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Concrete mathematics : a foundation for computer science

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378 DISCRETE PROBABILITY<br />

And we can also make an estimate of the variance, using the <strong>for</strong>mula<br />

\ix 1<br />

x: + x: + + ;y’n _ (X, + X2 + ‘. + X,)2<br />

n - l n(n-1)<br />

(8.20)<br />

The (n ~ 1) ‘s in this <strong>for</strong>mula look like typographic errors; it seems they should<br />

be n’s, as in (8.1g), because the true variance VX is defined by expected values<br />

in (8.15). Yet we get a better estimate with n - 1 instead of n here, because<br />

definition (8.20) implies that<br />

E(i/X) = VX.<br />

Here’s why:<br />

E(\;/X) =&E( tx: -<br />

k=l<br />

k=l<br />

1 n<br />

=n<br />

- l(x W2)<br />

k=l<br />

- k f f (E(Xi’lj#kl+ E(X')Lj=kl))<br />

j=l k=l<br />

= &(nE(X’) - k(nE(X’) +n(n- l)E(X)'))<br />

= E(X')-E(X)“ = VX<br />

; f f xjxk)<br />

j=l k=l<br />

(8.21)<br />

(This derivation uses the independence of the observations when it replaces<br />

E(XjXk) by (EX)‘[j fk] + E(X’)[j =k].)<br />

In practice, experimental results about a random variable X are usually<br />

obtained by calculating a sample mean & = iX and a sample standard de-<br />

viation ir =fi, and presenting the answer in the <strong>for</strong>m ‘ fi f b/,/i? ‘. For<br />

example, here are ten rolls of two supposedly fair dice:<br />

The sample mean of the spot sum S is<br />

fi = (7+11+8+5+4+6+10+8+8+7)/10 = 7.4;<br />

the sample variance is<br />

(72+112+82+52+42+62+102+82+82+72-10~2)/9z<br />

2.12

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