統計-Variance-Population-Sample

sample mean x = S(xi) / n
E(x) = E[S(xi) / n] = E[S(xi)] / n = S[E(xi)] / n = S(m) / n = n * m / n = population mean m
sample variance s2 = S[(xi - x)2] / (n – 1)

we would use n – 1 in the derivation of sample variance because E(s2) = population variance s2

population variance s2 = Var(x) = Var(xi) = E(xi2) – E2(xi) = E(xi2) – m2
E(xi2) = s2 + m2

population mean m = E(x) = E[S(xi) / n] = E[S(xi)] / n
E[S(xi)] = n * m

E[(Sxj)2 = E{S[S(xk * xm)]} = E[(x12 + x1 * x2 + x1 * x3 + … + x1 * xn) + (x2 * x1 + x22 + x2 * x3 + … + x2 * xn) + (x3 * x1 + x3 * x2 + x32 + … + x3 * xn) + … + (xn * x1 + xn * x2 + xn * x3 + … + xn2)] = E(x12 + x22 + x32 + … + xn2) + E[(x1 * x2 + x1 * x3 + … + x1 * xn-1 + x1 * xn) + (x2 * x1 + x2 * x3 + … + x2 * xn-1 + x2 * xn) + (x3 * x1 + x3 * x2 + … + x3 * xn-1 + x3 * xn) + … + (xn * x1 + xn * x2 + xn * x3 + … + xn * xn-1)] = S[E(xi2)] + {[E(x1) * E(x2) + E(x1) * E(x3) + … + E(x1) * E(xn-1) + E(x1) * E(xn)] + [E(x2) * E(x1) + E(x2) * E(x3) + … + E(x2) * E(xn-1) + E(x2) * E(xn)] + [E(x3) * E(x1) + E(x3) * E(x2) + … + E(x3) * E(xn-1) + E(x3) * E(xn)] + … + [E(xn) * E(x1) + E(xn) * E(x2) + E(xn) * E(x3) + … + E(xn) * E(xn-1)]} = n * (s2 + m2) + [(n2 – n) * E(xk) * E(xm)] = n * (s2 + m2) + [(n2 – n) * m2] = n * (s2 + n * m2)

E[xi * S(xj)] = E(xi * x1 + xi * x2 + … + xi * xi-1 + xi2 + xi * xi+1 + … + xi * xn-1 + xi * xn) = E(xi2) + [(n – 1) * E(xi) * E(xj)] = s2 + m2 + [(n – 1) * m2] = s2 + n * m2

E(s2) = E{S[(xi - x)2] / (n – 1)}
= E{S[(xi - x)2]} / (n – 1)
= S{E[(xi - x)2]} / (n – 1)
= S{E[(xi - Sxj / n)2]} / (n – 1)
= S{E[(n * xi - Sxj)2]} / [n2 * (n – 1)]
= S{E[n2 * xi2 + (Sxj)2 – 2 * n * xi * (Sxj)]} / [n2 * (n – 1)]
= S{E(n2 * xi2) + E[(Sxj)2]– E[2 * n * xi * (Sxj)]} / [n2 * (n – 1)]
= S[n2 * E(xi2) + n * (s2 + n * m2) – 2 * n * E(xi * Sxj)] / [n2 * (n – 1)]
= S[n2 * (s2 + m2) + n * (s2 + n * m2) – 2 * n * (s2 + n * m2)] / [n2 * (n – 1)]
= S [n * (n – 1) * s2] / [n2 * (n – 1)] = s2

2009-07-07 17:37:18 補充：
sorry, the fonts are not question in my answer

S = summation sign
s2 = population variance
m = population mean

2009-07-07 17:39:15 補充：
you could see my answer as follows:

http://www.yousendit.com/transfer.php?action=check_download&ufid=dVlvdFdRYTJ3NUpMWEE9PQ&key=f966ad7ff163c352d95c5270cc703a17a1d67206&bid=cmcwY05uTmFKV09Ga1E9PQ

It is only valid till 14/7/09

2009-07-07 17:42:12 補充：
variance is one of the mean to measure the dispersion of the samples.

Other method such as mean standard error = xi - sample mean

however, this method is not good, it could be positive or negative

When we sum all the error up, they could be cancel each others

2009-07-07 17:46:32 補充：
eg: samples are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10

mean = 5.5
if we don't put a square there, e.g. mean standard error = summation (xi - 5.5) = 0. however, obviously, the dispersion of the samples are not zero.

2009-07-07 17:46:39 補充：
hence, we put a square there, so that all of them are positive.
for this eg. sample variance = 9.1667

in order not to exggerate the effect due to the square, we put a square root and obtain a standard deviation = 3.028

1 因為用n-1的那個是無偏的。即E(S^2)=(Sigma)^2
2 要平方因為想計離散程度。若果只用(x_i-mu)相加後會是0。用絕對值不方便。所以用平方將全部變成正數。但因為平方完單位變成二次方﹐所以要開方將單位變回原狀。
3 Square (2次方)是否有開大作用? Square root(開方)是否有縮細作用? 不是。開大縮細是單位問題﹐用Square和Square root無法解決﹐只可以透過cv = Coefficient of variation = (標準差/平均數) 100解決﹐或者一開始將全部數據用同一單位量度。


2009-07-07 18:14:45 補充：
用n-1的那個是無偏的証明可以看Hogg, Robert V. Introduction to mathematical statistics 6th ed p.200