updated stochastic calculus and new files

.github/workflows/buildpdf.yml

@@ -180,8 +180,8 @@ jobs:
       - name: Compile - INEPDE
         uses: xu-cheng/latex-action@v2
         with:
-          root_file: Introduction_to_nonlinear_elliptic_PDEs.tex
-          working_directory: Mathematics/5th/Introduction_to_nonlinear_elliptic_PDEs/
+          root_file: Introduction_to_non_linear_elliptic_PDEs.tex
+          working_directory: Mathematics/5th/Introduction_to_non_linear_elliptic_PDEs/
       - name: Compile - LTLD
         uses: xu-cheng/latex-action@v2
         with:
@@ -273,7 +273,7 @@ jobs:
             Mathematics/5th/Advanced_probability/Advanced_probability.pdf
             Mathematics/5th/Advanced_topics_in_functional_analysis_and_PDEs/Advanced_topics_in_functional_analysis_and_PDEs.pdf
             Mathematics/5th/Introduction_to_evolution_PDEs/Introduction_to_evolution_PDEs.pdf
-            Mathematics/5th/Introduction_to_nonlinear_elliptic_PDEs/Introduction_to_nonlinear_elliptic_PDEs.pdf
+            Mathematics/5th/Introduction_to_non_linear_elliptic_PDEs/Introduction_to_non_linear_elliptic_PDEs.pdf
             Mathematics/5th/Limit_theorems_and_large_deviations/Limit_theorems_and_large_deviations.pdf
             Mathematics/5th/Stochastic_calculus/Stochastic_calculus.pdf
             main_physics.pdf

