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Errata: Change P(X(i) = -1 = P(X(i) = 1 ) = 1/2 to P(X(i) = -1 ) = P(X(i) = 1 ) = 1/2 For more, let v(i) be the variance of X(i). Then since E[X(i)] = 0
by graycat 13y ago
Errata:
Change
P(X(i) = -1 = P(X(i) = 1 ) = 1/2
to
P(X(i) = -1 ) = P(X(i) = 1 ) = 1/2
For more, let v(i) be the variance
of X(i). Then since E[X(i)] = 0
v(i) = E[ (X(i) - E[X(i)])^2 ]
= E[ (X(i))^2 ]
= (1/2) 1 + (1/2) 1
= 1
Since in the i.i.d. case the variance
of a sum is the sum of the variances,
the variance of S(n) is n. Then the
standard deviation of S(n) is sqrt(n)
so that the standard deviation of
(1/sqrt(n)) S(n)
is 1.
So, for large n, the density of
(1/sqrt(n)) S(n)
converges to Gaussian with expectation 0
and standard deviation 1 and
variance 1.
- graycat 13y agoErrata: Change Since in the i.i.d. case the variance of a sum is the sum of the variances, to Since in the independent case, and, hence, also the i.i.d., case, the variance of a sum is the sum of the variances,