some small changes
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10 changed files with 48 additions and 45 deletions
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@ -87,7 +87,7 @@ to an infinite number of random variables.
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Let $\bP_n, n \in \N$ be probability measures on $(\R^n, \cB(\R^n))$
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Let $\bP_n, n \in \N$ be probability measures on $(\R^n, \cB(\R^n))$
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which are \vocab{consistent},
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which are \vocab{consistent},
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then there exists a unique probability measure $\bP^{\otimes}$
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then there exists a unique probability measure $\bP^{\otimes}$
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on $(\R^\infty, B(R^\infty))$ (where $B(R^{\infty}$ has to be defined),
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on $(\R^\infty, B(R^\infty))$ (where $B(R^{\infty})$ has to be defined),
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such that
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such that
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\[
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\[
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\forall n \in \N, B_1,\ldots, B_n \in B(\R):
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\forall n \in \N, B_1,\ldots, B_n \in B(\R):
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@ -43,7 +43,8 @@ Now, for any $C \subseteq \R^n$ let $C^\ast \coloneqq C \times \R^{\infty}$.
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Note that $C \in \cB_n \implies C^\ast \in \cF_n$.
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Note that $C \in \cB_n \implies C^\ast \in \cF_n$.
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Thus $\cF_n = \{C^\ast : C \in \cB_n\}$.
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Thus $\cF_n = \{C^\ast : C \in \cB_n\}$.
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Define $\lambda_n : \cF_n : \to [0,1]$ by $\lambda_n(C^\ast) \coloneqq (\mu_1 \otimes \ldots \otimes \mu_n)(C)$.
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Define $\lambda_n : \cF_n : \to [0,1]$ by $\lambda_n(C^\ast) \coloneqq (\mu_1 \otimes \ldots \otimes \mu_n)(C)$.
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It is easy to see that $\lambda_{n+1} \defon{\cF_n} = \lambda_n$ (\vocab{consistency}).
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It is easy to see that $\lambda_{n+1} \defon{\cF_n} = \lambda_n$,
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i.e. the $\lambda_n$ form a consistent family.
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Recall the following theorem from measure theory:
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Recall the following theorem from measure theory:
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@ -73,16 +73,16 @@ In order to prove \autoref{thm2}, we need the following:
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We have
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We have
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\begin{IEEEeqnarray*}{rCl}
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\begin{IEEEeqnarray*}{rCl}
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&&\int_{A_i} (\underbrace{X_1 + \ldots + X_i}_C
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&&\int_{A_i} (\underbrace{X_1 + \ldots + X_i}_C
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+ \underbrace{X_{i+1} + \ldots + X_n}_D)^2 d \bP\\
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+ \underbrace{X_{i+1} + \ldots + X_n}_D)^2 \dif\bP\\
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&=& \int_{A_i} C^2 d\bP
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&=& \int_{A_i} C^2 \dif\bP
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+ \underbrace{\int_{A_i} D^2 d \bP}_{\ge 0}
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+ \underbrace{\int_{A_i} D^2 \dif\bP}_{\ge 0}
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+ 2 \int_{A_i} CD d\bP\\
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+ 2 \int_{A_i} CD \dif\bP\\
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&\ge& \int_{A_i} \underbrace{C^2}_{\ge \epsilon^2} d \bP
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&\ge& \int_{A_i} \underbrace{C^2}_{\ge \epsilon^2} \dif\bP
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+ 2 \int \underbrace{\One_{A_i} (X_1 + \ldots + X_i)}_E \underbrace{(X_{i+1} + \ldots + X_n)}_D d \bP\\
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+ 2 \int \underbrace{\One_{A_i} (X_1 + \ldots + X_i)}_E \underbrace{(X_{i+1} + \ldots + X_n)}_D \dif\bP\\
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&\ge& \int_{A_i} \epsilon^2 d\bP,
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&\ge& \int_{A_i} \epsilon^2 \dif\bP,
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\end{IEEEeqnarray*}
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\end{IEEEeqnarray*}
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since by the independence of $E$ and $D$,
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since by the independence of $E$ and $D$,
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and $\bE(X_{i+1}) = \ldots = \bE(X_n) = 0$ we have $\int D E d\bP = 0$.
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and $\bE(X_{i+1}) = \ldots = \bE(X_n) = 0$ we have $\int D E \dif\bP = 0$.
