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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