i=1}^{n-1}$	$H_{2}^{n}+C_{2}^{n}=n^{2}$

Some facts will be used in the proof
$[\sum _{j=1}^{n}(x_{j}-\mu )][\sum _{i=1}^{n}(x_{i}-\mu )]=0$
$[\sum _{j=1}^{n}(x_{j}-\mu )][\sum _{i=1}^{n}(x_{i}-\mu )]=(\sum _{j=i}^{n}\sum _{i=1}^{n}+\sum _{j=i+1}^{n}\sum _{i=1}^{n-1})(x_{i}-\mu )(x_{j}-\mu )$
$0=H_{2}^{n}\sigma _{\bar {x}}^{2}-{\frac {\sigma ^{2}}{2}}$
$H_{2}^{n}\sigma _{\bar {x}}^{2}={\frac {\sigma ^{2}}{2}}$

The proof can be started by the expression:

$H_{r}^{n}\sigma _{\bar {x}}^{2}=\sum _{r=q}^{n}...\sum _{k=j}^{n}\sum _{j=i}^{n}\sum _{i=1}^{n}({\frac {x_{i}+x_{j}+x_{k}...+x_{p}+x_{q}+x_{r}}{r}}-\mu )^{2}$