josia-notes/s23-probability-theory
1
0
Fork 0
Code Issues Pull requests Projects Releases Packages Wiki Activity Actions

removed whitespace

Browse source
This commit is contained in:
Josia Pietsch 2023-07-06 00:36:26 +02:00
parent 023e865eb4
commit 2139f9d465
Signed by untrusted user who does not match committer: josia
GPG key ID: E70B571D66986A2D
14 changed files with 83 additions and 84 deletions
Show stats Download patch file Download diff file Expand all files Collapse all files

1
inputs/a_1_counterexamples.tex
View file

@ -3,4 +3,3 @@
Exercise 4.3
10.2

26
inputs/lecture_10.tex
View file

@ -21,7 +21,7 @@ where $\mu = \bP X^{-1}$.
Note that the term on the RHS is integrable, as
\[
\lim_{t \to 0} \frac{e^{-\i t b} - e^{-\i t a}}{- \i t} \pi(t) = a - b
\]
\]
and note that $\phi(0) = 1$ and $|\phi(t)| \le 1$.
% TODO think about this
@ -45,7 +45,7 @@ where $\mu = \bP X^{-1}$.
\label{fact:intsinxx}
\[
\int_0^\infty \frac{\sin x}{x} dx = \frac{\pi}{2}
\]
\]
where the LHS is an improper Riemann-integral.
Note that the LHS is not Lebesgue-integrable.
It follows that
@ -65,7 +65,7 @@ where $\mu = \bP X^{-1}$.
Then $\bP$ has a continuous probability density given by
\[
f(x) = \frac{1}{2 \pi} \int_{\R} e^{-\i t x} \phi_{\R}(t) dt.
\]
\]
\end{theorem}
\begin{example}
@ -74,12 +74,12 @@ where $\mu = \bP X^{-1}$.
Then
\[
\phi_{\R}(t) = \int e^{\i t x} d \delta_0(x) = e^{\i t 0 } = 1
\]
\]
\item Let $\bP = \frac{1}{2} \delta_1 + \frac{1}{2} \delta_{-1}$.
Then
\[
\phi_{\R}(t) = \frac{1}{2} e^{\i t} + \frac{1}{2} e^{- \i t} = \cos(t)
\]
\]
\end{itemize}
\end{example}
\begin{refproof}{thm:lec10_3}
@ -93,16 +93,16 @@ where $\mu = \bP X^{-1}$.
Then
\[
|e^{-\i t x} \phi(t)| \le |\phi(t)|
\]
\]
and $\phi \in L^1$, hence $f(x_n) \to f(x)$
by the dominated convergence theorem.
\end{subproof}
We'll show that for all $a < b$ we have
\[
\bP\left( (a,b] \right) = \int_a^b (x) dx.\label{thm10_3eq1}
\]
\]
Let $F$ be the distribution function of $\bP$.
It is enough to prove \autoref{thm10_3eq1}
for all continuity points $a $ and $ b$ of $F$.
@ -123,8 +123,8 @@ However, Fourier analysis is not only useful for continuous probability density
Then
\[
\forall x \in \R ~ \bP\left( \{x\} \right) = \lim_{T \to \infty} \frac{1}{2 T} \int_{-T}^T e^{-\i t x } \phi(t) dt.
\]
\]
\end{theorem}
\begin{refproof}{bochnersformula}
We have
@ -140,10 +140,10 @@ However, Fourier analysis is not only useful for continuous probability density
1, &y = x,\\
0, &y \neq x.
\end{cases}
\]
\]
Hence
\begin{IEEEeqnarray*}{rCl}
\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)
\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)
\end{IEEEeqnarray*}
% TODO by dominated convergence?
\end{refproof}
@ -177,7 +177,7 @@ However, Fourier analysis is not only useful for continuous probability density
\begin{theorem}[Bochner's theorem]\label{bochnersthm}
The converse to \autoref{thm:lec_10thm5} holds, i.e.~ any
$\phi: \R \to \C$ satisfying (a) and (b) of \autoref{thm:lec_10thm5}
must be the Fourier transform of a probability measure $\bP$
must be the Fourier transform of a probability measure $\bP$
on $(\R, \cB(\R))$.
\end{theorem}
Unfortunately, we won't prove \autoref{bochnersthm} in this lecture.

