some small changes

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,\\
+        - \frac{\pi}{2} &\text{if }x < a,\\
-        0, &x = a,\\
+        0 &\text{if }x = a,\\
-        \frac{\pi}{2}, & \frac{\pi}{2}
+        \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\\
+            &\overset{\text{Fubini}}{=}&
-            &=& \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 \int_{-T}^T
-            &=& \lim_{T \to  \infty} \frac{1}{2 T }\int_{\R} \frac{2 \sin(T  (y-x))}{T (y-x)} \bP(\dif y)\\
+                 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}

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\\
+        \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(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,\\
+                  &=& \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)\\
+        &=& \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\\
+        &=& \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[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]\\
+    &=& \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.
+    &=& \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
     \[
-    \phi\left(\frac{t}{ \sqrt{n} }\right) = 1 - \frac{t^2}{2n} + o\left( \frac{t^2}{n} \right) \text{ as $n \to  \infty$}
+    \phi\left(\frac{t}{ \sqrt{n}}\right) = 1 - \frac{t^2}{2n} + o\left( \frac{t^2}{n} \right) \text{ as $n \to  \infty$}
     \]
 
 
     \[
-    \phi_{V_n}(t) = \left( \phi\left( \frac{t}{\sqrt{n} } \right) \right)^n =
+    \phi_{V_n}(t) = \left( \phi\left( \frac{t}{\sqrt{n}} \right) \right)^n
-    1 - \frac{t^2}{2 n } + o\left( \frac{t^2}{n} \right)^n \xrightarrow{n \to \infty} e^{-\frac{t^2}{2}},
+    = \left(1 - \frac{t^2}{2 n} + o\left( \frac{t^2}{n} \right)\right)^n
+      \xrightarrow{n \to \infty} e^{-\frac{t^2}{2}},
     \]
     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}