Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

My friend and I have a bet going about the definition of the Central Limit Theorem.

If we define an example as a number drawn at random from some probability density function where the function has a defined finite mean and variance. And we define a sample as a set of size N examples (with N>1).

Then, we take S samples and create a sampling distribution D over the means of each individual sample.

I am arguing that the Central Limit Theorem states that as the number of samples S approaches infinity, then the sampling distribution D will approximate a normal distribution.

My friend is arguing that the Central Limit Theorem states that given any number of samples S, sampling distribution D will not necessarily approximate a normal distribution, but as the number of examples per sample N approaches infinity, then D will approximate a normal distribution.

Who is right?

Update: I lost this bet.

share|improve this question
    
Is this a duplicate of this earlier question by the same OP? –  Dilip Sarwate Jun 19 '13 at 3:30
    
I believe they are different questions related to the Central Limit Theorem, I can delete that other one if people feel they are duplicates. –  kmosley Jun 19 '13 at 3:58
add comment

3 Answers

up vote 3 down vote accepted

I'd say that your friend is more correct, in that he/she correctly points to $N$ (sample size=number of values that are summed to compute the average,i.e. the "sample mean") as the thing that must tend to $\infty$ for the CLT to hold.

We have

$$S_N=\frac{X_1+X_2+ \cdots +X_N}{N}$$

Here, in our setting, the set $\{X_1, \ldots X_N\}$ is one sample, of size $N$; and $S_N$ is the sample-mean (=average) of that sample.

This $S_N$ is a random variable (informally, it takes different random values for each sample). What the CLT says is about the distribution of this $S_N$ as $N\to \infty$. Of course, if you were practically interested in checking/experiencing that $S_N$ (for some fixed $N$) is indeed approximately gaussian, you might want to draw many values of $S_N$ and, eg, draw an histogram; for this, you would need to draw a lot of samples (each of size $N$). But this has nothing to do with the asymptotics of the theorem.

share|improve this answer
    
Sorry are you saying that for some fixed N, if you draw enough samples of SN then the distribution of SNs will approximate a gaussian? Because I thought that is what I was stating. If you increase N arbitrarily, the SNs would more accurately reflect the true mean of the underlying distribution your taking from and thus would also form a gaussian, just one with a smaller standard deviation. –  kmosley Jun 19 '13 at 4:13
    
and just to be clear, in this question one value of SN is a sample, and one value of X is an example. –  kmosley Jun 19 '13 at 4:20
    
I wrote a tool that you can use to empirically confirm that for certain finite N values and random variable distributions, the sampling distribution won't always converge on a normal. There's another tool I didn't write here. –  Sebastian Goodman Jun 19 '13 at 7:28
    
@kmosley No, I'm not saying that. The distribution of $S_N$ will be only approximately a Gaussian, for $N\to \infty$ it will tend to a Gaussian. It's not about the number of samples (and $S_N$) at all. Think about this unrelated example: we have some r.v. $X$, and we are told that it's uniformly distributed in $[0,1]$. Would you say that this fact (just a property of a random variable) depends on how many samples of $X$ I draw? I wouldn't. It's a property of the random variable in itself. Certainly, to check that, you'd want to draw many samples of $X$, but that's other thing. –  leonbloy Jun 19 '13 at 12:01
add comment

Neither of you, but you are much less wrong than he is.

Wikipeadia states what the CLT is reasonably clearly in its first 2 paragraphs. Please note that there are several variants but sampling from the same (unchanging) population meets the requirements for the Classical CLT - specifically that each of the samples is independent and identically distributed.

You have only partially captured the criteria for it to be true, specifically it is not enough for the mean and variance to be "defined" - $\sigma=\infty$ is defined, but that represents a power law distribution and the sample distribution will aproach an $\alpha$-stable distribution, not a normal distribution. Other than that you are bang on the money.

Your friend is incorrect, his postulate is demonstrably incorrect by considering setting the sample size equal to the population. In this case each sample will have the population distribution, but the sampling distribution will become more and more normal as there are more and more samples. Try the experiment with the Standard Uniform Distribution or for a more dramatic impact, this one:

$$f(x)=\begin{cases} x^2 &-\frac{1}{2}\le x\le \frac{1}{2}\\ 0 &\text{otherwise} \end{cases}$$

share|improve this answer
    
I think by "more samples" you mean the sample size: the number of observations in the sample. I think your choice of words here may lead to confusion since the question seems to use "more samples" for the number of samples of a given size — a quantity that is not relevant to the CLT at all. –  Jyotirmoy Bhattacharya Jun 19 '13 at 3:27
add comment

Both the definitions are incorrect though your friend's is less so.

The number of samples is not relevant at all since the sampling distribution gives you the probability of obtaining different values of a sample statistic and is therefore a theoretical object derived from the random process which generates your sample. Taking many samples may help you learn about the sampling distribution but that is not germane to the Central Limit Theorem since the sampling distribution exists whether we know about it or not.

Under the usual assumptions (you should look them up in a textbook) the sample mean $\bar X$ tends to the population mean so the sampling distribution of the former shrinks to a distribution that puts all its weight on one point. The variable which has a limiting normal distribution as the sample size $N$ tends to infinity is $$(\sqrt{N})\bar X.$$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.