diff --git a/inputs/lecture_10.tex b/inputs/lecture_10.tex index 31be813..d032207 100644 --- a/inputs/lecture_10.tex +++ b/inputs/lecture_10.tex @@ -50,11 +50,11 @@ where $\mu = \bP X^{-1}$. Note that the LHS is not Lebesgue-integrable. It follows that \begin{IEEEeqnarray*}{rCl} - \lim_{T \to \infty} \int_0^T \frac{\sin(t(x-a))}{x} \dif t &=& + \lim_{T \to \infty} \int_0^T \frac{\sin(t(x-a))}{t} \dif t &=& \begin{cases} - - \frac{\pi}{2}, &x < a,\\ - 0, &x = a,\\ - \frac{\pi}{2}, & \frac{\pi}{2} + - \frac{\pi}{2} &\text{if }x < a,\\ + 0 &\text{if }x = a,\\ + \frac{\pi}{2}&\text{if } x > a. \end{cases} \end{IEEEeqnarray*} \end{fact} @@ -73,7 +73,7 @@ where $\mu = \bP X^{-1}$. \item Let $\bP = \delta_{0}$. Then \[ - \phi_{\bP}(t) = \int e^{\i t x} \dif \delta_0(x) = e^{\i t 0 } = 1 + \phi_{\bP}(t) = \int e^{\i t x} \delta_0(\dif x) = e^{\i t 0 } = 1 \] \item Let $\bP = \frac{1}{2} \delta_1 + \frac{1}{2} \delta_{-1}$. Then @@ -131,23 +131,24 @@ However, Fourier analysis is not only useful for continuous probability density We have \begin{IEEEeqnarray*}{rCl} RHS &=& \lim_{T \to \infty} \frac{1}{2 T} \int_{-T}^T e^{-\i t x} \int_{\R} e^{\i t y} \bP(\dif y) \\ - &\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))} \dif t\\ - &=& \lim_{T \to \infty} \frac{1}{2T} \int_{\R} \bP(\dif y) \int_{-T}^T \cos(t(y - x)) \dif t\\ - &=& \lim_{T \to \infty} \frac{1}{2 T }\int_{\R} \frac{2 \sin(T (y-x))}{T (y-x)} \bP(\dif y)\\ + &\overset{\text{Fubini}}{=}& + \lim_{T \to \infty} \frac{1}{2 T} \int_\R \int_{-T}^T + e^{-\i t (y - x)} \dif t \bP(\dif y)\\ + &=& \lim_{T \to \infty} \frac{1}{2 T} \int_\R \int_{-T}^T + \cos(t(y-x)) + \underbrace{\i \sin(t (y-x))}_{\text{odd}} + \dif t \bP(\dif y)\\ + &=& \lim_{T \to \infty} \frac{1}{2T}\int_{\R} + \int_{-T}^T \cos(t(y - x)) \dif t \bP(\dif y)\\ + &=& \lim_{T \to \infty} \frac{1}{2T}\int_{\R} + 2T \sinc(T(y-x)) + \footnote{$\sinc(x) = \begin{cases} + \frac{\sin(x)}{x} &\text{if } x \neq 0,\\ + 1 &\text{otherwise.} + \end{cases}$} \bP(\dif y)\\ + &\overset{\text{DCT}}{=}& \int_{\R}\lim_{T \to \infty} + \sinc(T(y-x)) \bP(\dif y)\\ + &=& \bP(\{x\}). \end{IEEEeqnarray*} - Furthermore - \[ - \lim_{T \to \infty} \frac{\sin(T(x-y)}{T (y- x)} = \begin{cases} - 1, &y = x,\\ - 0, &y \neq x. - \end{cases} - % TODO TODO TODO - \] - Hence - \begin{IEEEeqnarray*}{rCl} - \lim_{T \to \infty} \frac{1}{2 T }\int_{\R} \frac{2 \sin(T (y-x))}{T (y-x)} \bP(\dif y) &=& \bP\left( \{x\}\right) - \end{IEEEeqnarray*} - % TODO by dominated convergence? \end{refproof} \begin{theorem} % Theorem 5 @@ -160,7 +161,7 @@ However, Fourier analysis is not only useful for continuous probability density i.e.