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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?
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?
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