diff --git a/inputs/lecture_10.tex b/inputs/lecture_10.tex index 80aa817..40e082a 100644 --- a/inputs/lecture_10.tex +++ b/inputs/lecture_10.tex @@ -64,7 +64,7 @@ where $\mu = \bP X^{-1}$. Let $\bP \in M_1(\R)$ such that $\phi_\R \in L^1(\lambda)$. Then $\bP$ has a continuous probability density given by \[ - f(x) = \frac{1}{2 \pi} \int_{\R} e^{-\i t x} \phi_{\R(t) dt}. + f(x) = \frac{1}{2 \pi} \int_{\R} e^{-\i t x} \phi_{\R}(t) dt. \] \end{theorem} @@ -247,11 +247,13 @@ Unfortunately, we won't prove \autoref{bochnersthm} in this lecture. which is the distribution of $X \equiv 0$. But $F_n(0) \centernot\to F(0)$. \end{example} -\begin{theorem} +\begin{theorem} % Theorem 1 + \label{lec10_thm1} $X_n \xrightarrow{\text{dist}} X$ iff $F_n(t) \to F(t)$ for all continuity points $t$ of $F$. \end{theorem} \begin{theorem}[Levy's continuity theorem]\label{levycontinuity} + % Theorem 2 $X_n \xrightarrow{\text{dist}} X$ iff $\phi_{X_n}(t) \to \phi(t)$ for all $t \in \R$. \end{theorem} diff --git a/inputs/lecture_12.tex b/inputs/lecture_12.tex index acec514..3c8be54 100644 --- a/inputs/lecture_12.tex +++ b/inputs/lecture_12.tex @@ -139,19 +139,19 @@ First, we need to prove some properties of characteristic functions. - \int_{-\infty}^\infty t \cos(tx) \frac{1}{\sqrt{2 \pi} } e^{-\frac{x^2}{2}} \d x\\ &=& -t \phi_X(t) \end{IEEEeqnarray*} - Thus, for all $t \in \R$ + Thus, for all $t \in \R$ \[ (\log(\phi_X(t))' = \frac{\phi'_X(t)}{\phi_X(t)} = -t. - \] + \] Hence there exists $c \in \R$, such that \[ \log(\phi_X(t)) = -\frac{t^2}{2} + c. - \] + \] Since $\phi_X(0) = 1$, we obtain $c = 0$. Thus \[ \phi_X(t) = e^{-\frac{t^2}{2}}. - \] + \] \end{refproof} @@ -163,12 +163,12 @@ Now, we can finally prove the CLT: Let \[ Y_i \coloneqq \frac{X_i - \mu}{\sigma} - \] + \] i.e.~we normalize to $\bE[Y_1] = 0$ and $\Var(Y_1) = 1$. We need to show that \[ V_n \coloneqq \frac{S_n - n \mu}{ \sigma \sqrt{n}} = \frac{Y_1+ \ldots + Y_n}{\sqrt{n}} \xrightarrow{\omega, n\to \infty} \cN(0,1) % TODO - \] + \] Let $t \in \R$. Then \begin{IEEEeqnarray*}{rCl} @@ -181,7 +181,7 @@ Now, we can finally prove the CLT: We have \begin{IEEEeqnarray*}{rCl} \phi(s) &=& \phi(0) + \phi'(0) s + \frac{\phi''(0)}{2} s^2 + o(s^2), \text{as $s \to 0$}\\ - &=& 1 - \underbrace{\i \bE[Y_1] s}_{=0} + &=& 1 - \underbrace{\i \bE[Y_1] s}_{=0} - \bE[Y_1^2] \frac{s^2}{2} + o(s^2)\\ &=& 1 - \frac{s^2}{2} + o(s^2), \text{as $s \to $} \end{IEEEeqnarray*} @@ -189,13 +189,13 @@ Now, we can finally prove the CLT: Setting $s \coloneqq \frac{t}{\sqrt{n}}$ we obtain \[ \phi\left(\frac{t}{ \sqrt{n} }\right) = 1 - \frac{t^2}{2n} + o\left( \frac{t^2}{n} \right) \text{ as $n \to \infty$} - \] + \] \[ \phi_{V_n}(t) = \left( \phi\left( \frac{t}{\sqrt{n} } \right) \right)^n = (1 - \frac{t^2}{2 n } + o\left( \frac{t^2}{n} \right)^n \xrightarrow{n \to \infty} e^{-\frac{t^2}{2}}, - \] + \] where we have used the following: \begin{claim} @@ -207,7 +207,7 @@ Now, we can finally prove the CLT: We have shown that \[ \phi_n(t) \xrightarrow{n \to \infty} e^{-\frac{t^2}{2}} = \phi_N(t). - \] + \] Using \autoref{levycontinuity}, we obtain \autoref{clt}. \end{refproof} @@ -221,6 +221,3 @@ Now, we can finally prove the CLT: where $\langle t, X\rangle \coloneqq \sum_{i = 1}^d t_i X_i$. \end{remark} Exercise: Find out, which properties also hold for $d > 1$. - - - diff --git a/inputs/lecture_13.tex b/inputs/lecture_13.tex new file mode 100644 index 0000000..093d992 --- /dev/null +++ b/inputs/lecture_13.tex @@ -0,0 +1,109 @@ +% Lecture 13 2023-05 +%The difficult part is to show \autoref{levycontinuity}. +%This is the last lecture, where we will deal with independent random variables. + +We have seen, that +if $X_1, X_2,\ldots$ are i.i.d.