Normal Distribution Detailed

Property A: the moment generating function of a random variable x with normal distribution N(µ, σ) is

Moment generating function normal

Property B: The mean of a random variable x with norm distribution N(µ, σ) is µ and the standard deviation is σ

Proof: From Property A and Property 2 Advanced Properties of Probability Distributions,

image3220 image3221 image3222

But the variance is
image3223

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

image6062

Proof: By Property A

image6063

Property 1: If x has normal distribution N(μ, σ) then the linear transform y = ax + b, where a and  b are constants, has normal distribution N(aμ+b, aσ).

Proof: By Property 3 of Advanced Properties of Probability Distributions,

image3225

But the right side of the equation is the moment generating function for N(aμ + b, aσ), and so since probability distributions are uniquely determined by their moment generating functions (Theorem 1 of Advanced Properties of Probability Distributions), we conclude that y has distribution N(aμ + b, aσ).

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(μμ2σ) where

image365

Proof: Let y = x1 + x2. Since x1 and x2 are independent, by Property 4 of Advanced Properties of Probability Distributions

image3233

But the right side of the equation is the moment generating function for N(μ1 μ2σ) with σ as defined above, and so by Theorem 1 of Advanced Properties of Probability Distributions we conclude that y has distribution N(μ1 μ2σ).

2 Responses to Normal Distribution Detailed

  1. sanjay joshi says:

    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 ??

Leave a Reply

Your email address will not be published. Required fields are marked *