Proof of central limit theorem

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SUMMARY

The discussion focuses on proving the Central Limit Theorem using moment-generating functions (MGFs) without referencing the standard normal distribution. The user aims to demonstrate that the MGF of the average of independent variables converges to the MGF of a normal distribution, specifically showing that as n approaches infinity, the limit of the MGF is e^(ut + sigma^2t/2). The user calculates the MGF as [M(t/n)]^n and applies l'Hôpital's rule to evaluate the limit, but encounters an issue in deriving the additional term sigma^2/2 in the final result.

PREREQUISITES
  • Understanding of moment-generating functions (MGFs)
  • Familiarity with the Central Limit Theorem
  • Knowledge of l'Hôpital's rule for evaluating limits
  • Basic concepts of probability theory, particularly independent random variables
NEXT STEPS
  • Study the derivation of moment-generating functions for independent random variables
  • Learn about the application of l'Hôpital's rule in limit evaluation
  • Research the properties of the normal distribution and its moment-generating function
  • Explore alternative proofs of the Central Limit Theorem using characteristic functions
USEFUL FOR

Mathematicians, statisticians, and students studying probability theory who are interested in understanding the Central Limit Theorem and its proof using moment-generating functions.

skwey
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Hi I want to prove this using momentgenerating functions. I would like to do this without going into the standard normal distribution, just the normal distribution.

I would like to show that the momentgenerating function of
(x1+x2+x3...xn)/n--->e^(ut+sigma^2t/2) as n-->infinity.

x1, x2, x3,xn =independent variables with mean u and variance sigma^2
e^(ut+sigma^2t/2)=momentgenerating function of a normal distribution


1.calculating the moment generating function
This I get to be

[M(t/n)]^n where M(t) is the momentgenerating function of the variable x1 or x2 or xn. [M(t/n)]^n is the momentgenerating function of (x1+x2+x3..xn)/n

2.
Finding the limit as n->infinity.

I take the natural logarithm and get n*ln[M(t/n)]=ln[M(t/n)]/(1/n)
as n->infinity we get 0/0 since M(0)=1 and ln(1)=0

I then use l'Hopital to get:

M'(t/n)*t/M(t/n)

when n goes to infinity this goes to ut since M'(0)=u, but it should be ut+sigma^2/2

Does anyone see why I do not get the last part, what have I forgotten?
 
Last edited:
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The l'Hopital part:

ln[M(t/n)]' / (1/n) '
=[M'(t/n)*-t/n^2 / M(t/n)] / [-1/n^2]= M'(t/n)*t/M(t/n). As n-> infinity this becomes ut.
 

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