5.3 Central Limit Theorem

<< 5.2 Convergence in Distribution.md | 5.4 Extensions to Multivariate Distributions.md >>

Theorem 5.3.1: Central limit theorem

Let

• $X_1\dots X_n$ : Observations of a random sample, with
  • Common mean $\mu$
  • Common variance ${\sigma }^2\in \mathbb{P}$

If

Y_{n} & = \frac{ \sum_{i=1}^n X_{i} - n\mu }{\sqrt{ n }\sigma} \\ & = \frac{\bar{X}_{n}-\mu}{\sigma/\sqrt{ n }} \end{align} $$ Then $$ Y_{n}\xrightarrow D N(0,1) $$

Note

Recall that the notation $Y_n\stackrel{D}{\to }N(0,1)$ was introduced in convergence in distribution’s remark.

We often state the central limit theorem as:

$$\sqrt{n}(\bar{X}-\mu )\to N(0,{\sigma }^2)$$

One of the key applications of this theorem is for statistical inference, as shown in Example 5.3.1-5.3.6 in the source book.

Link to original

Exercise

Hogg & Craig 5th ed. 5.33.

Let

• $X_n$ : Random sample of size $n$
• ${\bar{X}}_n$ : Mean of $X_n$
• $X_n\sim \text{Gamma}(\alpha ,\beta )$
• $\alpha =\mu >0$
• $\beta =1$

Show that $\sqrt{n}({\bar{X}}_n-\mu )/\sqrt{{\bar{X}}_n}\stackrel{D}{\to }N(0,1)$

Answer

Diketahui $X_n$ memiliki mean $\mu =\alpha \beta$ dan variansi ${\sigma }^2=\alpha {\beta }^2$. Perhatikan bahwa $\alpha \beta =\alpha \cdot 1=\alpha$ dan $\alpha {\beta }^2=\alpha \cdot 1^2=\alpha$ sehingga $\mu ={\sigma }^2=\alpha$.

Berdasarkan law of large numbers, ${\bar{X}}_n\stackrel{P}{\to }\mu$. Sehingga berdasarkan Theorem 1,

$$\begin{array}{rl}& \sqrt{{\bar{X}}_n}\stackrel{P}{\to }\sqrt{\mu }\\ ⟹& \frac{\sqrt{{\bar{X}}_n}}{\sqrt{\mu }}\stackrel{P}{\to }1\end{array}$$

Berdasarkan central limit theorem, $\sqrt{n}(\bar{X}-\mu )/\sigma \stackrel{D}{\to }N(0,1)$. Sehingga berdasarkan Theorem 3,

$$\begin{array}{rl}& \sqrt{n}({\bar{X}}_n-\mu )/\sqrt{{\bar{X}}_n}\\ ⟹& \frac{\sqrt{n}({\bar{X}}_n-\mu )/\sqrt{\mu }}{\sqrt{{\bar{X}}_n}/\sqrt{\mu }}\stackrel{D}{\to }N(0,1)\end{array}$$

$\therefore$ $\sqrt{n}({\bar{X}}_n-\mu )/\sqrt{{\bar{X}}_n}\stackrel{D}{\to }N(0,1)$

Hogg & Craig 5th ed. 5.34.

Let

• $X_n$ : Random sample of size $n$
• $X_n\sim N(\mu ,{\sigma }^2)$
• ${\bar{X}}_n$ : Mean of $X_n$
• $S_n^2$ : Variance of $X_n$
• $T_n=({\bar{X}}_n-\mu )/\sqrt{S_n^2/(n-1)}$

Prove that $T_n\stackrel{D}{\to }N(0,1)$

Answer

Karena $X_i\sim N(\mu ,{\sigma }^2)$, berdasarkan central limit theorem: $\frac{\sqrt{n}({\bar{X}}_n-\mu )}{\sigma }\stackrel{D}{\to }N(0,1)$

Berdasarkan law of large numbers, $S_n^2\stackrel{P}{\to }{\sigma }^2$, sehingga berdasarkan Theorem 2, $\sqrt{S_n^2}\stackrel{P}{\to }\sigma$, dan berdasarkan Theorem 1, $\frac{\sqrt{S_n^2}}{\sigma }\stackrel{P}{\to }1$.

Perhatikan bahwa

$$\begin{array}{rl}{T}_n& =\frac{\sqrt{n}({\bar{X}}_n-\mu )}{\sqrt{S_n^2}}\\ & =\frac{\sqrt{n}({\bar{X}}_n-\mu )/\sigma }{\sqrt{S_n^2}/\sigma }\end{array}$$

sehingga berdasarkan Theorem 3, diperoleh

$$T_n\stackrel{D}{\to }N(0,1)$$