--- Raul Miller <[EMAIL PROTECTED]> wrote:
> On 6/28/07, John Randall <[EMAIL PROTECTED]> wrote:
> > X_i has the population distribution, so the variance of X_i is \sigma^2.
> > The parenthetical (population variance) is an explanatory note, not an
> > argument for \sigma^2.
>
> Ok, fair enough.
>
> That said, I believe I now understand your proof well enough to assert
> that your proof works just as well if I use the same calculation
> for sample standard deviation as I use for population standard
> deviation. (With no "n-1" adjustment for the variance when working
> with samples).
>
> In other words:
>
> Let $\mu$ and $\sigma$ be the population mean and standard deviation.
> Fix the sample size $n$. Let $\bar X$ be the sample mean, $S^2$ the
> sample variance given by
>
> $$ S^2=(1/n)\sum (X_i-\bar X)^2$$
>
> It is elementary to show $E(\bar X)=\mu$. We now show
> $E(S^2)=\sigma^2$.
This looks like Law of Large Numbers applied--
however, not for baised estimate of the second
central moment above. (See experiment.)
...
>
> I've checked my work on this -- I built myself a numerical
> model representing most of the above statements and
> have tested both versions of it [yours and mine] against
> several different populations and sample sizes.
It would be good to see the model.
Here's a simple and straightforward Monte Carlo
experiment running biased and unbiased estimates
head-to-head.
In fact it show that
E((1/n)\sum (X_i-\bar X)^2) is (n-1)/n \sigma^2
and
E(1/(n-1)\sum (X_i-\bar X)^2) is \sigma^2
NB. =========================================================
load 'stats'
samp=: ?@(# #) { ]
bvar=: var * <:@# % #
PS=: 100000
testvar=: 4 : 0
V=. B=. 0
D=. normalrand PS
for_i. i.x do.
d=. y samp D
V=. V + var d
B=. B + bvar d
end.
(V%x),(B%x),bvar D
)
NB. =========================================================
10000 testvar 20 NB. 10000 runs with 20-item samples
0.993172 0.943514 0.994428
0.943514 * 20%19 NB. compensate bias
0.993173
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