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Hence
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Hence
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\[
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\[
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@ -122,7 +122,7 @@ In order to prove \autoref{thm2}, we need the following:
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is a Cauchy sequence iff $a(\omega) = 0$.
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is a Cauchy sequence iff $a(\omega) = 0$.
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We want to show that $\bP[a(\omega) > 0] = 0$.
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We want to show that $\bP[a(\omega) > 0] = 0$.
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For this, it suffices to show that $\bP(a(\omega) > \epsilon] = 0$
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For this, it suffices to show that $\bP[a(\omega) > \epsilon] = 0$
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for all $\epsilon > 0$.
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for all $\epsilon > 0$.
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For a fixed $\epsilon > 0$, we obtain:
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For a fixed $\epsilon > 0$, we obtain:
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\begin{IEEEeqnarray*}{rCl}
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\begin{IEEEeqnarray*}{rCl}
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@ -13,8 +13,8 @@
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of numbers converge:
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of numbers converge:
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\begin{itemize}
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\begin{itemize}
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\item $\sum_{n \ge 1} \bP(|X_n| > C)$,
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\item $\sum_{n \ge 1} \bP(|X_n| > C)$,
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\item $\sum_{n \ge 1} \underbrace{\int_{|X_n| \le C} X_n d\bP}_{\text{\vocab{truncated mean}}}$,
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\item $\sum_{n \ge 1} \underbrace{\int_{|X_n| \le C} X_n \dif\bP}_{\text{\vocab{truncated mean}}}$,
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\item $\sum_{n \ge 1} \underbrace{\int_{|X_n| \le C} X_n^2 d\bP - \left( \int_{|X_n| \le C} X_n d\bP \right)^2}_{\text{\vocab{truncated variance} }}$.
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\item $\sum_{n \ge 1} \underbrace{\int_{|X_n| \le C} X_n^2 \dif\bP - \left( \int_{|X_n| \le C} X_n \dif\bP \right)^2}_{\text{\vocab{truncated variance} }}$.
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\end{itemize}
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\end{itemize}
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Then $\sum_{n \ge 1} X_n$ converges almost surely.
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Then $\sum_{n \ge 1} X_n$ converges almost surely.
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\item Suppose $\sum_{n \ge 1} X_n$ converges almost surely.
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\item Suppose $\sum_{n \ge 1} X_n$ converges almost surely.
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@ -37,8 +37,8 @@ For the proof we'll need a slight generalization of \autoref{thm2}:
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Since the $X_n$ are independent, the $Y_n$ are independent as well.
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Since the $X_n$ are independent, the $Y_n$ are independent as well.
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Furthermore, the $Y_n$ are uniformly bounded.
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Furthermore, the $Y_n$ are uniformly bounded.
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By our assumption, the series
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By our assumption, the series
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$\sum_{n \ge 1} \int_{|X_n| \le C} X_n d\bP = \sum_{n \ge 1} \bE[Y_n]$
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$\sum_{n \ge 1} \int_{|X_n| \le C} X_n \dif\bP = \sum_{n \ge 1} \bE[Y_n]$
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and $\sum_{n \ge 1} \int_{|X_n| \le C} X_n^2 d\bP - \left( \int_{|X_n| \le C} X_n d\bP \right)^2 = \sum_{n \ge 1} \Var(Y_n)$
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and $\sum_{n \ge 1} \int_{|X_n| \le C} X_n^2 \dif\bP - \left( \int_{|X_n| \le C} X_n \dif\bP \right)^2 = \sum_{n \ge 1} \Var(Y_n)$
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converges.
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converges.
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By \autoref{thm4} it follows that $\sum_{n \ge 1} Y_n < \infty$
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By \autoref{thm4} it follows that $\sum_{n \ge 1} Y_n < \infty$
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almost surely.
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almost surely.
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@ -76,10 +76,10 @@ For the proof we'll need a slight generalization of \autoref{thm2}:
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We have
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We have
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\begin{IEEEeqnarray*}{rCl}
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\begin{IEEEeqnarray*}{rCl}
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\bE(Y_n) &=& \int_{|X_n| \le C} X_n d \bP + C \bP(|X_n| \ge C),\\
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\bE(Y_n) &=& \int_{|X_n| \le C} X_n \dif\bP + C \bP(|X_n| \ge C),\\
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\bE(Z_n) &=& \int_{|X_n| \le C} X_n d \bP - C \bP(|X_n| \ge C).