~ \[\forall t_1,\ldots, t_n \in \R, (c_1,\ldots,c_n) \in \C^n ~ \sum_{j,k = 1}^n c_j \overline{c_k} \phi(t_j - t_k) \ge 0 \] - (equivalently, the matix $(\phi(t_j- t_k))_{j,k}$ is positive definite. + Equivalently, the matrix $(\phi(t_j- t_k))_{j,k}$ is positive definite. \end{enumerate} \end{theorem} \begin{refproof}{thm:lec_10thm5} @@ -231,7 +232,9 @@ Unfortunately, we won't prove \autoref{bochnersthm} in this lecture. since $f$ is bounded. Hence $\bP_n \implies \delta_0$. \item $\bP_n \coloneqq \frac{1}{\sqrt{2 \pi n}} e^{-\frac{x^2}{2n}}$. - This ``converges'' towards the $0$-measure, which is not a probability measure. Hence $\bP_n$ does not converge weakly. + This ``converges'' towards the $0$-measure, + which is not a probability measure. + Hence $\bP_n$ does not converge weakly. (Exercise) % TODO \end{itemize} \end{example} diff --git a/inputs/lecture_12.tex b/inputs/lecture_12.tex index 1fa9a32..350c82b 100644 --- a/inputs/lecture_12.tex +++ b/inputs/lecture_12.tex @@ -94,7 +94,7 @@ First, we need to prove some properties of characteristic functions. For arbitrary $h \in \R$, we have \begin{IEEEeqnarray*}{rCl} |e^{\i t X} \frac{e^{\i h X}}{h}| &\le & \left| \frac{1}{h} \left( e^{\i h X} - 1 \right)\right|\\ - &\overset{\text{\autoref{charfprop:c1}}}{\le }& \left|\frac{1}{h} \i h X\right| = |X|. + &\overset{\text{\autoref{charfprop:c1}}}{\le}& \left|\frac{1}{h} \i h X\right| = |X|. \end{IEEEeqnarray*} Thus the dominated convergence theorem can be applied and we obtain \[ @@ -123,9 +123,9 @@ First, we need to prove some properties of characteristic functions. \begin{refproof}{lec12_2} We have \begin{IEEEeqnarray*}{rCl} - \phi_X(t) &=& \frac{1}{\sqrt{2 \pi} } \int_{-\infty}^\infty e^{\i t x} e^{-\frac{x^2}{2}} \dif x\\ - &=& \frac{1}{\sqrt{2 \pi} } \int_{-\infty}^\infty (\cos(tx) + \i \sin(tx)) e^{-\frac{x^2}{2}} \dif x\\ - &=& \frac{1}{\sqrt{2 \pi} } \int_{-\infty}^\infty \cos(t x) e^{-\frac{x^2}{2}} \dif x,\\ + \phi_X(t) &=& \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty e^{\i t x} e^{-\frac{x^2}{2}} \dif x\\ + &=& \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty (\cos(tx) + \i \sin(tx)) e^{-\frac{x^2}{2}} \dif x\\ + &=& \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty \cos(t x) e^{-\frac{x^2}{2}} \dif x,\\ \end{IEEEeqnarray*} since, as $x \mapsto \sin(tx)$ is odd and $x \mapsto e^{-\frac{x^2}{2}}$ is even, their product is odd, wich gives that the integral is $0$. @@ -134,15 +134,15 @@ First, we need to prove some properties of characteristic functions. \phi'_X(t) &=& \bE[\i X e^{\i t X}] \\ &=& \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty \i x \left( \cos(t x) + \i \sin(tx) \right) e^{-\frac{x^2}{2}} \dif x\\ &=& \frac{1}{\sqrt{2 \pi}} \left( \i \int_{-\infty}^\infty x \cos(tx) \right) e^{-\frac{x^2}{2}} \dif x\\ - &=& \frac{1}{\sqrt{2 \pi} } \left(\underbrace{\i \int_{-\infty}^\infty x \cos(tx) e^{-\frac{x^2}{2}} \dif x}_{= 