6
inputs/lecture_11.tex
View file

@ -21,7 +21,7 @@ For intuition, watch \url{https://3blue1brown.com/lessons/clt}.
\begin{example}
We throw a fair die $n = 100$ times and denote the sum of the faces
by $S_n \coloneqq X_1 + \ldots + X_n$, where $X_1,\ldots, X_n$
by $S_n \coloneqq X_1 + \ldots + X_n$, where $X_1,\ldots, X_n$
are i.i.d.~and uniformly distributed on $\{1,\ldots,6\}$.
Then $\bE[S_n] = 350$ and $\sqrt{\Var(S_n)} = \sigma \approx 17.07$.
\todo{Missing pictures}
@ -32,7 +32,7 @@ For intuition, watch \url{https://3blue1brown.com/lessons/clt}.
\end{question}
By definition, $\Var(X) = \bE[(X- \bE(X))^2]$, hence $\sqrt{\Var(X)}$
can be interpreted as a distance.
One could also define $\Var(X)$ to be $\bE[|X - \bE(X)|]$ but this is not
One could also define $\Var(X)$ to be $\bE[|X - \bE(X)|]$ but this is not
well behaved.
@ -95,7 +95,7 @@ If $S_n \sim \Bin(n,p)$ and $[a,b] \subseteq \R$, we have
With this in mind, a better approximation is
\[
\bP[S \le 25] = \bP[S \le 25.5] \approx \Phi\left( \frac{5.5}{\sqrt{10} } \right) \approx 0.9541.
\]
\]
\end{example}
\begin{example}

8
inputs/lecture_13.tex
View file

@ -23,7 +23,7 @@ if $X_1, X_2,\ldots$ are i.i.d.~with $ \mu = \bE[X_1]$,
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]
@ -34,7 +34,7 @@ if $X_1, X_2,\ldots$ are i.i.d.~with $ \mu = \bE[X_1]$,
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}
@ -54,7 +54,7 @@ details can be found in the notes.\notes
if
\[
\lim_{a \to \infty} \sup_{n \in \N} \bP[|X_n| > a] = 0.
\]
\]
\end{definition}
\begin{example}+[Exercise 8.1]
\todo{Copy}
@ -194,7 +194,7 @@ for all $f \in C_b$ and $x \to e^{\i t x}$ is continuous and bounded.
It suffices to show that
\[
\frac{A}{2} \left| \int_{-\frac{2}{A}}^{\frac{2}{A}} \phi_n(t) dt\right| - 1 \ge 1 - 2\epsilon
\]
\]
or
\[
1 - \frac{A}{4} \left|\int_{-\frac{2}{A}}^{\frac{2}{A}} \phi_n(t) dt \right| \le \epsilon,

6
inputs/lecture_14.tex
View file

@ -74,7 +74,7 @@ We now want to generalize this to arbitrary random variables.
then
\[
\bE[X | \cG] = (\omega \mapsto \bE[X])
\]
\]
is a constant random variable.
\end{remark}
@ -128,13 +128,13 @@ and then do the harder proof.
which is a contradiction, since
\[
\bE[(Z - Z') \One_{Z - Z' > \frac{1}{n}}] \ge \frac{1}{n} \bP[ Z - Z' > \frac{1}{n}] > 0.
\]
\]
\bigskip
Existence of $\bE(X | \cG)$ for $X \in L^2$:
Let $H = L^2(\Omega, \cF, \bP)$
Let $H = L^2(\Omega, \cF, \bP)$
and $K = L^2(\Omega, \cG, \bP)$.
$K$ is closed, since a pointwise limit of $\cG$-measurable