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\bE(Z_n) &=& \int_{|X_n| \le C} X_n \dif\bP - C \bP(|X_n| \ge C).
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\end{IEEEeqnarray*}
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\end{IEEEeqnarray*}
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Since $\bE(Y_n) + \bE(Z_n) = 2 \int_{|X_n| \le C} X_n d\bP$
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Since $\bE(Y_n) + \bE(Z_n) = 2 \int_{|X_n| \le C} X_n \dif\bP$
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the second series converges,
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the second series converges,
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and since
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and since
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$\bE(Y_n) - \bE(Z_n)$ converges, the first series converges.
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$\bE(Y_n) - \bE(Z_n)$ converges, the first series converges.
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@ -160,8 +160,7 @@ More formally:
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and uniform boundedness was not used.
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and uniform boundedness was not used.
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The idea of `` $\implies$ '' will lead to coupling. % TODO ?
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The idea of `` $\implies$ '' will lead to coupling. % TODO ?
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\end{remark}
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\end{remark}
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A proof of \autoref{thm5} can be found in the notes.\notes
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% TODO Proof of thm5 in the notes
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\begin{example}[Application of \autoref{thm5}]
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\begin{example}[Application of \autoref{thm5}]
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The series $\sum_{n} \frac{1}{n^{\frac{1}{2} + \epsilon}}$
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The series $\sum_{n} \frac{1}{n^{\frac{1}{2} + \epsilon}}$
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does not converge for $\epsilon < \frac{1}{2}$.
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does not converge for $\epsilon < \frac{1}{2}$.
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@ -106,6 +106,7 @@ We have
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\item We have $\phi(0) = 1$.
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\item We have $\phi(0) = 1$.
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\item $|\phi(t)| \le \int_{\R} |e^{\i t x} | \bP(dx) = 1$.
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\item $|\phi(t)| \le \int_{\R} |e^{\i t x} | \bP(dx) = 1$.
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\end{itemize}
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\end{itemize}
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\todo{Properties of characteristic functions}
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\begin{remark}
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\begin{remark}
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Suppose $(\Omega, \cF, \bP)$ is an arbitrary probability space and
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Suppose $(\Omega, \cF, \bP)$ is an arbitrary probability space and
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@ -9,7 +9,7 @@ We consider $(\R, \cB(\R))$.
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By $M_1 (\R)$ we denote the set of all probability measures on $\left( \R, \cB(\R) \right)$.
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By $M_1 (\R)$ we denote the set of all probability measures on $\left( \R, \cB(\R) \right)$.
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\end{notation}
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\end{notation}
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For all $\bP \in M_1(\R)$ we define $\phi_{\bP}(t) = \int_{\R} e^{\i t x}d\bP(x)$.
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For all $\bP \in M_1(\R)$ we define $\phi_{\bP}(t) = \int_{\R} e^{\i t x}\dif\bP(x)$.
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If $X: (\Omega, \cF) \to (\R, \cB(\R))$ is a random variable, we write
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If $X: (\Omega, \cF) \to (\R, \cB(\R))$ is a random variable, we write
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$\phi_X(t) \coloneqq \bE[e^{\i t X}] = \phi_{\mu}(t)$,
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$\phi_X(t) \coloneqq \bE[e^{\i t X}] = \phi_{\mu}(t)$,
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where $\mu = \bP X^{-1}$.
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where $\mu = \bP X^{-1}$.
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@ -27,15 +27,15 @@ where $\mu = \bP X^{-1}$.
