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I know that the Central Limit Theorem states that if a random sample of size $n$ is drawn from iid random variables $X_1, \ldots, X_n$, then the variable $$ Z = \frac{\hat x - E(X) }{{\rm sd}(\hat x)}\text{ is }\mathcal N(0,1) $$ as $n$ is big enough (where $\hat x$ denotes the sample average).

My question is: What about $\hat x$, does it also converge to a normal distribution?

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marked as duplicate by whuber Sep 1 at 16:31

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For intuition about the CLT, please see stats.stackexchange.com/questions/3734. The short answer (given by laws of large numbers) is no: $\hat x$ converges to the common expectation of the $X_i$. –  whuber Aug 31 at 16:28
    
Do you mean what does the value $\hat x$ converge to, or what does the distribution of $\hat x$ converge to? –  gung Aug 31 at 16:46

1 Answer 1

The central limit theorem says the variable $Z$ converges in distribution to the standard normal.

$$Z=\frac{\hat{x}-E(X)}{\text{sd}(\hat{x})} \overset{D}{\rightarrow} N(0,1)$$

Rearrange this at a fixed value of $n$ (number of samples) to get, approximately:

$$\hat{x} \sim N(E(X),\text{Var}(\hat{x})) $$

where $\text{Var}(\hat{x})=\frac{\text{Var}(X)}{n}$ is the variance of the sample mean. You might interpret this as the distribution of $\hat{x}$ tends towards being normal with mean $E(X)$ and variance $\frac{\text{Var}(X)}{n}$.

For large values of $n$, the variance will tend to $0$. which means that the normal distribution of $\hat{x}$ will actually converge to a point mass distribution at $E(X)$. Convergence in distribution to a constant implies convergence in probability to that constant. So

$$\hat{x} \overset{P}{\rightarrow} E(X)$$

Of course this is just the weak law of large numbers. (The strong law stating that the convergence in almost sure.)

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@DilipSarwate Thanks for pointing that out. I changed that line to be approximate. –  Comp_Warrior Sep 1 at 10:59
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You may want to take a look at this answer of mine which says pretty much the same things. –  Dilip Sarwate Sep 1 at 16:59

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