Property A: the moment-generating function of a random variable x with normal distribution N(µ,σ2) is
Property B: The mean of a random variable x with norm distribution N(µ,σ2) is µ and the standard deviation is σ.
Proof: From Property A and Property 2 Advanced Properties of Probability Distributions,
But the variance is
and so the standard deviation is σ.
Property C: the moment-generating function of a random variable x with standard normal distribution N(0, 1) is
Proof: By Property A
Property 1: If x has a normal distribution N(μ,σ2) then the linear transform y = ax + b, where a and b are constants, has the normal distribution N(aμ+b, a2σ2).
Proof: By Property 3 of Advanced Properties of Probability Distributions,
But the right side of the equation is the moment-generating function for N(aμ + b, a2σ2), and so since probability distributions are uniquely determined by their moment-generating functions (Property 1 of Advanced Properties of Probability Distributions), we conclude that y has distribution N(aμ + b, a2σ2).
Property 2: If x1 and x2 are independent random variables, and x1 has normal distribution N(μ1, σ1) and x2 has normal distribution N(μ2, σ2) then x1 + x2 has normal distribution N(μ1 + μ2, σ2) where
Proof: Let y = x1 + x2. Since x1 and x2 are independent, by Property 4 of Advanced Properties of Probability Distributions
But the right side of the equation is the moment generating function for N(μ1 + μ2, σ2) with σ as defined above, and so by Theorem 1 of Advanced Properties of Probability Distributions, we conclude that y has distribution N(μ1 + μ2, σ2).
References
Wikipedia (2013) Normal distribution
https://en.wikipedia.org/wiki/Normal_distribution
Soch, J. (2020) Proof: Moment-generating function of the normal distribution. The book of statistical proofs
https://statproofbook.github.io/P/norm-mgf.html
if x1 —- N1 ( mu1 , sigma1) and X2 —- N2(mu2, sigma2) then
x1 +x2 — N1+N2 ( ( mu1 + mu2 ) , ( root (sigma1^2 + sigma2^2)
But in the above proof N 1 + N2 is not mentioned .
Does it mean that If Standard Deviation of the large population ( of normal distribution) is suppose sigma . then the standard deviations of the samples of the said population , are sigma 1, sigma 2 ,…… sigma n and means are mu-1, mu-2 , mu3 …….mu-n
is the central limit theorem is derived from this property of normal distribution ?? I think so.
Am I correct ??
This proof depends on the fact that properties about distributions can often be proved based on the moment generating functions of those distributions, namely Theorem 1 of https://real-statistics.com/general-properties-of-distributions/properties-probability-distributions-detailed/.
No this property is not a result of the Central Limit Theorem.
In fact, the standard deviations of the samples of a population are not sigma 1, sigma 2 ,…… sigma n; the means are approximately equal to mu-1, mu-2 , mu3 …….mu-n
Charles