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How do I prove the Central Limit Theorem for Sample Means?
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in Data Science & Statistics by Platinum (138,124 points) | 48 views

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Proof of the Central Limit Theorem


Suppose \(X_{1}, \ldots, X_{n}\) are i.i.d. random variables with mean 0 , variance \(\sigma_{x}^{2}\) and Moment Generating Function (MGF) \(M_{x}(t)\). Note that this assumes an MGF exists, which is not true of all random variables.
Let \(S_{n}=\sum_{i=1}^{n} X_{i}\) and \(Z_{n}=S_{n} / \sqrt{n \sigma_{x}^{2}}\). Then
\[
M_{S_{n}}(t)=\left(M_{x}(t)\right)^{n} \text { and } M_{Z_{n}}(t)=\left(M_{x}\left(\frac{t}{\sigma_{x} \sqrt{n}}\right)\right)^{n} .
\]
Using Taylor's theorem, we can write \(M_{x}(s)\) as
\[
M_{x}(s)=M_{x}(0)+s M_{x}^{\prime}(0)+\frac{1}{2} s^{2} M_{x}^{\prime \prime}(0)+e_{s},
\]
where \(e_{s} / s^{2} \rightarrow 0\) as \(s \rightarrow 0\).
\(M_{x}(0)=1\), by definition, and with \(E\left(X_{i}\right)=0\) and \(\operatorname{Var}\left(X_{i}\right)=\sigma_{x}^{2}\), we know \(M_{x}^{\prime}(0)=0\) and \(M_{x}^{\prime \prime}(0)=\sigma_{x}^{2}\). So
\[
M_{x}(s)=1+\frac{\sigma_{x}^{2}}{2} s^{2}+e_{s} .
\]
Letting \(s=t /\left(\sigma_{x} \sqrt{n}\right)\), we have \(s \rightarrow 0\) as \(n \rightarrow \infty\), and
\[
M_{Z_{n}}(t)=\left(1+\frac{\sigma_{x}^{2}}{2}\left(\frac{t}{\sigma_{x} \sqrt{n}}\right)^{2}+e_{n}\right)^{n}=\left(1+\frac{t^{2}}{2 n}+e_{n}\right)^{n}
\]
where \(n \sigma_{x}^{2} e_{n} / t^{2} \rightarrow 0\) as \(n \rightarrow \infty\).
If \(a_{n} \rightarrow a\) as \(n \rightarrow \infty\), it can be shown that
\[
\lim _{n \rightarrow \infty}\left(1+\frac{a_{n}}{n}\right)^{n}=e^{a} .
\]
It follows that
\[
\lim _{n \rightarrow \infty} M_{Z_{n}}(t)=\lim _{n \rightarrow \infty}\left(1+\frac{t^{2} / 2+n e_{n}}{n}\right)^{n}=e^{t^{2} / 2},
\]
which is the MGF of a standard Normal. If the MGF exists, then it uniquely defines the distribution. Convergence in MGF implies that \(Z_{n}\) converges in distribution to \(N(0,1)\).
The practical application of this theorem is that, for large \(n\), if \(Y_{1}, \ldots, Y_{n}\) are independent with mean \(\mu_{y}\) and variance \(\sigma_{y}^{2}\), then
\[
\sum_{i=1}^{n}\left(\frac{Y_{i}-\mu_{y}}{\sigma_{y} \sqrt{n}}\right) \dot{\sim} N(0,1), \quad \text { or } \quad \bar{Y} \dot{\sim} N\left(\mu_{y}, \sigma_{y}^{2} / n\right) .
\]
How large is "large" depends on the distribution of the \(Y_{i}\) 's. If Normal, then \(n=1\) is large enough. As the distribution becomes less Normal, larger values of \(n\) are needed.

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