Why to worry about biases?
For last 30 years I am using bootstrapping as the really practical method
for statistical estimates. Method works with ANY statistical estimate and
ANY distribution.
If you really need variance analysis (ANOVA) only then read about degrees
of freedom and F-distribution.
Bootstrapping discovers often some hidden structure of your data you never
knew that it did exist at all. So it is really useful.
WbR-s
Leo Võhandu
> On 6/27/07, John Randall <[EMAIL PROTECTED]> wrote:
>> This is trivial. Suppose X1,...,Xn are independent random variables,
>> and a1,...,an are numbers. Then
>>
>> E(a1 X1+...+an Xn)=a1 E(X1)+...+an E(Xn),
>>
>> that is, E is linear (indeed affine) on independent variables. This
>> just follows from the linearity of summation or integration.
>
> Ok, but I do not think that works for standard deviation, because
> standard deviation is not a linear operation. Then again, I still do
> not understand your proof, so this may be a moot point..
>
> Anyways, I'm still stuck working through your earlier post:
>
> Let's take the roll of a (six sided) die, and a sample of that
> happens to be the values 1 2 3. You asserted
> $E(S^2)=E((1/n-1)\sum (X_i-\bar X)^2$
>
> The expected value for X_i-\bar X is a random value from the
> set _2.5 _1.5 _0.5 0.5 1.5 2.5. The expected value for
> (X_i-\bar X)^2 is a random value from the set 0.25 2.25 6.25.
> So the right hand side of that assertion seems
> to be, for this case, one of the possibilities for
> 0.5*+/(?3#3){0.25 2.25 6.25
> If I assign an equal chance to each of these possibilities, I
> get:
> mean=:+/ % #
> mean 0.5*+/"1(3 3 3#:i.27){"1]0.25 2.25 6.25
> 4.375
>
> Or, if you are comfortable with a more brute-force approach,
> mu=:3.5
> mean 0.5*+/"1 (6 6 6#:i.6^3){"1]*:1 2 3 4 5 6-mu
> 4.375
>
> But for the left hand side of that assertion ($E(S^2)$), I get:
> require'stats'
> mean *:@stddev"1 (6 6 6#:i.6^3){"1]1 2 3 4 5 6
> 2.91667
>
> In other words, I get a number less than 3 for the left
> side of that assertion and a number greater than 4
> for the right side.
>
> So, either I have made a mistake which I have not spotted
> (unfortunately, this is all too possible for my comfort) or proofs based
> on this assertion must be invalid.
>
> If you're not sick of me pestering you on this subject, could you take a
> look over the above and tell me where you think we differ?
>
> Thanks,
>
> --
> Raul
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