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We have
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We have
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\begin{IEEEeqnarray*}{rCl}
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\begin{IEEEeqnarray*}{rCl}
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&&\lim_{T \to \infty} \frac{1}{2 \pi} \int_{-T}^T \int_{\R} \frac{e^{-\i t b}- e^{-\i t a}}{-\i t} e^{\i t x} dt d \bP(x)\\
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&&\lim_{T \to \infty} \frac{1}{2 \pi} \int_{-T}^T \int_{\R} \frac{e^{-\i t b}- e^{-\i t a}}{-\i t} e^{\i t x} dt \dif\bP(x)\\
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&\overset{\text{Fubini for $L^1$}}{=}& \lim_{T \to \infty} \frac{1}{2 \pi} \int_{\R} \int_{-T}^T \frac{e^{-\i t b}- e^{-\i t a}}{-\i t} e^{\i t x} dt d \bP(x)\\
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&\overset{\text{Fubini for $L^1$}}{=}& \lim_{T \to \infty} \frac{1}{2 \pi} \int_{\R} \int_{-T}^T \frac{e^{-\i t b}- e^{-\i t a}}{-\i t} e^{\i t x} dt \dif\bP(x)\\
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&=& \lim_{T \to \infty} \frac{1}{2 \pi} \int_{\R} \int_{-T}^T \frac{e^{\i t (b-x)}- e^{\i t (x-a)}}{-\i t} dt d \bP(x)\\
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&=& \lim_{T \to \infty} \frac{1}{2 \pi} \int_{\R} \int_{-T}^T \frac{e^{\i t (b-x)}- e^{\i t (x-a)}}{-\i t} dt \dif\bP(x)\\
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&=& \lim_{T \to \infty} \frac{1}{2 \pi} \int_{\R} \underbrace{\int_{-T}^T \left[ \frac{\cos(t (x-b)) - \cos(t(x-a))}{-\i t}\right] dt d \bP(x)}_{=0 \text{, as the function is odd}}
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&=& \lim_{T \to \infty} \frac{1}{2 \pi} \int_{\R} \underbrace{\int_{-T}^T \left[ \frac{\cos(t (x-b)) - \cos(t(x-a))}{-\i t}\right] dt \dif\bP(x)}_{=0 \text{, as the function is odd}}
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\\&&
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\\&&
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+ \lim_{T \to \infty} \frac{1}{2\pi} \int_{\R}\int_{-T}^T \frac{\sin(t ( x - b)) - \sin(t(x-a))}{-t} dt d\bP(x)\\
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+ \lim_{T \to \infty} \frac{1}{2\pi} \int_{\R}\int_{-T}^T \frac{\sin(t ( x - b)) - \sin(t(x-a))}{-t} dt \dif\bP(x)\\
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&=& \lim_{T \to \infty} \frac{1}{\pi} \int_\R \int_{0}^T \frac{\sin(t(x-a)) - \sin(t(x-b))}{t} dt d\bP(x)\\
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&=& \lim_{T \to \infty} \frac{1}{\pi} \int_\R \int_{0}^T \frac{\sin(t(x-a)) - \sin(t(x-b))}{t} dt \dif\bP(x)\\
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&\overset{\substack{\text{\autoref{fact:intsinxx},}\\\text{dominated convergence}}}{=}& \frac{1}{\pi} \int -\frac{\pi}{2} \One_{x < a} + \frac{\pi}{2} \One_{x > a }
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&\overset{\substack{\text{\autoref{fact:intsinxx},}\\\text{dominated convergence}}}{=}& \frac{1}{\pi} \int -\frac{\pi}{2} \One_{x < a} + \frac{\pi}{2} \One_{x > a }
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- (- \frac{\pi}{2} \One_{x < b} + \frac{\pi}{2} \One_{x > b}) d\bP(x)\\
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- (- \frac{\pi}{2} \One_{x < b} + \frac{\pi}{2} \One_{x > b}) \dif\bP(x)\\
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&=& \frac{1}{2} \bP(\{a\} ) + \frac{1}{2} \bP(\{b\}) + \bP((a,b))\\
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&=& \frac{1}{2} \bP(\{a\} ) + \frac{1}{2} \bP(\{b\}) + \bP((a,b))\\
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&=& \frac{F(b) + F(b-)}{2} - \frac{F(a) - F(a-)}{2}
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&=& \frac{F(b) + F(b-)}{2} - \frac{F(a) - F(a-)}{2}
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\end{IEEEeqnarray*}
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\end{IEEEeqnarray*}
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@ -129,10 +129,10 @@ However, Fourier analysis is not only useful for continuous probability density
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\begin{refproof}{bochnersformula}
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\begin{refproof}{bochnersformula}
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We have
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We have
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\begin{IEEEeqnarray*}{rCl}
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\begin{IEEEeqnarray*}{rCl}
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RHS &=& \lim_{T \to \infty} \frac{1}{2 T} \int_{-T}^T e^{-\i t x} \int_{\R} e^{\i t y} d \bP(y) \\
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RHS &=& \lim_{T \to \infty} \frac{1}{2 T} \int_{-T}^T e^{-\i t x} \int_{\R} e^{\i t y} \dif\bP(y) \\
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&\overset{\text{Fubini}}{=}& \lim_{T \to \infty} \frac{1}{2 T} \int_\R \bP(dy) \int_{-T}^T \underbrace{e^{-\i t (y - x)}}_{\cos(t ( y - x)) + \i \sin(t (y-x))} dt\\