32
inputs/lecture_15.tex
View file

@ -25,7 +25,7 @@ We want to derive some properties of conditional expectation.
Then
\[
\int_A X \dif \bP \ge \frac{1}{n}\bP(A) + \int_A Y \dif \bP,
\]
\]
contradicting property (b) from \autoref{conditionalexpectation}.
\end{proof}
@ -61,7 +61,7 @@ We want to derive some properties of conditional expectation.
However it follows that
\[
\int_G \bP[X | \cG] \dif \bP \le -\frac{1}{n} \bP[G] < 0 \le \int_G X \dif \bP.
\]
\]
\end{proof}
\begin{theorem}[Conditional monotone convergence theorem]
\label{ceprop5}
@ -77,11 +77,11 @@ We want to derive some properties of conditional expectation.
we have
\[
\bE[X_n | \cG] \overset{\text{a.s.}}{\ge } 0
\]
\]
and
\[
\bE[X_n | \cG] \uparrow \text{a.s.}
\]
\]
(consider $X_{n+1} - X_n$ ).
Define $Z \coloneqq \limsup_{n \to \infty} Z_n$.
@ -91,7 +91,7 @@ We want to derive some properties of conditional expectation.
Take some $G \in \cG$.
We know by (b) % TODO REF
that $\bE[Z_n \One_G] = \bE[X_n \One_G]$.
The LHS increases to $\bE[Z \One_G]$ by the monotone
The LHS increases to $\bE[Z \One_G]$ by the monotone
convergence theorem.
Again by MCT, $\bE[X_n \One_G]$ increases to
$\bE[X \One_G]$.
@ -105,7 +105,7 @@ We want to derive some properties of conditional expectation.
Then
\[
\bE[ \liminf_{n \to \infty} X_n | \cG] \le \liminf_{n \to \infty} \bE[X_n | \cG].
\]
\]
\end{theorem}
\begin{proof}
\notes
@ -117,7 +117,7 @@ We want to derive some properties of conditional expectation.
Suppose $|X_n(\omega)| < X(\omega)$ a.e.~
and $\int |X| \dif \bP < \infty$.
Then $X_n(\omega) \to X\left( \omega \right) \implies \bE[ X_n | \cG] \to \bE[X | \cG]$.
\end{theorem}
\begin{proof}
\notes
@ -144,7 +144,7 @@ For conditional expectation, we have
such that
\[
c(x) = \sup_n(a_n x + b_n).
\]
\]
\end{fact}
\begin{refproof}{cjensen}
By \autoref{convapprox}, $c(x) \ge a_n X + b_n$
@ -153,14 +153,14 @@ For conditional expectation, we have
\[
\bE[c(X) | \cG] \ge a_n \bE[X | \cG] + \bE[b_n | \cG]
= a_n \bE[X | \cG] + b_n \text{ a.s.}
\]
\]
for all $n$.
Using that a countable union of sets o f measure zero has measure zero,
we conclude that a.s~this happens simultaneously for all $n$.
Hence
\[
\bE[c(X) | \cG] \ge \sup_n (a_n \bE[X | \cG] + b_n) \overset{\text{\autoref{convapprox}}}{=} c(\bE(X | \cG)).
\]
\]
\end{refproof}
Recall
@ -170,7 +170,7 @@ Recall
Then
\[
\bE(X Y) \le \underbrace{\bE(|X|^p)^{\frac{1}{p}}}_{\text{\reflectbox{$\coloneqq$}} \|X\|_{L^p}} \bE(|Y|^q)^{\frac{1}{q}}.
\]
\]
\end{fact}
\begin{theorem}[Conditional Hölder's inequality]
@ -181,7 +181,7 @@ Recall
Then
\[
\bE(X Y | \cG) \le \bE(|X|^p | \cG)^{\frac{1}{p}} \bE(|Y|^q | \cG)^{\frac{1}{q}}.
\]
\]
\end{theorem}
\begin{proof}
\todo{Exercise}
@ -194,7 +194,7 @@ Recall
Then
\[
\bE\left[\bE[X | \cG] \mid \cH\right] \overset{\text{a.s.}}{=} \bE[X | \cH].
\]
\]
\end{theorem}
\begin{proof}
By definition, $\bE[\bE[X | \cG] | \cH]$ is $\cH$-measurable.
@ -214,7 +214,7 @@ Recall
If $Y$ is $\cG$-measurable and bounded, then
\[
\bE[YX| \cG] \overset{\text{a.s.}}{=} Y \bE[X | \cG].
\]
\]
\end{theorem}
\begin{proof}
Assume w.l.o.g.~$X \ge 0$.
@ -239,13 +239,13 @@ Assume $Y = \One_B$, then $Y$ simple, then take the limit (using that $Y$ is bou
then
\[
\bE[X | \sigma(\cG, \cH)] \overset{\text{a.s.}}{=} \bE[X | \cG].
\]
\]
In particular, if $X$ is independent of $\cG$,
then
\[
\bE[X | \cG] \overset{\text{a.s.}}{=} \bE[X].
\]
\]
\end{theorem}
\begin{example}[Martingale property of the simple random walk]