Mathematics/4th/Stochastic_processes/Stochastic_processes.tex

@@ -157,7 +157,7 @@
   \end{definition}
   \subsubsection{Galton-Watson process}
   \begin{model}\label{SP:galtonwatsonModel}
-    Let $(X_n)$, $n\in\NN\cup\{0\}$ be a sequence of discrete random vairables representing the number of new individuals of a certain population at the $n$-th generation. Suppose they are defined as $$X_{n+1}=\sum_{k=1}^{X_n}Z_{n+1}^{(k)}$$ and $X_0=1$. Here $Z_{n+1}^{(k)}$ has support $\NN\cup\{0\}$ $\forall n,k$ and represent the number of descendants (to the next generation) of the $k$-th individual of the $n$-th generation. Suppose that $Z_{n+1}^{(k)}\sim Z$ are \iid and independent of $(X_n)$. We would like to study the probability $\rho$ of extinction of this population: $$\rho=\Prob(\{X_n=0:\text{for some $n\in\NN$}\})=\Prob\left(\bigcup_{n=1}^\infty\{X_n=0\}\right)$$
+    Let $(X_n)$, $n\in\NN\cup\{0\}$ be a sequence of discrete random variables representing the number of new individuals of a certain population at the $n$-th generation. Suppose they are defined as $$X_{n+1}=\sum_{k=1}^{X_n}Z_{n+1}^{(k)}$$ and $X_0=1$. Here $Z_{n+1}^{(k)}$ has support $\NN\cup\{0\}$ $\forall n,k$ and represent the number of descendants (to the next generation) of the $k$-th individual of the $n$-th generation. Suppose that $Z_{n+1}^{(k)}\sim Z$ are \iid and independent of $(X_n)$. We would like to study the probability $\rho$ of extinction of this population: $$\rho=\Prob(\{X_n=0:\text{for some $n\in\NN$}\})=\Prob\left(\bigcup_{n=1}^\infty\{X_n=0\}\right)$$
   \end{model}
   \begin{lemma}\label{SP:lemmaGaltonWatson}
     Let $(Z_n)$ be a sequence of \iid random variables distributed as $Z$ with support $\NN\cup\{0\}$, and $N$ be a random variable also with support $\NN\cup\{0\}$ and independent to $(Z_n)$. Let $X=\sum_{k=1}^NZ_k$. Then, $\forall s\in[-1,1]$ we have: $$g_X(s)=g_N(g_Z(s))$$
@@ -1271,14 +1271,14 @@
   \end{theorem}
   \subsection{Brownian motion}
   \subsubsection{Gaussian processes}
-  \begin{proposition}\label{SP:gaussian_vector}
-    Let $\vf{x}\in \RR^n$ be a random vector. Then, $\vf{x}$ is a \emph{gaussian vector}, that is it distributes as an $n$-dimensional normal, if and only if there exists $k\in\NN$, $\vf{A}\in\mathcal{M}_{n\times k}(\RR)$, $\vf{z}\in\RR^k$ with \iid components distributed as $N(0,1)$, and $\vf\mu\in\RR^n$ such that: $$\vf{x}=\vf{A}\vf{z}+\vf\mu$$
+  \begin{proposition}\label{SP:Gaussian_vector}
+    Let $\vf{x}\in \RR^n$ be a random vector. Then, $\vf{x}$ is a \emph{Gaussian vector}, that is it distributes as an $n$-dimensional normal, if and only if there exists $k\in\NN$, $\vf{A}\in\mathcal{M}_{n\times k}(\RR)$, $\vf{z}\in\RR^k$ with \iid components distributed as $N(0,1)$, and $\vf\mu\in\RR^n$ such that: $$\vf{x}=\vf{A}\vf{z}+\vf\mu$$
   \end{proposition}
   \begin{definition}
-    A stochastic process ${(X_t)}_{t\geq 0}$ is called a \emph{gaussian process} if for all $t_1,\ldots,t_n\geq 0$ the random vector $(X_{t_1},\ldots,X_{t_n})$ is gaussian.
+    A stochastic process ${(X_t)}_{t\geq 0}$ is called a \emph{Gaussian process} if for all $t_1,\ldots,t_n\geq 0$ the random vector $(X_{t_1},\ldots,X_{t_n})$ is Gaussian.
   \end{definition}
   \begin{definition}
-    Let ${(X_t)}_{t\geq 0}$ be a gaussian process.
+    Let ${(X_t)}_{t\geq 0}$ be a Gaussian process.
     Then, the \emph{mean function} is defined as:
     $$
       \function{\mu}{[0,\infty)}{\RR}{t}{\Exp(X_t)=:\mu_t}
@@ -1300,7 +1300,7 @@
     The Brownian motion is said to be \emph{standard} if $\sigma=1$.
   \end{definition}
   \begin{proposition}
-    Let $B:={(B_t)}_{t\geq 0}$ be a standard Brownian motion. Then, $B$ is a gaussian process with mean function $\mu_t=0$ and covariance function $C(s,t)=\min(s,t)$.
+    Let $B:={(B_t)}_{t\geq 0}$ be a standard Brownian motion. Then, $B$ is a Gaussian process with mean function $\mu_t=0$ and covariance function $C(s,t)=\min(s,t)$.
   \end{proposition}
   \begin{proof}
     Let $0< t_1<\cdots<t_n$. We can write the vector $\vf{b}:=\transpose{(B_{t_1},\ldots,B_{t_n})}$ as:
@@ -1318,13 +1318,13 @@
         B_{t_n} - B_{t_{n-1}}
       \end{pmatrix}
     $$
-    And so $\vf{b}$ is gaussian because is a linear combination of gaussian. Now in the general, let $s_1, \ldots, s_n\geq 0$. We can write any vector $(B_{s_1}, \ldots, B_{s_n})$ as a linear transformation of the vector $(B_{t_1}, \ldots, B_{t_n})$ with $0< t_1<\cdots<t_n$. On the other hand, it is clear that $\mu_t=\Exp(B_t)=\Exp(B_t-B_0)=0$ and if $s\leq t$: $$\Exp(B_sB_t)=\Exp(B_s(B_t-B_s))+\Exp(B_s^2)=\mu_s\mu_{t-s}+s=s$$
+    And so $\vf{b}$ is Gaussian because is a linear combination of Gaussian. Now in the general, let $s_1, \ldots, s_n\geq 0$. We can write any vector $(B_{s_1}, \ldots, B_{s_n})$ as a linear transformation of the vector $(B_{t_1}, \ldots, B_{t_n})$ with $0< t_1<\cdots<t_n$. On the other hand, it is clear that $\mu_t=\Exp(B_t)=\Exp(B_t-B_0)=0$ and if $s\leq t$: $$\Exp(B_sB_t)=\Exp(B_s(B_t-B_s))+\Exp(B_s^2)=\mu_s\mu_{t-s}+s=s$$
   \end{proof}
   \begin{proposition}
-    Let $B:={(B_t)}_{t\geq 0}$ be a gaussian process with $B_0=0$, mean function $\mu_t=0$ and covariance function $C(s,t)=\min(s,t)$. Then, $B$ is a standard Brownian motion.
+    Let $B:={(B_t)}_{t\geq 0}$ be a Gaussian process with $B_0=0$, mean function $\mu_t=0$ and covariance function $C(s,t)=\min(s,t)$. Then, $B$ is a standard Brownian motion.
   \end{proposition}
   \begin{proof}
-    Since gaussian uncorrelated variables are independent, it suffices to show that the covariance matrix of $(B_{t_1}, B_{t_2}-B_{t_1}, \ldots, B_{t_n}-B_{t_{n-1}})$ is:
+    Since Gaussian uncorrelated variables are independent, it suffices to show that the covariance matrix of $(B_{t_1}, B_{t_2}-B_{t_1}, \ldots, B_{t_n}-B_{t_{n-1}})$ is:
     $$
       \begin{pmatrix}
         t_1    & 0       & \cdots & 0            \\

