"Dirk Van de moortel" <[EMAIL PROTECTED]> wrote
in message news:[EMAIL PROTECTED]
>
> If you already know the mean mu and think it is reliable,
> my first idea would be to throw away all the data above
> mu, and then for each value v < mu, I would "add to the
> sample" a new value 2mu-v (which is symmetrical w.r.t.
> the mean), and use this manipulated sample to estimate
> the variance ;-)
>
> Just a thought of course...
Just under above assumption,
I now introduce easy solution, without adding dummy values in to the set, by
using some manipulation of simple formulations. Let us assume {x} is
original set, and {x1| x1=m-a, a>0} and {x2| x2=m+a, a>0} are special sets,
which are sets of values smaller than mean m and larger, respectively. From
the original assumption, original x2 is useless, so we use virtual ~x2 = m+a
= m + (m - x1) = 2*m - x1, which was already noted by Dirk. Then, the
variance can be denoted as
s^2 = E[x^2] - m^2
= (1/2) * { E[x1^2] + E[x2^2] } - m^2
= (1/2) * { E[x1^2] + E[~x2^2] } - m^2
= (1/2) * { E[x1^2] + E[(2*m-x1)^2] } - m^2
= E[x1^2] - m*(2*E[x1] - m),
or simply s^2 = E[a^2].
Before using this solution, it should be kept in mind of the assumption that
you exactly know mean and are perfectly sure that all values smaller than
mean are never corrupted.
BR,
--
James K. ([EMAIL PROTECTED], http://home.naver.com/txdiversity)
- Any remarks, proposal and/or indicator to text would be greatly
respected.
- Private opinions: These are not the opinions from my affiliation.
.
.
=================================================================
Instructions for joining and leaving this list, remarks about the
problem of INAPPROPRIATE MESSAGES, and archives are available at:
. http://jse.stat.ncsu.edu/ .
=================================================================