18
inputs/lecture_16.tex
View file

@ -5,7 +5,7 @@
\begin{refproof}{ceroleofindependence}
Let $\cH$ be independent of $\sigma(\sigma(X), \cG)$.
Then for all $H \in \cH$, we have that $\One_H$
and any random variable measurable with respect to either $\sigma(X)$
and any random variable measurable with respect to either $\sigma(X)$
or $\cG$ must be independent.
It suffices to consider the case of $X \ge 0$.
@ -55,7 +55,7 @@ The Radon Nikodym theorem is the converse of that:
Suppose
\[
\forall A \in \cF . ~ \mu(A) = 0 \implies \nu(A) = 0.
\]
\]
Then
\begin{enumerate}[(1)]
\item there exists $Z: \Omega \to [0, \infty)$ measurable,
@ -70,7 +70,7 @@ The Radon Nikodym theorem is the converse of that:
\begin{definition}
Whenever the property $\forall A \in \cF, \mu(A) = 0 \implies \nu(A) = 0$
holds for two measures $\mu$ and $\nu$,
we say that $\nu$ is \vocab{absolutely continuous}
we say that $\nu$ is \vocab{absolutely continuous}
w.r.t.~$\mu$.
This is written as $\nu \ll \mu$.
\end{definition}
@ -81,7 +81,7 @@ of conditional expectation:
Let $(\Omega, \cF, \bP)$ as always, $X \in L^1(\bP)$ and $\cG \subseteq \cF$.
It suffices to consider the case of $X \ge 0$.
For all $G \in \cG$, define $\nu(G) \coloneqq \int_G X \dif \bP$.
Obviously, $\nu \ll \bP$ on $\cG$.
Obviously, $\nu \ll \bP$ on $\cG$.
Then apply \autoref{radonnikodym}.
\end{proof}
@ -89,7 +89,7 @@ of conditional expectation:
\begin{refproof}{radonnikodym}
We will only sketch the proof.
A full proof can be found in the official notes.
\paragraph{Step 1: Uniqueness} \notes
\paragraph{Step 2: Reduction to the finite measure case}
\notes
@ -114,7 +114,7 @@ of conditional expectation:
\item For all $f \in \cC$, we have
\[
\int_\Omega f \dif \mu \le \nu(\Omega) < \infty.
\]
\]
\end{enumerate}
Define $\alpha \coloneqq \sup \{ \int f \dif \mu : f \in \cC\} \le \nu(\Omega) < \infty$.
@ -142,7 +142,7 @@ of conditional expectation:
Then $\lambda(A) = 0$ since $\mu$ is finite.
Assume the claim does not hold.
Then there must be some $k \in \N$, $A \in \cF$
Then there must be some $k \in \N$, $A \in \cF$
such that $\lambda(A) - \frac{1}{k} \mu(A) > 0$.
Fix this $A$ and $k$.
Then $A$ satisfies condition (i) of being good,
@ -187,8 +187,8 @@ Typically $\cF_n = \sigma(X_1, \ldots, X_n)$ for a sequence of random variables.
if it is adapted to $\cF_n$ but
\[
\bE[X_{n+1} | \cF_n] \overset{\text{a.s.}}{\ge} X_n.
\]
It is called a \vocab{super-martingale}
\]
It is called a \vocab{super-martingale}
if it is adapted but $\bE[X_{n+1} | \cF_n] \overset{\text{a.s.}}{\le} X_n$.
\end{definition}
\begin{corollary}

