From 57ed1484c707b3403261a454d8bdb58ffc6b74d3 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?V=C3=ADctor?= Date: Mon, 19 Jun 2023 11:59:25 +0200 Subject: [PATCH] updated numericla integ --- .../Numerical_methods/Numerical_methods.tex | 2 +- .../Images/courant-friedrichs-lewy.pdf | Bin 59534 -> 51474 bytes .../Images/courant-friedrichs-lewy.tex | 2 +- ...tion_of_partial_differential_equations.tex | 49 +++- .../Stochastic_processes.tex | 231 ++++++++++++------ 5 files changed, 195 insertions(+), 89 deletions(-) diff --git a/Mathematics/2nd/Numerical_methods/Numerical_methods.tex b/Mathematics/2nd/Numerical_methods/Numerical_methods.tex index 964245c..3537b61 100644 --- a/Mathematics/2nd/Numerical_methods/Numerical_methods.tex +++ b/Mathematics/2nd/Numerical_methods/Numerical_methods.tex @@ -571,7 +571,7 @@ &\hspace*{4cm}\leq \max_{1\leq i\leq n}\sum_{j=1}^n\abs{a_{ij}}= A_\infty \end{split} \end{align*} - And taking $\vf{v}=\vf{e}_{j_0}$ and $\vf{u}=(\sign{a_{i_01}}, \ldots, \sign{a_{i_0n}})$ we have that $\norm{\vf{A}v}_1=A_1$ and $\norm{\vf{A}u}_\infty=A_\infty$. So $\norm{\vf{A}}_1=A_1$ and $\norm{\vf{A}}_\infty=A_\infty$. Now, let's do the $\norm{\cdot}_2$ norm. Observe that $\transpose{\vf{A}}\vf{A}$ is symmetric, and therefore it diagonalizes in an orthonormal basis of eigenvectors $\vf{v}_1, \ldots, \vf{v}_n$ with eigenvalues $\lambda_1, \ldots, \lambda_n$. Note that for each of these eigenvectors we have: + And taking $\vf{v}=\vf{e}_{j_0}$ and $\vf{u}=(\sign{a_{i_01}}, \ldots, \sign{a_{i_0n}})$ we have that $\norm{\vf{Av}}_1=A_1$ and $\norm{\vf{Au}}_\infty=A_\infty$. So $\norm{\vf{A}}_1=A_1$ and $\norm{\vf{A}}_\infty=A_\infty$. Now, let's do the $\norm{\cdot}_2$ norm. Observe that $\transpose{\vf{A}}\vf{A}$ is symmetric, and therefore it diagonalizes in an orthonormal basis of eigenvectors $\vf{v}_1, \ldots, \vf{v}_n$ with eigenvalues $\lambda_1, \ldots, \lambda_n$. 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z&jsD@up(6D9z4qOj^xxjPO$M?`)7T@7?m_PfcbMKfeAnlW!#L@Loi4CW7u~JQR!q)`yoXwxK z>s$_h3mHQMCD<_x>(a9RRIz^Ud7(QGFwxxVKyA8g(V#!iNOiAZNOOvQTx@xeU*?E$RwFknj-NHx^G!_WI3b}!VpdiTo znw5_a1%lRuARtyixUo0c2ZGl6C%FK*p@X@AH#9WV!fF_xG?4p5u+Jel0~`T`!)a*| z&;-NtkpJJZKjm*)+P5<8(}xD+-~W}wU%wD)^%Z+H6Ck?N8BFA2<)Q)ymP`S})k~ f%vC==z4N1$ for some eigenvalue $a_i$ and let $\vf{u}_0(x)=\vf{c}\indi{\{\abs{x}> \frac{1}{\abs{\lambda}}\}}$ with $\vf{c}=(c_1,\ldots,c_q)$ and $c_i\ne 0$. As shown in figure \mcref{NIPDE:courant-friedrichs-lewy_fig} the numerical solution at $\vf{u}(1,0)$ will have the $i$-th component always equal to 0, whereas in general the exact solution won't be. + It suffices to study only the case $q=1$. Suppose $\abs{a\lambda}>1$ for some eigenvalue $a$ of $\vf{A}$ and let $\vf{u}_0(x)=\vf{c}\indi{\{\abs{x}> \frac{1}{\abs{\lambda}}\}}$ with $\vf{c}=(c_1,\ldots,c_q)$ and $c_i\ne 0$. As shown in figure \mcref{NIPDE:courant-friedrichs-lewy_fig}, by the form of the scheme, the numerical solution at $(t,x)=(1,0)$, $v_0^n$, will only depend on $v_m^0$ with $\abs{m}\leq n$. But taking $n$ such that $kn=1$, we have that $\abs{m}h\leq nk/\lambda\leq 1/\lambda$. So $v_0^n$ will depend on $x$ for $\abs{x}\leq \frac{1}{\lambda}<\abs{a}$. Thus, in general we will have the numerical solution equal to 0, whereas the exact solution will not be. \begin{figure}[H] \centering \includestandalone[mode=image|tex, width=\linewidth]{Images/courant-friedrichs-lewy} - \caption{Finite difference scheme (blue) versus the characteristic lines (in red). The arrows inward a bullet come from the points from which it depends.