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&\overset{\text{Fubini}}{=}& \lim_{T \to \infty} \frac{1}{2 T} \int_\R \bP(dy) \int_{-T}^T \underbrace{e^{-\i t (y - x)}}_{\cos(t ( y - x)) + \i \sin(t (y-x))} dt\\
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&=& \lim_{T \to \infty} \frac{1}{2T} \int_{\R} d\bP(y) \int_{-T}^T \cos(t(y - x)) dt\\
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&=& \lim_{T \to \infty} \frac{1}{2T} \int_{\R} \dif\bP(y) \int_{-T}^T \cos(t(y - x)) dt\\
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&=& \lim_{T \to \infty} \frac{1}{2 T }\int_{\R} \frac{2 \sin(T (y-x)}{T (y-x)} d \bP(y)\\
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&=& \lim_{T \to \infty} \frac{1}{2 T }\int_{\R} \frac{2 \sin(T (y-x)}{T (y-x)} \dif\bP(y)\\
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\end{IEEEeqnarray*}
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\end{IEEEeqnarray*}
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Furthermore
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Furthermore
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\[
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\[
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\]
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\]
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Hence
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Hence
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\begin{IEEEeqnarray*}{rCl}
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\begin{IEEEeqnarray*}{rCl}
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\lim_{T \to \infty} \frac{1}{2 T }\int_{\R} \frac{2 \sin(T (y-x)}{T (y-x)} d \bP(y) &=& \bP\left( \{x\}\right)
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\lim_{T \to \infty} \frac{1}{2 T }\int_{\R} \frac{2 \sin(T (y-x)}{T (y-x)} \dif\bP(y) &=& \bP\left( \{x\}\right)
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\end{IEEEeqnarray*}
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\end{IEEEeqnarray*}
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% TODO by dominated convergence?
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% TODO by dominated convergence?
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\end{refproof}
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\end{refproof}
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For part (b) we have:
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For part (b) we have:
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\begin{IEEEeqnarray*}{rCl}
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\begin{IEEEeqnarray*}{rCl}
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\sum_{j,k} c_j \overline{c_k} \phi(t_j - t_k) &=& \sum_{j,k} c_j \overline{c_k} \int_\R e^{\i (t_j - t_k) x} d \bP(x)\\
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\sum_{j,k} c_j \overline{c_k} \phi(t_j - t_k) &=& \sum_{j,k} c_j \overline{c_k} \int_\R e^{\i (t_j - t_k) x} \dif\bP(x)\\
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&=& \int_{\R} \sum_{j,k} c_j \overline{c_k} e^{\i t_j x} \overline{e^{\i t_k x}} d\bP(x)\\
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&=& \int_{\R} \sum_{j,k} c_j \overline{c_k} e^{\i t_j x} \overline{e^{\i t_k x}} \dif\bP(x)\\
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&=& \int_{\R}\sum_{j,k} c_j e^{\i t_j x} \overline{c_k e^{\i t_k x}} d\bP(x)\\
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&=& \int_{\R}\sum_{j,k} c_j e^{\i t_j x} \overline{c_k e^{\i t_k x}} \dif\bP(x)\\
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&=& \int_{\R} \left| \sum_{l} c_l e^{\i t_l x}\right|^2 \ge 0
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&=& \int_{\R} \left| \sum_{l} c_l e^{\i t_l x}\right|^2 \ge 0
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\end{IEEEeqnarray*}
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\end{IEEEeqnarray*}
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\end{refproof}
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\end{refproof}
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@ -187,7 +187,7 @@ Unfortunately, we won't prove \autoref{bochnersthm} in this lecture.
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\label{def:weakconvergence}
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\label{def:weakconvergence}
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We say that $\bP_n \subseteq M_1(\R)$ \vocab[Convergence!weak]{converges weakly} towards $\bP \in M_1(\R)$ (notation: $\bP_n \implies \bP$), iff
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We say that $\bP_n \subseteq M_1(\R)$ \vocab[Convergence!weak]{converges weakly} towards $\bP \in M_1(\R)$ (notation: $\bP_n \implies \bP$), iff
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\[
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\[
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\forall f \in C_b(\R)~ \int f d\bP_n \to \int f d\bP.
