0
$\begingroup$

I have a bit confusion of understanding. The central limit theorem in statistics states that, given a sufficiently large sample size, the sampling distribution of the mean for a variable will approximate a normal distribution. But the definition of sampling distribution is the probability distribution of a statistic that is obtained from many random samples of a specific population.

I don't get it and see a contradiction of how getting a sample distribution from one large sample as CLT states while sampling distribution definition says it needs more than one sample?

$\endgroup$
0

1 Answer 1

0
$\begingroup$

You are confused only about the high level description of the CLT which may be using terminology that you are misinterpreting. You may also need to clarify what you mean by "one large sample." Recall that the number of samples $n$ is the number of iid random variables. A "large sample size" means $n$ is large (recall CLT needs $n\rightarrow\infty$).


CLT: If $\{X_i\}_{i=1}^{\infty}$ are iid with $E[X_i]=m$ and $Var(X_i)=\sigma^2$, where $0<\sigma^2<\infty$, then define $$G_n=\frac{1}{\sqrt{n \sigma^2}}\sum_{i=1}^n(X_i-m) \quad \forall n \in \{1, 2, 3, ...\}$$ The CLT says the distribution of $G_n$ approaches the $N(0,1)$ distribution as $n\rightarrow\infty$.

Equivalent expression using sample mean: Define the sample mean over $n$ samples by $$M_n=\frac{1}{n}\sum_{i=1}^nX_i \quad \forall n \in \{1, 2, 3, ...\}$$ We can write $G_n$ in terms of only $M_n$ by
$$G_n=\left(\sqrt{\frac{n}{\sigma^2}}\right)(M_n-m)$$ This expression for $G_n$ hides the iid random variables $\{X_i\}_{i=1}^{\infty}$. You can use it if you prefer, but do not forget that the iid assumptions are crucial to the CLT.

Usage: In practice, when $n$ is "large enough" the CLT is used to justify the following distribution approximations: \begin{align} &G_n \approx N(0,1)\\ &M_n \approx N\left(m, \frac{\sigma^2}{n}\right) \quad (Eq. *) \end{align} these approximations have exact values for the mean and variances (for example, we exactly have $E[M_n]=m$ and $Var(M_n)=\sigma^2/n$ for all $n$). It is the Gaussian distribution that is the approximation. The number of samples $n$ needed to ensure a good CLT approximation would depend on the second and third moments of the iid random variables $X_i$ (see, for example, the Berry-Esseen theorem). However, those parameters are often unknown. Most practical applications of the CLT assume at least 10-20 iid samples, with the understanding that if we use more iid samples then we get a better approximation to the Gaussian distribution.


Other problems in statistics consider the iid random variables $\{X_i\}_{i=1}^{\infty}$ being grouped into multiple batches, with each batch containing a "large" number of the iid random variables (say, 10-20 per batch). This is the "method of batch means." This is used in certain student-t results that are exact results when the underlying random variables are iid Gaussian with unknown mean and variance (in which case we would not need the grouping into batches), but are only approximate results when we approximate the sample mean of each batch as having a Gaussian distribution with an unknown mean and variance. This approximation for each batch is motivated by (Eq. *) with $n$ being the number of iid random variables in the batch.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .