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central limit theorem

1 Statement

Let \(X_1,...,X_n\) be a collection of \(n\) random variables, where \(X_1,...,X_n\) all have mean \(\mu\) and variance \(\sigma^2\). Let \(\bar{X}_n = \frac{1}{n}\sum_{i=1}^{n}X_i\). Then, \[ \frac{\sqrt{n}}{\sigma} (\bar{X}_n - \mu) \underset{n\rightarrow\infty}{\overset{d}{\longrightarrow}} \mathcal{N}(0,1) \]

That is, \(\frac{\sqrt{n}}{\sigma}(\bar{X}_n - \mu)\) converges in distribution to \(\mathcal{N}(0,1)\) as \(n\) grows. This means that samples from \(\frac{\sqrt{n}}{\sigma}(\bar{X}_n - \mu)\) will form an approximate histogram of \(\mathcal{N}(0,1)\) and this approximation will get better as \(n\) increases.

2 some notes on standardization

Where does the \(\sqrt{n}\) come from? Well, we are trying to get to something that has unit variance.

  • An \(X_i\) has variance \(\sigma^2\).
  • \(\sum X_i\) has variance \(n\sigma^2\).
  • \(\frac{1}{n}\sum X_i\) has variance \(\frac{1}{n}\sigma^2\)
  • \(\sqrt{n}\frac{1}{n}\sum X_i\) has variance \(\sigma^2\)
  • \(\sqrt{n}\frac{1}{n\sigma}\sum X_i\) has variance \(1\)

3 delta method

Roughly: If \(Y_n - \theta\) converges in distribution to a normal random variable then \(g(Y_n) - g(\theta)\) also converges in distribution to a normal random variable provided \(g'(\theta)\) exists and \(g'(\theta) \neq 0\)

3.1 usage

want to use central limit theorem for function of random variable

4 Resources

Created: 2021-09-14 Tue 21:44