diff --git a/inputs/a_1_counterexamples.tex b/inputs/a_1_counterexamples.tex index 1660476..b451952 100644 --- a/inputs/a_1_counterexamples.tex +++ b/inputs/a_1_counterexamples.tex @@ -3,4 +3,3 @@ Exercise 4.3 10.2 - diff --git a/inputs/lecture_10.tex b/inputs/lecture_10.tex index d7113c7..adddc04 100644 --- a/inputs/lecture_10.tex +++ b/inputs/lecture_10.tex @@ -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. diff --git a/inputs/lecture_11.tex b/inputs/lecture_11.tex index 7fc4e23..4be3db3 100644 --- a/inputs/lecture_11.tex +++ b/inputs/lecture_11.tex @@ -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} diff --git a/inputs/lecture_13.tex b/inputs/lecture_13.tex index 85574c4..d31db12 100644 --- a/inputs/lecture_13.tex +++ b/inputs/lecture_13.tex @@ -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, diff --git a/inputs/lecture_14.tex b/inputs/lecture_14.tex index 0d8a1f0..78716af 100644 --- a/inputs/lecture_14.tex +++ b/inputs/lecture_14.tex @@ -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 diff --git a/inputs/lecture_15.tex b/inputs/lecture_15.tex index db70cfa..23fc935 100644 --- a/inputs/lecture_15.tex +++ b/inputs/lecture_15.tex @@ -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] diff --git a/inputs/lecture_16.tex b/inputs/lecture_16.tex index 6171945..3db4926 100644 --- a/inputs/lecture_16.tex +++ b/inputs/lecture_16.tex @@ -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} diff --git a/inputs/lecture_17.tex b/inputs/lecture_17.tex index 97f69fc..f2ee73e 100644 --- a/inputs/lecture_17.tex +++ b/inputs/lecture_17.tex @@ -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} diff --git a/inputs/lecture_18.tex b/inputs/lecture_18.tex index b505b05..16dcb16 100644 --- a/inputs/lecture_18.tex +++ b/inputs/lecture_18.tex @@ -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} diff --git a/inputs/lecture_19.tex b/inputs/lecture_19.tex index 69e1f0a..f1935a6 100644 --- a/inputs/lecture_19.tex +++ b/inputs/lecture_19.tex @@ -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}]\\ diff --git a/inputs/lecture_21.tex b/inputs/lecture_21.tex index 0197fcd..faf1325 100644 --- a/inputs/lecture_21.tex +++ b/inputs/lecture_21.tex @@ -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} diff --git a/inputs/lecture_5.tex b/inputs/lecture_5.tex index 6a3ddc3..a466324 100644 --- a/inputs/lecture_5.tex +++ b/inputs/lecture_5.tex @@ -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} diff --git a/inputs/lecture_6.tex b/inputs/lecture_6.tex index c42c22d..ed2e6f1 100644 --- a/inputs/lecture_6.tex +++ b/inputs/lecture_6.tex @@ -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} diff --git a/inputs/prerequisites.tex b/inputs/prerequisites.tex index c51bf7c..f16a38d 100644 --- a/inputs/prerequisites.tex +++ b/inputs/prerequisites.tex @@ -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}