16
inputs/lecture_17.tex
View file

@ -11,7 +11,7 @@
\begin{goal}
What about a ``gambling strategy''?
Consider a stochastic process $(X_n)_{n \in \N}$.
Note that the increments $X_{n+1} - X_n$ can be thought of as the win
@ -27,7 +27,7 @@
while
\[
Y_n \coloneqq \sum_{j=1}^n C_j(X_j - X_{j-1})
\]
\]
defines the cumulative win process.
\end{goal}
\begin{lemma}
@ -70,11 +70,11 @@ It follows that the monotonic limit $U_\infty([a,b]) \coloneqq \lim_{N \to \inf
\label{lec17l1}
\[
\{\omega | \liminf_{N \to \infty} Z_N(\omega) < a < b <
\limsup_{N \to \infty} Z_N(\omega)\} \subseteq
\limsup_{N \to \infty} Z_N(\omega)\} \subseteq
\{\omega: U^{Z}_\infty([a,b])(\omega) = \infty\}
\]
\]
for every sequence of measurable functions $(Z_n)_{n \ge 1}$.
\end{lemma}
\begin{lemma} % 2
\label{lec17l2}
@ -111,7 +111,7 @@ It follows that the monotonic limit $U_\infty([a,b]) \coloneqq \lim_{N \to \inf
by the monotone convergence theorem
\[
\bE(U_n([a,b])] \uparrow \bE[U_\infty([a,b])].
\]
\]
\end{proof}
Assume now, that our process $(X_n)_{n \ge 1}$ is a supermartingale
@ -119,7 +119,7 @@ bounded in $L^1(\bP)$.
Let
\[
\Lambda \coloneqq \{\omega | X_n(\omega) \text{ does not converge to anything in $[-\infty,\infty]$}\}.
\]
\]
We have
\begin{IEEEeqnarray*}{rCl}
\Lambda &=& \{\omega | \liminf_N X_N(\omega) < \limsup_N X_N(\omega)\}\\
@ -160,7 +160,7 @@ The second part follows from
\end{claim}
\begin{subproof}
We need to show $\sup_n \bE(|X_n|) < \infty$.
Since the supermartingale is non-negative, we have $\bE[|X_n|] = \bE[X_n]$
Since the supermartingale is non-negative, we have $\bE[|X_n|] = \bE[X_n]$
and since it is a supermartingale $\bE[X_n] \le \bE[X_0]$.
\end{subproof}