} + \caption{Finite difference scheme (blue) versus the characteristic lines (red). The arrows inward a bullet come from the points from which it depends.} \label{NIPDE:courant-friedrichs-lewy_fig} \end{figure} \end{proof} @@ -198,11 +204,12 @@ \end{align*} \end{proof} \begin{proposition} - Consider the pde of \mcref{NIPDE:traffic}. Then: + Consider the pde of \mcref{NIPDE:traffic} with $\lambda =k/h=\const$ Then: \begin{itemize} \item The FTFS scheme is stable if and only if $a\lambda\in [-1,0]$. \item The FTBS scheme is stable if and only if $a\lambda\in [0,1]$. \item The FTCS scheme is always unstable. + \item The BTCS scheme is unconditionally stable. \item The Lax-Friedrichs scheme is stable if and only if $\abs{a\lambda}\leq 1$. \end{itemize} \end{proposition} @@ -252,8 +259,38 @@ The Crank-Nicolson scheme is a one-step method that has order of consistency 2, and it is unconditionally stable. \end{proposition} \begin{sproof} - An easy check shows that: - $$P_{k,h}\phi-R_{k,h}P\phi=\O{{(k+h)}^2}+\O{\frac{k^3}{h}}$$ + Let $P=\pdv{}{t}+a\pdv{}{x}$. Let's start with the consistency. Using $\phi=\phi(t,x)=v_m^n$ we can simplify the first term as: + $$ + \frac{\phi(t+k,x)-\phi}{k}=\phi_t+\frac{k}{2}\phi_{tt}+\O{k^2} + $$ + For the second term note that: + \begin{align*} + \begin{split} + \phi(t+k,x+h) & =\phi(t+k,x)+ h\phi_x(t+k,x)+\\&\hspace{2cm}+\frac{h^2}{2}\phi_{xx}(t+k,x)+\O{h^3} + \end{split} \\ + \begin{split} + -\phi(t+k,x-h) & =-\phi(t+k,x)+ h\phi_x(t+k,x)-\\&\hspace{2cm}-\frac{h^2}{2}\phi_{xx}(t+k,x)+\O{h^3} + \end{split} \\ + \phi(t,x+h) & =\phi + h\phi_x+\frac{h^2}{2}\phi_{xx}+\O{h^3} \\ + -\phi(t,x-h) & =-\phi + h\phi_x-\frac{h^2}{2}\phi_{xx}+\O{h^3} + \end{align*} + Summing these equations and multiplying by $\frac{a}{4h}$ we get: + $$ + \frac{a}{2}[\phi_x+\phi_x(t+k,x)]+\O{h^2}\!=\!a\phi_x+\frac{a}{2}k\phi_{xt}+\O{h^2}+\O{k^2} + $$ + Thus: + $$ + P_{k,h}\phi=\phi_t+a\phi_x+\frac{k}{2}\phi_{tt}+\frac{a}{2}k\phi_{xt}+\O{k^2}+\O{h^2} + $$ + On the other hand: + \begin{align*} + R_{k,h}P\phi & =\frac{\phi_t(t+k,x)+a\phi_x(t+k,x)+\phi_t+a\phi_x}{2} \\ + & =\phi_t+a\phi_x+\frac{1}{2}k\phi_{tt}+\frac{a}{2}k\phi_{xt}+\O{k^2} + \end{align*} + Finally: + $$ + P_{k,h}\phi-R_{k,h}P\phi=\O{k^2}+\O{h^2} + $$ For the stability, substitute $v_m^n=g^n\exp{\ii m\theta}$ in the scheme. Simplifying we get: $$ g=\frac{1+\frac{a\lambda\ii}{2}\sin\theta}{1-\frac{a\lambda\ii}{2}\sin\theta} diff --git a/Mathematics/4th/Stochastic_processes/Stochastic_processes.tex b/Mathematics/4th/Stochastic_processes/Stochastic_processes.tex index 2279638..06b7c0f 100644 --- a/Mathematics/4th/Stochastic_processes/Stochastic_processes.tex +++ b/Mathematics/4th/Stochastic_processes/Stochastic_processes.tex @@ -145,12 +145,16 @@ \captionof{table}{Probability-generating functions of common distributions.