central limit theorem basic question

leo nidas

New Member
I have a basic question concerning the central limit theorem

let the 2 by one vectors x1, x2, ..., xn come from a population with mean μ=[μ1 μ2]' , and variance Σ=[σ11 σ12;σ21 σ22].

Τhe clt states that sqrt(n)(Xbar-μ)~Ν(0,Σ) where Xbar=(x1+x2+...+xn)/n = (Σx(i))/n and because I had a problem typing in latex the tilde "~" will be "converge in distribution" throughout this post.

So, can I say the following?:

1. sqrt(n)(Xbar-μ)~Ν(0,Σ) =>
2. (Xbar-μ)~Ν(0,Σ/n) =>
3. Xbar~N(μ, Σ/n) =>
4. (Σx(i))/n~N(μ, Σ/n) =>
5. Σx(i)~N(μ, nΣ).

So are the above 5 steps correct? Can I claim that the sum at last step has approximately that normal? I think there is a problem with that "n" in the variance. As n goes to infinity then the variance goes to infinity? Then we have no convergence i think?! Where is my mistake?

Generally can I say somthing about the distribution of that sum?

BGM

TS Contributor
When you keep summing up independent random variables, the variance must increase.
Thats why in CLT you need to "standardize" the variance.
and when formally using CLT, we cannot state the last equation like that.
(Just like we do not directly manipulate the indeterminate forms of limit in mathematics)
You can also see CLT can be used to explain the normality of the error term in law
of large number.

Loosely speaking, we shall see the normality of the sum. You may use the last equation
as an approximation (not mathematically rigorous I think).
But just be careful that when the variance is very large, even
the underlying distribution is normal, you may not be able to see the distribution like a
normal (bell shape curve). The distribution may just looks like a flat line.

For example, you try to sum up the independent bernoulli trials. If you do not standardize
it, the range of the resulting binomial random variable will increase as n is increasing.
If you stardardize it, you will see an approximately bell curve in [0, 1].