28
inputs/lecture_18.tex
View file

@ -3,7 +3,7 @@
Recall our key lemma \ref{lec17l3} for supermartingales from last time:
\[
(b-a) \bE[U_N([a,b])] \le \bE[(X_n - a)^-].
\]
\]
What happens for submartingales?
If $(X_n)_{n \in \N}$ is a submartingale, then $(-X_n)_{n \in \N}$ is a supermartingale.
@ -40,12 +40,12 @@ Hence the same holds for submartingales, i.e.
By the SLLN, we have
\[
\frac{1}{n} \sum_{k=1}^{n} Z_k \xrightarrow{a.s.} \bE[Z_1] = zp - 1.
\]
\]
Hence
\[
\left(\frac{X_n}{x}\right)^{\frac{1}{n}} = u^{\frac{1}{n} \sum_{k=1}^n Z_k}
\xrightarrow{a.s.} u^{zp -1}.
\]
\]
Since $(X_n)_{n \ge 0}$ is a martingale, we must have $\bE[u^{Z_1}] = 1$.
Hence $2p - 1 < 0$, because $u > 1$.
@ -67,7 +67,7 @@ consider $L^2$.
\begin{fact}[Martingale increments are orthogonal in $L^2$ ]
\label{martingaleincrementsorthogonal}
Let $(X_n)_n$ be a martingale
and let $Y_n \coloneqq X_n - X_{n-1}$
and let $Y_n \coloneqq X_n - X_{n-1}$
denote the \vocab{martingale increments}.
Then for all $m \neq n$ we have that
\[
@ -86,7 +86,7 @@ consider $L^2$.
Then
\[
2 \bE[X^2] + 2 \bE[Y^2] = \bE[(X+Y)^2] + \bE[(X-Y)^2].
\]
\]
\end{fact}
\begin{theorem}\label{martingaleconvergencel2}
@ -96,23 +96,23 @@ consider $L^2$.
Then there is a random variable $X_\infty$ such that
\[
X_n \xrightarrow{L^2} X_\infty.
\]
\]
\end{theorem}
\begin{proof}
Let $Y_n \coloneqq X_n - X_{n-1}$ and write
\[
X_n = \sum_{j=1}^{n} Y_j.
\]
\]
We have
\[
\bE[X_n^2] = \bE[X_0^2] + \sum_{j=1}^{n} \bE[Y_j^2]
\]
\]
by \autoref{martingaleincrementsorthogonal}
% (this is known as the \vocab{parallelogram identity}). % TODO how exactly is this used here?
In particular,
\[
\sup_n \bE[X_n^2] < \infty \iff \sum_{j=1}^{\infty} \bE[Y_j^2] < \infty.
\]
\]
Since $(X_n)_n$ is bounded in $L^2$,
there exists $X_\infty$ such that $X_n \xrightarrow{\text{a.s.}} X_\infty$
@ -128,7 +128,7 @@ consider $L^2$.
limit
\[
\sum_{j \ge n + 1} \xrightarrow{n\to \infty} 0
\]
\]
we get $\bE[(X_\infty - X_n)^2] \xrightarrow{n\to \infty} 0$.
\end{proof}
@ -192,8 +192,8 @@ In order to prove \autoref{dooblp}, we first need
where
\[
E_j = \{|X_1| \le \ell, |X_2| \le \ell, \ldots, |X_{j-1}| \le \ell, |X_j| \ge \ell\}.
\]
Then
\]
Then
\begin{equation}
\bP[E_j] \overset{\text{Markov}}{\le } \frac{1}{\ell} \int_{E_j} |X_j| \dif \bP
\label{lec18eq2star}
@ -211,7 +211,7 @@ In order to prove \autoref{dooblp}, we first need
\begin{equation}
\bE[\One_{E_j} (|X_n| - |X_j|)] \ge 0. \label{lec18eq3star}
\end{equation}
Now
\begin{IEEEeqnarray*}{rCl}
\bP(E) &=& \sum_{j=1}^n \bP(E_j)\\
@ -221,6 +221,6 @@ In order to prove \autoref{dooblp}, we first need
This proves the first part.
For the second part, we apply the first part and
For the second part, we apply the first part and
\autoref{dooplplemma} (choose $Y \coloneqq X_n^\ast$).
\end{refproof}

6
inputs/lecture_19.tex
View file

@ -179,7 +179,7 @@ However, some subsets can be easily described, e.g.
(1) $\implies$ (2)
$X_n \xrightarrow{L^1} X \implies X_n \xrightarrow{\bP} X$
$X_n \xrightarrow{L^1} X \implies X_n \xrightarrow{\bP} X$
by Markov's inequality.
Fix $\epsilon > 0$.
@ -195,7 +195,7 @@ However, some subsets can be easily described, e.g.
Then by \autoref{lec19f4} part (a) it follows that
\[
\int_{|X_n| > k} |X_n| \dif \bP \le \underbrace{\int |X - X_n| \dif \bP}_{< \epsilon} + \int_{|X_n| > k} |X| \dif \bP \le 2 \epsilon.
\]
\]
\end{proof}
\subsection{Martingale Convergence Theorems in \texorpdfstring{$L^p, p \ge 1$}{$Lp, p >= 1$}}
@ -211,7 +211,7 @@ Let $(\Omega, \cF, \bP)$ as always and let $(\cF_n)_n$ always be a filtration.
It is clear that $(\bE[X | \cF_n])_n$ is adapted to $(\cF_n)_n$.
Let $X_n \coloneqq \bE[X | \cF_n]$.
Consider
Consider
\begin{IEEEeqnarray*}{rCl}
\bE[X_n - X_{n-1} | \cF_{n-1}]
&=& \bE[\bE[X | \cF_n] - \bE[X | \cF_{n-1}] | \cF_{n-1}]\\

