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The Central Limit Theorem states for a sequence of i.i.d. random variables $\{X_i\}$, $$\frac{\overline{X} - \mu}{\sigma/\sqrt{n}} \to N(0,1)$$ in distribution as $n \to \infty$. I saw in some lecture notes that this implies $$\overline{X} \to N\left(\mu,\frac{\sigma^2}{n}\right).$$ But does that statement make any sense? The right side is a function of $n$. What let's us move the $\sqrt{n}$ and derive this implication? I can see it heuristically and how it can be used in statistics exercises (use a large value of $n$).

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  • $\begingroup$ I believe you multiply by $\sqrt n$ rather than divide. $\endgroup$ – Zach466920 Jul 15 '15 at 17:14
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    $\begingroup$ @Zach466920 that's if you are summing $X_i$ and subtracting $n\mu$. I'm explicitly working with the sample mean. $\endgroup$ – user217285 Jul 15 '15 at 17:17
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    $\begingroup$ You're right, it doesn't make sense since the righthand side depends on $n$. $\endgroup$ – Stefan Hansen Jul 15 '15 at 17:20
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    $\begingroup$ Indeed it does not make sense because a limit cannot depend on $n$ (and I might have gone as far as explaining this on the site a few times, some of them raising very vocal oppositions). $\endgroup$ – Did Jul 15 '15 at 17:26
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    $\begingroup$ The CLT says that $Y_n\to N(0,1)$ in distribution, for the random variables $Y_n$ you know. This means that $P(Y_n\leqslant y)\to\Phi(y)$ for every fixed $y$. This says nothing about $P(Y_n\leqslant y_n)$ when $y_n$ depends on $n$. $\endgroup$ – Did Jul 15 '15 at 17:37
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Taken literally, anything that says $\displaystyle\left[\lim_{n\to\infty} (\cdots\cdots) = \text{something depending on }n\right]$ is wrong, as is anything that says $\displaystyle \left[ (\cdots\cdots) \to\text{something depending on }n\text{ as }n\to\infty\right]$. However, some authors adopt a convention according to which $\displaystyle\left[ (\cdots\cdots) \to N\left( \mu, \frac{\sigma^2} n \right) \right]\text{ as }n\to\infty$ means the same thing as $$ \frac{\bar X-\mu}{\sigma/\sqrt n} \to AN(0,1), $$ where "$AN$" means "asymptotically normal". The expression $\sigma^2/n$ emphasizes the rate of convergence.

The question says "any sequence of i.i.d. random variables". That's not quite right: the most usual form assumes they have finite variance. If they all have the standard Cauchy distribution, for example, then the conclusion is false: the sample mean $\bar X$ still lacks a finite variance, and in fact has the same distribution as does any one of the random variables being averaged.

Someone said in the comments under the question "I guess they just meant $\bar X \sim N(\mu,\sigma^2/n)$." That, however, is not correct unless the distibution you start with is normal, and then it's not a limit theorem at all: it's not about what happens as $n\to\infty$.

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  • $\begingroup$ In an applied statistics course (i.e. working with nice data and all CLT assumptions are satisfied), is it meaningful to say $\overline{X} \sim N\left(\mu,\frac{\sigma^2}{n}\right)$ for large $n$? $\endgroup$ – user217285 Jul 15 '15 at 18:34
  • $\begingroup$ @Nitin : One can say that's approximately true for large $n$, and approximately true may be good enough that one can neglect the difference. One thing that is exact is that the expected value is $\mu$ and the variance is $\sigma^2/n$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Jul 16 '15 at 0:07
  • $\begingroup$ That's obtained by summing the expected values and variances and dividing by n right? I can't believe I didn't see that. $\endgroup$ – user217285 Jul 16 '15 at 0:27
  • $\begingroup$ @Nitin : Correct. ${}\qquad{}$ $\endgroup$ – Michael Hardy Jul 16 '15 at 0:41

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