~with $ \mu = \bE[X_1]$, + $\sigma^2 = \Var(X_1)$, + then $\frac{\sum_{i=1}^{n} (X_i - \mu)}{\sigma \sqrt{n} } \xrightarrow{(d)} \cN(0,1)$. + +\begin{question} + What happens if $X_1, X_2,\ldots$ are independent, but not identically distributed? Do we still have a CLT? +\end{question} + +\begin{theorem}[Lindeberg CLT] + \label{lindebergclt} + 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 $S_n = \sqrt{\sum_{i=1}^{n} \sigma_i^2}$ + and assume that $\lim_{n \to \infty} \frac{1}{S_n^2} \bE\left[(X_i - \mu_i)^2 \One_{|X_i - \mu_i| > \epsilon \S_n}\right] = 0$ for all $\epsilon > 0$ + (\vocab{Lindeberg condition}, ``The truncated variance is negligible compared to the variance.''). + + Then the CLT holds, i.e.~ + \[ + \frac{\sum_{i=1}^n (X_i - \mu_i)}{S_n} \xrightarrow{(d)} \cN(0,1). + \] +\end{theorem} + +\begin{theorem}[Lyapunov condition] + \label{lyapunovclt} + Let $X_1, X_2,\ldots$ be independent, $\mu_i = \bE[X_i] < \infty$, + $\sigma_i^2 = \Var(X_i) < \infty$ + and $S_n \coloneqq \sqrt{\sum_{i=1}^n \sigma_i^2}$. + Then, assume that, for some $\delta > 0$, + \[ + \lim_{n \to \infty} \sum_{i=1}^{n} \bE[(X_i - \mu_i)^{2 + \delta}] = 0 + \] + (\vocab{Lyapunov condition}). + Then the CLT holds. +\end{theorem} +\begin{remark} + The Lyapunov condition implies the Lindeberg condition. + (Exercise). +\end{remark} +We will not prove the \autoref{linebergclt} or \autoref{lyapunovclt} +in this lecture. However, they are quite important. + +We will now sketch the proof of \autoref{levycontinuity}, +details can be found in the notes.\todo{Complete this} +A generalized version of \autoref{levycontinuity} is the following: +\begin{theorem}[A generalized version of Levy's continuity \autoref{levycontinuity}] + \label{genlevycontinuity} + Suppose we have random variables $(X_n)_n$ such that + $\bE[e^{\i t X_n}] \xrightarrow{n \to \infty} \phi(t)$ for all $t \in \R$ + for some function $\phi$ on $\R$. + Then the following are equivalent: + \begin{enumerate}[(a)] + \item The distribution of $X_n$ is \vocab[Distribution!tight]{tight} (dt. ``straff''), + i.e.~$\lim_{a \to \infty} \sup_{n \in \N} \bP[|X_n| > a] = 0$. + \item $X_n \xrightarrow{(d)} X$ for some real-valued random variable $X$. + \item $\phi$ is the characteristic function of $X$. + \item $\phi$ is continuous on all of $\R$. + \item $\phi$ is continuous at $0$. + \end{enumerate} +\end{theorem} +\begin{example} + Let $Z \sim \cN(0,1)$ and $X_n \coloneqq n Z$. + We have $\phi_{X_n}(t) = \bE[[e^{\i t X_n}] = e^{-\frac{1}{2} t^2 n^2} \xrightarrow{n \to \infty} \One_{\{t = 0\} }$. + $\One_{\{t = 0\}}$ is not continuous at $0$. + By \autoref{genlevycontinuity}, $X_n$ can not converge to a real-valued + random variable. + + Exercise: $X_n \xrightarrow{(d)} \overline{X}$, + where $\bP[\overline{X} = \infty] = \frac{1}{2} = \bP[\overline{X} = -\infty]$. + + Similar examples are $\mu_n \coloneqq \delta_n$ and + $\mu_n \coloneqq \frac{1}{2} \delta_n + \frac{1}{2} \delta_{-n}$. +\end{example} + +\begin{example} + Suppose that $X_1, X_2,\ldots$ are i.d.d.~with $\bE[X_1] = 0$. + Let $\sigma^2 \coloneqq \Var(X_i)$. + Then the distribution of $\frac{S_n}{\sigma \sqrt{n}}$ is tight: + + \begin{IEEEeqnarray*}{rCl} + \bE\left[ \left( \frac{S_n}{\sqrt{n} }^2 \right)^2 \right] &=& + \frac{1}{n} \bE[ (X_1+ \ldots + X_n)^2]\\ + &=& \sigma^2 + \end{IEEEeqnarray*} + For $a > 0$, by Chebyshev's inequality, % TODO + we have + \[ + \bP\left[ \left| \frac{S_n}{\sqrt{n}} \right| > a \right] \leq \frac{\sigma^2}{a^2} \xrightarrow{a \to \infty} 0. + \] + verifying \autoref{genlevycontinuity}. +\end{example} + +\begin{example} + Suppose $C$ is a random variable which is Cauchy distributed, i.e.~$C$ + has probability distribution $f_C(x) = \frac{1}{\pi} \frac{1}{1 + x^2}$. + + We know that $\bE[|C|] = \infty$. + + We have $\phi_C(t) = \bE[e^{\i t C}] = e^{-|t|}$. + Suppose $C_1, C_2, \ldots, C_n$ are i.i.d.~Cauchy distributed + and let $S_n \coloneqq C_1 + \ldots + C_n$. + + Exercise: $\phi_{S_n}(t) = e^{-|t|} = \phi_{C_1}(t)$, thus $S_n \sim C$. +\end{example} + +