2
inputs/lecture_21.tex
View file

@ -108,7 +108,7 @@ is the unique solution to this problem.
\begin{subproof}
\todo{TODO}
% We have $\sigma(\One_{X_{n+1} \in B}) \subseteq \sigma(X_{n}, \xi_{n+1})$.
% $\sigma(X_1,\ldots,X_{n-1})$
% $\sigma(X_1,\ldots,X_{n-1})$
% is independent of $\sigma( \sigma(\One_{X_{n+1} \in B}), X_n)$.
% Hence the claim follows from \autoref{ceroleofindependence}.
\end{subproof}

4
inputs/lecture_5.tex
View file

@ -49,7 +49,7 @@ For the proof of (b) we need the following general result:
\end{theorem}
\begin{proof}
\end{proof}
\begin{question}
@ -60,7 +60,7 @@ This does not hold. Consider for example $X_n = \frac{1}{n^2} \delta_n + \frac{1
\begin{refproof}{lln}
\begin{enumerate}
\item[(b)]
\item[(b)]
\end{enumerate}
\end{refproof}

10
inputs/lecture_6.tex
View file

@ -21,7 +21,7 @@
Then $a_1 + \ldots + a_n = (S_1 - S_0) + 2(S_2 - S_1) + 3(S_3 - S_2) +
\ldots + n (S_n - S_{n-1})$.
Thus $a_1 + \ldots + a_n = n S_n - (S1 $ % TODO
\end{subproof}
The SLLN follows from the claim.
\end{refproof}
@ -50,12 +50,12 @@ We need the fol]
(By the independence of $X_1,\ldots, X_n$ and therefore that of $E$ and $D$ and $\bE(X_{i+1}) = \ldots = \bE(X_n) = 0$ we have $\int D E d\bP = 0$.)
% TODO
\end{proof}
\begin{refproof}{thm2}
% TODO
\end{refproof}
@ -67,7 +67,7 @@ We need the fol]
Let $S_n \coloneqq \sum_{i=1}^n X_i$.
For all $t > 0$ let \[
N_t \coloneqq \sup \{n : S_n \le t\}.
\]
\]
Then $\frac{N_t}{t} \xrightarrow{a.s.} \frac{1}{m}$ as $t \to \infty$.
\end{theorem}
@ -92,7 +92,7 @@ We need the fol]
By definition, we have $S_{N_t} \le t \le S_{N_t + t}$.
Then $\frac{S_{N_t}}{N_t} \le \frac{t}{N_t} \le S_{N_t + 1}{N_t} \le \frac{S_{N_t + 1}}{N_t + 1} \cdot \frac{N_t + 1}{N_t}$.
Hence $\frac{t}{N_t} \to m$.
\end{proof}

4
inputs/prerequisites.tex
View file

@ -98,7 +98,7 @@ from the lecture on stochastic.
\end{claim}
\begin{subproof}
We can use the same counterexample as in c).
$\bP[\lim_{n \to \infty} X_n = 0] \ge \bP[X_n = 0] = 1 - \frac{1}{n} \to 0$.
We have already seen, that $X_n$ does not converge in $L_1$.
\end{subproof}
@ -166,7 +166,7 @@ We used Chebyshev's inequality. Linearity of $\bE$, $\Var(cX) = c^2\Var(X)$ and
Then
\[
\lim_{n \to \infty} \int_{\R} f(x) \cos(n x) \lambda(\dif x) = 0.
\]
\]
\end{theorem}