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why binomial prob. can be tranform to normal distribution??
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This is actually the result of Central Limit Theory :- " For large values of n, the distributions of the count X and the sample proportion p are approximately normal. This result follows from the Central Limit Theorem. The mean and variance for the approximately normal distribution of X are np and np(1-p), identical to the mean and variance of the binomial(n,p) distribution. Similarly, the mean and variance for the approximately normal distribution of the sample proportion are p and (p(1-p)/n)." - quoted from http://www.stat.yale.edu/Courses/1997-98/101/binom.htm Proof of Central Limit Theorem :- Characteristic function of any random variable Y can be written as f(t) = 1 - t^2/2 + o(t^2), t -> 0 (by Taylor expansion) where o(t^2) refers to any function that goes to zero more rapidly than t^2. Let Yi = (Xi - mean)/stdev, the standardised value of Xi. Then the standardised mean of the observations of Xi, X2...Xn is Zn = (n * sample mean - n *n mean)/(stdev * root of n ) = sum of Yi / root of n So the characteristic function of Z is given by [ f( t / root of n ) ]^n = [1-t^2/(2n) + o(t^2/n) ]^n -> e^(-t^2/2), n -> infinity By uniqueness of characteristic function (Levy continuity theorem), any distribution tends to normal distributions for large n (and relatively small p).
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why binomial prob. can be tranform to normal distribution??
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X~B(n,p)=>X~N(u, var^1/2)??最佳解答:
This is actually the result of Central Limit Theory :- " For large values of n, the distributions of the count X and the sample proportion p are approximately normal. This result follows from the Central Limit Theorem. The mean and variance for the approximately normal distribution of X are np and np(1-p), identical to the mean and variance of the binomial(n,p) distribution. Similarly, the mean and variance for the approximately normal distribution of the sample proportion are p and (p(1-p)/n)." - quoted from http://www.stat.yale.edu/Courses/1997-98/101/binom.htm Proof of Central Limit Theorem :- Characteristic function of any random variable Y can be written as f(t) = 1 - t^2/2 + o(t^2), t -> 0 (by Taylor expansion) where o(t^2) refers to any function that goes to zero more rapidly than t^2. Let Yi = (Xi - mean)/stdev, the standardised value of Xi. Then the standardised mean of the observations of Xi, X2...Xn is Zn = (n * sample mean - n *n mean)/(stdev * root of n ) = sum of Yi / root of n So the characteristic function of Z is given by [ f( t / root of n ) ]^n = [1-t^2/(2n) + o(t^2/n) ]^n -> e^(-t^2/2), n -> infinity By uniqueness of characteristic function (Levy continuity theorem), any distribution tends to normal distributions for large n (and relatively small p).
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