Mathematics/5th/Advanced_probability/Advanced_probability.tex

@@ -75,9 +75,9 @@
     $$
   \end{proposition}
   \begin{definition}[Product measure]
-    Let $(E,\mathcal{E},\mu)$ and $(F,\mathcal{F},\nu)$ be two measurable spaces. We define the \emph{product measure} $\mu\times\nu$ on $(E\times F,\mathcal{E}\otimes\mathcal{F})$ as:
+    Let $(E,\mathcal{E},\mu)$ and $(F,\mathcal{F},\nu)$ be two measurable spaces. We define the \emph{product measure} $\mu\otimes\nu$ on $(E\times F,\mathcal{E}\otimes\mathcal{F})$ as:
     $$
-      \forall A\in\mathcal{E}, B\in\mathcal{F}, \quad \mu\times\nu(A\times B):=\mu(A)\nu(B)
+      \forall A\in\mathcal{E}, B\in\mathcal{F}, \quad \mu\otimes\nu(A\times B):=\mu(A)\nu(B)
     $$
   \end{definition}
   \begin{definition}
@@ -86,14 +86,14 @@
   \begin{theorem}[Fubini]
     Let $(E,\mathcal{E},\mu)$ and $(F,\mathcal{F},\nu)$ be two $\sigma$-finite measurable spaces and $f:E\times F\to \RR$ be a measurable function. Then, the following are equivalent:
     \begin{enumerate}
-      \item $f$ is integrable with respect to $\mu\times\nu$.
+      \item $f$ is integrable with respect to $\mu\otimes\nu$.
       \item $\displaystyle\int_E{\left(\int_F{\abs{f(x,y)}\dd{\nu(y)}}\right)\dd{\mu(x)}}<\infty$.
       \item $\displaystyle\int_F{\left(\int_E{\abs{f(x,y)}\dd{\mu(x)}}\right)\dd{\nu(y)}}<\infty$.
     \end{enumerate}
     And if any of the above holds, then:
     \begin{align*}
-      \int_{E\times F}{f\dd{(\mu\times\nu)}} & =\int_E{\left(\int_F{f(x,y)\dd{\nu(y)}}\right)\dd{\mu(x)}} \\
-                                             & =\int_F{\left(\int_E{f(x,y)\dd{\mu(x)}}\right)\dd{\nu(y)}}
+      \int_{E\times F}{f\dd{(\mu\otimes\nu)}} & =\int_E{\left(\int_F{f(x,y)\dd{\nu(y)}}\right)\dd{\mu(x)}} \\
+                                              & =\int_F{\left(\int_E{f(x,y)\dd{\mu(x)}}\right)\dd{\nu(y)}}
     \end{align*}
   \end{theorem}
   \begin{definition}

