Central Limit Theorem and Standardized Sums

[caffeine]

I don't think I *really* understand the Central Limit Theorem.

Suppose we have a set of n independent random variables \{X_i\} with the same distribution function, same finite mean, and same finite variance. Suppose we form the sum S_n = \sum_{i=1}^n X_i. Suppose I want to know the probability that S_n is between a and b. In other words, I want to know P(b > S_n > a).

The central limit theorem uses a standardized sum:


P\left(b > \frac{ \sum_{i=1}^n X_i - n\mu}{\sqrt{n}\sigma} > a\right)
= \frac{1}{\sqrt{2\pi}} \int_a^b e^{-y^2/2} \, dy


What is the relationship between what I want:

P(b > S_n > a)

and what the central limit theorem tells me about:

P\left(b > \frac{ \sum_{i=1}^n X_i - n\mu}{\sqrt{n}\sigma} > a\right)


How can they possibly be equal? If they are equal, how is that possible? And if they're not equal, how would I get what I want?

[HallsofIvy]

I don't think I *really* understand the Central Limit Theorem.

Suppose we have a set of n independent random variables \{X_i\} with the same distribution function, same finite mean, and same finite variance. Suppose we form the sum S_n = \sum_{i=1}^n X_i. Suppose I want to know the probability that S_n is between a and b. In other words, I want to know P(b > S_n > a).

The central limit theorem uses a standardized sum:


P\left(b > \frac{ \sum_{i=1}^n X_i - n\mu}{\sqrt{n}\sigma} > a\right)
= \frac{1}{\sqrt{2\pi}} \int_a^b e^{-y^2/2} \, dy

No, the Central Limit Theorem does not use a standardized sum. (At least it doesn't have to. I can't speak for whatever text book you are using.)

Essentially, the Central Limit Theorem says that if we consider all possible samples of size n from a distribution having finite mean, \mu, and finite standard deviation, \sigma, then their sum will be approximately normally distributed with mean n\mu and standard distribution \sqrt{n}\sigma- and the larger n is the better that approximation will be.

What is the relationship between what I want:

P(b > S_n > a)

and what the central limit theorem tells me about:

P\left(b > \frac{ \sum_{i=1}^n X_i - n\mu}{\sqrt{n}\sigma} > a\right)


How can they possibly be equal? If they are equal, how is that possible? And if they're not equal, how would I get what I want?
Well, they are not equal- they are approximately equal.

Assuming that your base distribution has mean \mu and standard deviation \sigma, then Sn is approximately normally distributed with mean n\sigma and standard deviation \sqrt{n}\sigma. You then convert from that to the standard normal distribution (with mean 0 and standard deviation 1) as you probably have learned before:
z= \frac{x- \mu}{\sigma}
In particular, take x= Sn, a< S_n< b becomes a- n\mu< S_n-n\mu< b-n\mu and then
\frac{a-n\mu}{\sqrt{n}\sigma}< \frac{S_n- n\mu}{\sqrt{n}\sigma}< \frac{b- n\mu}{\sqrt{n}\sigma}

Look up those values in a standardized normal distribution table.

[ssd]

Essentially, the Central Limit Theorem says that if we consider all possible samples of size n from a distribution having finite mean, \mu, and finite standard deviation, \sigma, then their sum will be approximately normally distributed with mean n\mu and standard distribution \sqrt{n}\sigma- and the larger n is the better that approximation will be.


This is famous as Lindeberg-Levy CLT. I am not sure but I have a doubt about the terms "all possible samples".... essentially, for a sequence of independently and identically distributed random variables with same mean \mu and same finite sd \sigma, the quantity (sum of n variables - n\mu)/\sqrt{n}\sigma should asymptotically follow normal distribution with mean 0 and standard deviation 1(belive "distribution" is a typing mistake).

[HallsofIvy]

Yes, I meant "all possible samples" for a specific distribution- therefore "independently and identically distributed"

You are right that I mean "standard deviation", not "standard distribution"!