"James K." <[EMAIL PROTECTED]> wrote in message news:[EMAIL PROTECTED] > "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].
Indeed, s^2 = E[ (m-x1)^2 ]. Nice :-) Dirk Vdm . . ================================================================= Instructions for joining and leaving this list, remarks about the problem of INAPPROPRIATE MESSAGES, and archives are available at: . http://jse.stat.ncsu.edu/ . =================================================================