Mathematics/5th/Stochastic_calculus/Stochastic_calculus.tex

@@ -3,6 +3,87 @@
 \begin{document}
 \changecolor{SC}
 \begin{multicols}{2}[\section{Stochastic calculus}]
-
+  Along the document we assume that we work in a probability space $(\Omega,\mathcal{F},\Prob)$ and that all the random variables are defined on this space.
+  \subsection{Preliminaries}
+  \subsubsection{Stochastic processes}
+  \begin{proposition}
+    A stochastic process $X={(X_t)}_{t\in \TT}$ is Gaussian if and only if $\forall n\in\NN$, $\forall t_1,\ldots,t_n\in\TT$, $\forall \lambda_1,\ldots,\lambda_n\in\RR$,
+    $$
+      Z:=\lambda_1 X_{t_1}+\cdots+\lambda_n X_{t_n}
+    $$
+    is a Gaussian random variable. In particular, we have:
+    $$
+      \Exp(\exp{\ii Z})=\exp{\ii \Exp(Z)-\frac{1}{2}\Var(Z)}
+    $$
+  \end{proposition}
+  \begin{remark}
+    A stochastic process $X={(X_t)}_{t\in \TT}$ can also be viewed as a single random variable taking values in $\RR^{\TT}$, equipped with the product $\sigma$-algebra $\displaystyle \bigotimes_{t\in\TT}\mathcal{B}(\RR)$.
+  \end{remark}
+  \begin{proposition}
+    Let $m:\TT\to\RR$ be a measurable function and $\gamma:\TT^2\to\RR$ be a symmetric positive-definite function. Then, there exists a Gaussian process ${(X_t)}_{t\in \TT}$ such that $\Exp(X_t)=m(t)$ and $\cov(X_s,X_t)=\gamma(s,t)$.
+  \end{proposition}
+  \begin{definition}
+    Let ${(X_t)}_{t\in \TT}$, ${(Y_s)}_{s\in \SS}$ be two stochastic processes. We say that they are \emph{jointly Gaussian} if the concatenated process $({(X_t)}_{t\in \TT},{(Y_s)}_{s\in \SS})$ is Gaussian.
+  \end{definition}
+  \begin{lemma}\label{lemma:indep_joint_gauss}
+    Two jointly Gaussian stochastic processes ${(X_t)}_{t\in \TT}$, ${(Y_s)}_{s\in \SS}$ are independent if and only if $\forall t\in\TT$, $\forall s\in\SS$, $\cov(X_t,Y_s)=0$.
+  \end{lemma}
+  \begin{proposition}
+    Two stochastic processes ${(X_t)}_{t\in \TT}$, ${(Y_s)}_{s\in \SS}$ are independent if and only if $\forall n\in\NN$, $\forall t_1,\ldots,t_n\in\TT$, $\forall s_1,\ldots,s_n\in\SS$ and $\forall f,g:\RR^n\to\RR$ bounded and measurable functions, we have:
+    \begin{multline*}
+      \Exp(f(X_{t_1},\ldots,X_{t_n})g(Y_{s_1},\ldots,Y_{s_n}))=\\=
+      \Exp(f(X_{t_1},\ldots,X_{t_n}))\Exp(g(Y_{s_1},\ldots,Y_{s_n}))
+    \end{multline*}
+  \end{proposition}
+  \subsubsection{Brownian motion}
+  \begin{theorem}[Strong law of large numbers for Brownian motion]
+    Let ${(B_t)}_{t\geq 0}$ be a Brownian motion. Then:
+    $$
+      \frac{B_t}{t}\underset{t\to\infty}{\overset{\text{a.s.}}{\longrightarrow}}0
+    $$
+  \end{theorem}
+  \begin{proof}
+    We already now that the process $s\to sB_{1/s} \indi{s>0}$ is a Brownian motion. In particular, we must have continuity at $0=B_0$.
+  \end{proof}
+  \begin{theorem}[Markov property for Brownian motion]