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\forall f \in C_b(\R)~ \int f \dif\bP_n \to \int f \dif\bP.
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\]
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\]
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Where
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Where
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\[
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\[
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@ -197,12 +197,13 @@ Unfortunately, we won't prove \autoref{bochnersthm} in this lecture.
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In analysis, this is also known as $\text{weak}^\ast$ convergence.
|
In analysis, this is also known as $\text{weak}^\ast$ convergence.
|
||||||
\end{definition}
|
\end{definition}
|
||||||
\begin{remark}
|
\begin{remark}
|
||||||
This notion of convergence makes $M_1(\R)$ a separable metric space. We can construc a metric on $M_1(\R)$ that turns $M_1(\R)$ into a complete
|
This notion of convergence makes $M_1(\R)$ a separable metric space.
|
||||||
|
We can construct a metric on $M_1(\R)$ that turns $M_1(\R)$ into a complete
|
||||||
and separable metric space:
|
and separable metric space:
|
||||||
|
|
||||||
Consider the sets
|
Consider the sets
|
||||||
\[
|
\[
|
||||||
\{\bP \in M_1(\R): \forall i=1,\ldots,n ~ \int f d \bP - \int f_i d\bP < \epsilon \}
|
\{\bP \in M_1(\R): \forall i=1,\ldots,n ~ \int f \dif\bP - \int f_i \dif\bP < \epsilon \}
|
||||||
\]
|
\]
|
||||||
for any $f,f_1,\ldots, f_n \in C_b(\R)$.
|
for any $f,f_1,\ldots, f_n \in C_b(\R)$.
|
||||||
These sets form a basis for the topology on $M_1(\R)$.
|
These sets form a basis for the topology on $M_1(\R)$.
|
||||||
|
@ -211,13 +212,13 @@ Unfortunately, we won't prove \autoref{bochnersthm} in this lecture.
|
||||||
\begin{example}
|
\begin{example}
|
||||||
\begin{itemize}
|
\begin{itemize}
|
||||||
\item Let $\bP_n = \delta_{\frac{1}{n}}$.
|
\item Let $\bP_n = \delta_{\frac{1}{n}}$.
|
||||||
Then $\int f d \bP_n = f(\frac{1}{n}) \to f(0) = \int f d \delta_0$
|
Then $\int f \dif\bP_n = f(\frac{1}{n}) \to f(0) = \int f d \delta_0$
|
||||||
for any continuous, bounded function $f$.
|
for any continuous, bounded function $f$.
|
||||||
Hence $\bP_n \to \delta_0$.
|
Hence $\bP_n \to \delta_0$.
|
||||||
\item $\bP_n \coloneqq \delta_n$ does not converge weakly,
|
\item $\bP_n \coloneqq \delta_n$ does not converge weakly,
|
||||||
as for example
|
as for example
|
||||||
\[
|
\[
|
||||||
\int \cos(\pi x) d\bP_n(x)
|
\int \cos(\pi x) \dif\bP_n(x)
|
||||||
\]
|
\]
|
||||||
does not converge.
|
does not converge.
|
||||||
|
|
||||||
|
@ -225,7 +226,7 @@ Unfortunately, we won't prove \autoref{bochnersthm} in this lecture.
|
||||||
Let $f \in C_b(\R)$ arbitrary.
|
Let $f \in C_b(\R)$ arbitrary.
|
||||||
Then
|
Then
|
||||||
\[
|
\[
|
||||||
\int f d\bP_n = \frac{1}{n}(n) + (1 - \frac{1}{n}) f(0) \to f(0)
|
\int f \dif\bP_n = \frac{1}{n}(n) + (1 - \frac{1}{n}) f(0) \to f(0)
|
||||||
\]
|
\]
|
||||||
since $f$ is bounded.
|
since $f$ is bounded.
|
||||||
Hence $\bP_n \implies \delta_0$.
|
Hence $\bP_n \implies \delta_0$.