0} + \int_{-\infty}^\infty - \sin(t x) e^{-\frac{x^2}{2}} \dif x\right)\\ - &=& \int_{-\infty}^\infty \underbrace{\sin(tx)}_{y(x)} \underbrace{ \frac{1}{\sqrt{2 \pi} }(-x) e^{\i\frac{x^2}{2}}}_{f'(x)} \dif x\\ + &=& \frac{1}{\sqrt{2 \pi}} \left(\underbrace{\i \int_{-\infty}^\infty x \cos(tx) e^{-\frac{x^2}{2}} \dif x}_{= 0} + \int_{-\infty}^\infty - \sin(t x) e^{-\frac{x^2}{2}} \dif x\right)\\ + &=& \int_{-\infty}^\infty \underbrace{\sin(tx)}_{y(x)} \underbrace{ \frac{1}{\sqrt{2 \pi}}(-x) e^{\i\frac{x^2}{2}}}_{f'(x)} \dif x\\ &=& \underbrace{[ \sin(tx) \frac{1}{\sqrt{2 \pi} e^{-\frac{x^2}{2}}}]_{x=-\infty}^\infty}_{=0} - - \int_{-\infty}^\infty t \cos(tx) \frac{1}{\sqrt{2 \pi} } e^{-\frac{x^2}{2}} \dif x\\ + - \int_{-\infty}^\infty t \cos(tx) \frac{1}{\sqrt{2 \pi}} e^{-\frac{x^2}{2}} \dif x\\ &=& -t \phi_X(t) \end{IEEEeqnarray*} Thus, for all $t \in \R$ \[ - (\log(\phi_X(t))' = \frac{\phi'_X(t)}{\phi_X(t)} = -t. + (\log(\phi_X(t)))' = \frac{\phi'_X(t)}{\phi_X(t)} = -t. \] Hence there exists $c \in \R$, such that \[ @@ -174,9 +174,9 @@ Now, we can finally prove the CLT: Then \begin{IEEEeqnarray*}{rCl} \phi_{V_n}(t) &=& \bE[e^{\i t Y_n}]\\ - &=& \bE[e^{\i t \left( \frac{Y_1 + \ldots + Y_n}{\sqrt{n} } \right) }] \\ - &=& \bE\left[e^{\i t \frac{Y_1}{\sqrt{n}}}\right] \cdot \ldots \cdot \bE\left[e^{\i t \frac{Y_n}{\sqrt{n} }}\right]\\ - &=& \left( \phi\left(\frac{t}{\sqrt{n} }\right) \right)^n. + &=& \bE[e^{\i t \left( \frac{Y_1 + \ldots + Y_n}{\sqrt{n}} \right)}] \\ + &=& \bE\left[e^{\i t \frac{Y_1}{\sqrt{n}}}\right] \cdot \ldots \cdot \bE\left[e^{\i t \frac{Y_n}{\sqrt{n}}}\right]\\ + &=& \left( \phi\left(\frac{t}{\sqrt{n}}\right) \right)^n. \end{IEEEeqnarray*} where $\phi(t) \coloneqq \phi_{Y_1}(t)$. @@ -191,13 +191,14 @@ Now, we can finally prove the CLT: Setting $s \coloneqq \frac{t}{\sqrt{n}}$ we obtain \[ - \phi\left(\frac{t}{ \sqrt{n} }\right) = 1 - \frac{t^2}{2n} + o\left( \frac{t^2}{n} \right) \text{ as $n \to \infty$} + \phi\left(\frac{t}{ \sqrt{n}}\right) = 1 - \frac{t^2}{2n} + o\left( \frac{t^2}{n} \right) \text{ as $n \to \infty$} \] \[ - \phi_{V_n}(t) = \left( \phi\left( \frac{t}{\sqrt{n} } \right) \right)^n = - 1 - \frac{t^2}{2 n } + o\left( \frac{t^2}{n} \right)^n \xrightarrow{n \to \infty} e^{-\frac{t^2}{2}}, + \phi_{V_n}(t) = \left( \phi\left( \frac{t}{\sqrt{n}} \right) \right)^n + = \left(1 - \frac{t^2}{2 n} + o\left( \frac{t^2}{n} \right)\right)^n + \xrightarrow{n \to \infty} e^{-\frac{t^2}{2}}, \] where we have used the following: @@ -209,7 +210,7 @@ Now, we can finally prove the CLT: We have shown that \[ - \phi_n(t) \xrightarrow{n \to \infty} e^{-\frac{t^2}{2}} = \phi_N(t). + \phi_n(t) \xrightarrow{n \to \infty} e^{-\frac{t^2}{2}} = \phi_{\cN(0,1)}(t). \] Using \autoref{levycontinuity}, we obtain \autoref{clt}. \end{refproof}