+    Let $B={(B_t)}_{t\geq 0}$ be a Brownian motion and $a\geq 0$ fixed. Then, the Brownian motion ${(B_{t+a}-B_a)}_{t\geq 0}$ is independent of ${(B_s)}_{s\in [0,a]}$.
+  \end{theorem}
+  \begin{proof}
+    The processes ${(B_s)}_{s\in[0,a]}$ and ${(B_{t+a}-B_a)}_{t\geq 0}$ are jointly Gaussian, because their coordinates are linear combinations of coordinates of the same Gaussian process $B$. Thus, by \mcref{lemma:indep_joint_gauss} it reduces to compute the following correlation:
+    $$
+      \cov(B_s,B_{t+a}-B_a)=s\wedge(t+a)-s\wedge a=0
+    $$
+  \end{proof}
+  \begin{remark}
+    Recall that $s\wedge t:=\min(s,t)$ and $s\vee t:=\max(s,t)$.
+  \end{remark}
+  \subsubsection{Martingales}
+  From now on, we will assume that we work in a filtered probability space $(\Omega,\mathcal{F},\Prob,{(\mathcal{F}_t)}_{t\geq 0})$.
+  \begin{proposition}
+    Let ${(B_t)}_{t\geq 0}$ be a Brownian motion. Then, the following processes are martingales ${(M_t)}_{t\geq 0}$ with respect to the filtration induced by ${(B_t)}_{t\geq 0}$:
+    \begin{itemize}
+      \item $M_t=B_t$
+      \item $M_t=B_t^2-t$
+      \item $M_t=\exp{\theta B_t-\frac{1}{2}\theta^2t}$, for any fixed $\theta\in\RR$.
+    \end{itemize}
+  \end{proposition}
+  \begin{proposition}
+    Let $A\subseteq \RR$ be a closed set and $X={(X_t)}_{t\geq 0}$ be an adapted continuous process. Then, the \emph{hitting time} of $A$ by $X$, defined as:
+    $$
+      T_A:=\inf\{t\geq 0:X_t\in A\}
+    $$
+    is a stopping time.
+  \end{proposition}
+  \begin{proof}
+    Using the continuity of $X$ and the fact that $A$ is closed, one can easily check that:
+    $$
+      \{T_A \leq t\}=\bigcap_{k\in\NN}\bigcup_{s\in[0,t]\cap\QQ}\left\{d(X_s,A)\leq\frac{1}{k}\right\}
+    $$
+    Now, $\left\{d(X_s,A)\leq\frac{1}{k}\right\}\in \mathcal{F}_s\subseteq \mathcal{F}_t$ because $X$ is adapted and $z\mapsto d(z,A)$ is measurable.
+    Thus, $\{T_A \leq t\}\in \mathcal{F}_t$ because it is a countable union and intersection of events in $\mathcal{F}_t$.
+  \end{proof}
+  \begin{theorem}[Doob's optional sampling theorem]
+    Let ${(M_t)}_{t\geq 0}$ be a continuous martingale and $T$ be a stopping time. Then, the \emph{stopped process} $M^T:={(M_{t\wedge T})}_{t\geq 0}$ is a continuous martingale. In particular, $\forall t\geq 0$, $\Exp(M_{t\wedge T})=\Exp(M_0)$. If $M^T$ is uniformly integrable and $T\overset{\text{a.s.}}{\leq}\infty$, then taking $t\to\infty$ we have $\Exp(M_T)=\Exp(M_0)$.
+  \end{theorem}
 \end{multicols}
 \end{document}

preamble_formulas.sty

@@ -282,7 +282,8 @@
 \newcommand{\RR}{\ensuremath{\mathbb{R}}} % set of real numbers
 \newcommand{\CC}{\ensuremath{\mathbb{C}}} % set of complex numbers
 \newcommand{\KK}{\ensuremath{\mathbb{K}}} % a general field
-
+\newcommand{\TT}{\ensuremath{\mathbb{T}}} % time for stochastic processes
+\renewcommand{\SS}{\ensuremath{\mathbb{S}}} % other things
 
 %%% TOPOLOGY
 \DeclareMathOperator{\Int}{Int} % interior set