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Chapter5. The Central Limit Theorem

5-27 The Central Limit Theorem

In this section,

\[P(N \in A) = \frac{ 1 }{ \sqrt{2 \pi} } \int_{A} e^{- x^{2}/2} \ dx .\]

Theorem 27.1

If

\[S_{n} := \sum_{i=1}^{n} X_{i} ,\] \[\frac{ S_{n} - nc }{ \sigma \sqrt{n} } \overset{d}{\rightarrow} N ,\]

$\mu$ is said to be complete if

\[N \in \mathcal{A}, \ \mi(N) = 0, \ \Rightarrow \ \forall S \subseteq N, \ S \in \mathcal{A}\]

Lemma 1

\[\abs{ z_{1} \cdots z_{m} - w_{1} \cdots w_{m} } \le \sum_{k=1}^{m} \abs{ z_{k} - w_{k} } .\]

proof

Let $m = 2$.

\[\begin{eqnarray} (z_{1} - w_{1}) z_{2} + w_{1} (z_{2} - w_{2}) & = & z_{1}z_{2} - w_{1}z_{2} + w_{1}z_{2} - z_{1}z_{2} \nonumber \\ & = & z_{1}z_{2} - w_{1}w_{2} \nonumber \end{eqnarray}\]

Thus,

\[\begin{eqnarray} \abs{ z_{1}z_{2} - w_{1}w_{2} } & \le & \abs{ (z_{1} - w_{1}) z_{2} } + \abs{ w_{1} (z_{2} - w_{2}) } \nonumber \\ & \le & \abs{ (z_{1} - w_{1}) } + \abs{ (z_{2} - w_{2}) } . \nonumber \end{eqnarray}\]

Suppose that the statement holds up to $m = n - 1$. Then we prove the statement in case of $m = n$.

\[\begin{eqnarray} (z_{1} - w_{1}) z_{2}\cdots z_{m} + w_{1} (z_{2} \cdots z_{m} - w_{2} \cdots w_{m}) & = & z_{1}\cdots z_{m} - w_{1}z_{2} \cdots z_{m} + w_{1}z_{2} \cdots z_{M} - w_{1}w_{2} \cdots w_{m} \nonumber \\ & = & z_{1} \cdots z_{m} - w_{1} \cdots w_{m} \nonumber \end{eqnarray}\]

Thus,

\[\begin{eqnarray} \abs{ z_{1} \cdots z_{m} - w_{1} \cdots w_{m} } & \le & \abs{ (z_{1} - w_{1}) z_{2}\cdots z_{m} } + \abs{ w_{1} (z_{2} \cdots z_{m} - w_{2} \cdots w_{m}) } \nonumber \\ & \le & \abs{ (z_{1} - w_{1}) } + \abs{ (z_{2} \cdots z_{m} - w_{2} \cdots w_{m}) } \nonumber \\ & \le & \abs{ (z_{1} - w_{1}) } + \sum_{i=2}^{m} \abs{ (z_{i} - w_{i}) } . \nonumber \end{eqnarray}\]
$\Box$