|
||||||
|
|
|
@ -124,7 +124,7 @@ If $S_n \sim \Bin(n,p)$ and $[a,b] \subseteq \R$, we have
|
||||||
&\approx& \Phi(0.01 \sqrt{\frac{n}{p(1-p)}}) - \Phi(-0.01 \sqrt{\frac{n}{p(1-p)}})\\
|
&\approx& \Phi(0.01 \sqrt{\frac{n}{p(1-p)}}) - \Phi(-0.01 \sqrt{\frac{n}{p(1-p)}})\\
|
||||||
&=& 2\Phi(0.01 \sqrt{\frac{n}{p(1-p)}}) - 1\\
|
&=& 2\Phi(0.01 \sqrt{\frac{n}{p(1-p)}}) - 1\\
|
||||||
\end{IEEEeqnarray*}
|
\end{IEEEeqnarray*}
|
||||||
Hence, we want $\Phi(0.01 \sqrt{\frac{n}{p(1-p}}) \approx \frac{1.95}{2}$,
|
Hence, we want $\Phi(0.01 \sqrt{\frac{n}{p(1-p)}}) \approx \frac{1.95}{2}$,
|
||||||
i.e.~$n = (1.96)^2 100^2 p\cdot (1-p)$
|
i.e.~$n = (1.96)^2 100^2 p\cdot (1-p)$
|
||||||
We have $p\cdot (1-p) \le \frac{1}{4}$,
|
We have $p\cdot (1-p) \le \frac{1}{4}$,
|
||||||
thus $n \approx (1.96)^2 \cdot 100^2 \cdot \frac{1}{4} = 9600$ suffices.
|
thus $n \approx (1.96)^2 \cdot 100^2 \cdot \frac{1}{4} = 9600$ suffices.
|
||||||
|
|
|
@ -223,3 +223,4 @@ Now, we can finally prove the CLT:
|
||||||
where $\langle t, X\rangle \coloneqq \sum_{i = 1}^d t_i X_i$.
|
where $\langle t, X\rangle \coloneqq \sum_{i = 1}^d t_i X_i$.
|
||||||
\end{remark}
|
\end{remark}
|
||||||
Exercise: Find out, which properties also hold for $d > 1$.
|
Exercise: Find out, which properties also hold for $d > 1$.
|
||||||
|
\todo{TODO}
|
||||||
|
|
|
@ -16,7 +16,7 @@ if $X_1, X_2,\ldots$ are i.i.d.~with $ \mu = \bE[X_1]$,
|
||||||
Assume $X_1, X_2, \ldots,$ are independent (but not necessarily identically distributed) with $\mu_i = \bE[X_i] < \infty$ and $\sigma_i^2 = \Var(X_i) < \infty$.
|
Assume $X_1, X_2, \ldots,$ are independent (but not necessarily identically distributed) with $\mu_i = \bE[X_i] < \infty$ and $\sigma_i^2 = \Var(X_i) < \infty$.
|
||||||
Let $S_n = \sqrt{\sum_{i=1}^{n} \sigma_i^2}$
|
Let $S_n = \sqrt{\sum_{i=1}^{n} \sigma_i^2}$
|
||||||
and assume that
|
and assume that
|
||||||
\[\lim_{n \to \infty} \frac{1}{S_n^2} \bE\left[(X_i - \mu_i)^2 \One_{|X_i - \mu_i| > \epsilon \S_n}\right] = 0\]
|
\[\lim_{n \to \infty} \frac{1}{S_n^2} \bE\left[(X_i - \mu_i)^2 \One_{|X_i - \mu_i| > \epsilon S_n}\right] = 0\]
|
||||||
for all $\epsilon > 0$
|
for all $\epsilon > 0$
|
||||||
(\vocab{Lindeberg condition}\footnote{``The truncated variance is negligible compared to the variance.''}).
|
(\vocab{Lindeberg condition}\footnote{``The truncated variance is negligible compared to the variance.''}).
|
||||||
|
|
||||||
|
|
|
@ -87,7 +87,7 @@ we need the following theorem, which we won't prove here:
|
||||||
via
|
via
|
||||||
\begin{IEEEeqnarray*}{rCl}
|
\begin{IEEEeqnarray*}{rCl}
|
||||||
L^p &\longrightarrow & (L^q)^\ast \\
|
L^p &\longrightarrow & (L^q)^\ast \\
|
||||||
f &\longmapsto & (g \mapsto \int g f \dif d\bP)
|
f &\longmapsto & (g \mapsto \int g f \dif\bP)
|
||||||
\end{IEEEeqnarray*}
|
\end{IEEEeqnarray*}
|
||||||
|
|
||||||
We also have $(L^1)^\ast \cong L^\infty$,
|
We also have $(L^1)^\ast \cong L^\infty$,
|
||||||
|
|
Reference in a new issue