3 ms·
Here's what's going on, what will illustrate the central limit theorem (CLT) and what will not: For positive integer n and for i = 1, 2, ..., n, let real valu
by graycat 13y ago
Here's what's going on, what will
illustrate the central limit theorem (CLT)
and what will not:
For positive integer n and for i =
1, 2, ..., n, let real valued random
variable X(i) be so that
P(X(i) = -1 = P(X(i) = 1 ) = 1/2
Assume that {X(1), ..., X(n)}
is an independent set of random
variables.
Easily the expectation
E[X(i)] = 0.
Let real valued random variable
S(n) = X(1) + X(2) + ... + X(n)
Easily E[S(n)] = 0.
For n = 4, S(n) can take on values
-4, -3, ..., 0, 1, ..., 4.
The animation in the OP should
converge to the density of
S(n) where n = 4.
By the CLT, as n approaches
infinity, the density of
( 1/sqrt(n) ) S(n)
will be Gaussian with expectation 0.
By the strong law of large numbers,
as n approaches infinity, the
probability
random variable
(1/n) S(n)
converges to 0 is 1.
For increasing n, S(n)
is a martingale and
is a discrete approximation to
Brownian motion. As n approaches
infinity, the density of
S(n) approaches 0 everywhere on
the real line and in its limit
is not a density.
- graycat 13y agoErrata: 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,