Dear Reza, the proof goes as follows:
your question: why does one set empirical variance = s^2 = (sum(xi-xm)^2)/(n-1), where xm := sum(xi)/n (i=1...n), the estimated mean, is equivalent to asking: is s^2 an unbiased estimator of the variance of the parent distribution, E(s^2) = Var(x) ? (E = expectation value) First, remember, Var(x) := E((x-Ex)^2) = E(x^2) - (Ex)^2. Now, (n-1)*s^2 = sum(xi-xm)^2 = (calculate the square) = (sum(xi^2)-sum(xm^2)) = (sum(xi^2) - n*xm^2) Next take the expectation value of this, (n-1)*E(s^2) = n*(E(x^2)-E(xm^2)) We know from the central limit theorem that E(xm) = E(x), and Var(xm) = Var(x)/n. Therefore, (n-1)*E(s^2) = n*(E(x^2) - (Ex)^2 + (Exm)^2 - E(xm^2)) ....(the 2nd and 3rd term cancel) .... = n*((E(x^2)-(Ex)^2) - (E(xm^2)-(Exm)^2))= = n*(Var(x) - Var(xm)) = n*Var(x)*(1-(1/n)) = (n-1) * Var(x), or E(s^2) = Var(x) q.e.d. There are different (very similar) versions of this proof, this one follows closely Roger Barlow, Statistics, John Wiley & Sons 1989 (chapter 5.2.2.), which I find a good introduction into basic statistics. best regards, Peter >Dear Experts >> Sorry may be the question is so basic .After searching my statistics >books >> to find an answer with no great success, could you please explain me >why we >> consider degree of freedom as n-1 in calculating variance. Thanks for >your >> kind advises. ================================================================= Dr. Peter Bossew Division of Physics and Biophysics, University of Salzburg, Austria home: A-1090 Vienna, Austria, Georg Sigl-Gasse 13/11, ph: +43-1-3177627 telefonino: +43-650-8625623 [EMAIL PROTECTED] [EMAIL PROTECTED] =================================================================
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