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Related to this question. The central limit theorem proves that the distribution of the average of $n$ uniform distributions (say, on the unit interval) converges in probability to a normal distribution. I've seen the proof, and it makes sense. But conceptually, it doesn't make sense to me. The average of $n$ values in the unit interval lies again in the unit interval. So how can their be a non-zero probability that the average lies outside of the unit interval? Shouldn't it converge to a probability distribution function that is only non-zero in the unit interval?

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  • $\begingroup$ Where does the central theorem say what you claim it says? $\endgroup$ – Dilip Sarwate Dec 21 '12 at 22:31
  • $\begingroup$ @DilipSarwate Hence the "stupid"... :p $\endgroup$ – Brian Rushton Dec 21 '12 at 22:48
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You are confusing the law of large numbers and the central limit theorem. The law of large numbers asserts: $$ \operatorname{plim}\limits_{n \to \infty} \frac{1}{n} \left(X_1 + X_2 + \cdots + X_n\right) = \mathbb{E}(X) $$ Notice that $\mathbb{E}(X)$ is going to belong to the unit interval in accord with your intuition.

The central limit theorem states: $$ \frac{ \frac{1}{n} \left(X_1 + X_2 + \cdots + X_n\right) - \mathbb{E}(X) }{ \sqrt{ \frac{\mathbb{Var}(X)}{n}}} \operatorname{\stackrel{d}{\longrightarrow}}\limits_{n\to\infty} Z $$ where $Z$ is the standard normal random variable.

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  • $\begingroup$ That's what I was looking for. It was the factor in the denominator I was missing. On Wikipedia, they phrased the same idea in a more confusing way. Thanks! $\endgroup$ – Brian Rushton Dec 21 '12 at 22:47
  • $\begingroup$ @Learner Thanks, I have corrected the names and added links. $\endgroup$ – Sasha Dec 22 '12 at 3:33

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