} \end{center} \subsection{Discrete-time Markov chains} + \subsubsection{Stochastic processes} \begin{definition}[Stochastic process] Let $T\subseteq \RR^n$ be a set, $(E,\mathcal{E})$ be a measurable space and $(\Omega,\mathcal{A},\Prob)$ be a probability space. A \emph{stochastic process} on $(\Omega,\mathcal{A},\Prob)$ with \emph{parameter set} $T$ and \emph{state space} $(E,\mathcal{E})$ is a family of random variables ${\{X_t\}}_{t\in T}$ from $(\Omega,\mathcal{A})$ to $(E,\mathcal{E})$. That is, $X_t:\Omega\to E$ satisfies ${X_t}^{-1}(B)\in\mathcal{A}$ for all $B\in\mathcal{E}$ and all $t\in T$. \end{definition} \begin{remark} In general, we wil consider stochastic processes with parameter sets $T=\NN,\NN\cup\{0\},\ZZ,\RR,\RR_{\geq 0}$ and state spaces $(\NN\cup\{0\},\mathcal{P}(\NN \cup \{0\}))$ or $(\RR,\mathcal{B}(\RR))$. \end{remark} + \begin{definition} + Let ${(X_t)}_{t\in T}$, ${(Y_t)}_{t\in T}$ be two stochastic processes defined on the same probability space $(\Omega,\mathcal{A},\Prob)$. We say that ${(X_t)}_{t\in T}$ and ${(Y_t)}_{t\in T}$ are \emph{independent} if $\forall n,k\in\NN$ and all $t_1,\ldots,t_n,s_1,\ldots,s_k\in T$ we have that the random vectors $(X_{t_1},\ldots,X_{t_n})$ and $(Y_{s_1},\ldots,Y_{s_k})$ are independent. + \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)$$ @@ -567,8 +571,31 @@ with the convention that $\tau_i^1=\tau_i$ and $\tau_i^0=0$. Moreover, we define the time difference $T_i^k:=\tau_i^k-\tau_i^{k-1}$. \end{definition} \begin{lemma} - Let $(X_n)$ be a time-homogeneous Markov chain. Then, $\tau_i^k$ is a stopping time $\forall k\in\NN$ and moreover $T_i^k$ are \iid random variables distributed as $\tau_i$. + Let $(X_n)$ be a time-homogeneous Markov chain. Then, $\tau_i^k$ is a stopping time $\forall k\in\NN$ and moreover $T_i^k$ are \iid random variables distributed as $\tau_i$ with respect to the probability $\Prob_i$. \end{lemma} + \begin{proof} + We need to check that $\forall m_1,\ldots,m_k\in\NN$: + $$ + \Prob_i(T_i^1=m_1,\ldots,T_i^k=m_k)=\Prob_i(\tau_i=m_1)\cdots\Prob_i(\tau_i=m_k) + $$ + We expand the left-hand side using the \mnameref{P:compound}. Now we examine each term of the product, which have the form: + $$ + p_{\ell}:=\Prob_i(T_i^\ell=m_\ell\mid T_i^1=m_1,\ldots,T_i^{\ell-1}=m_{\ell-1}) + $$ + We have that: + \begin{equation*} + p_\ell=\Prob_i(\tau_i^{\ell}-\tau_i^{\ell-1}=m_\ell\mid A) + \end{equation*} + where $A=\{X_0=i,X_1\ne i, \ldots, X_{m_1-1}\ne i,X_{m_1}=i, X_{m_1+1}\ne i,\ldots, X_{m_1+\cdots+m_{\ell-1}}=i\}$. So, by the \mnameref{SP:substitutionPrinciple} we have: + \begin{align*} + \begin{split} + p_\ell&=\Prob_i(X_{m_1+\cdots+m_\ell}=i,X_{m_1+\cdots+m_\ell-1}\ne i,\ldots,\\ + &\hspace{4cm}X_{m_1+\cdots+m_{\ell-1}+1}\ne i\mid A) + \end{split} \\ + & =\Prob_i(X_{m_\ell}=i,X_{m_\ell-1}\ne i,\ldots,X_{1}\mid X_{0}=i) \\ + & =\Prob(\tau_i=m_\ell) + \end{align*} + \end{proof} \begin{proposition}\label{SP:recurrence} Let $(X_n)$ be a time-homogeneous Markov chain and $i\in I$. Then: $$ @@ -632,7 +659,7 @@ & =\sum_{m=1}^n \Prob_i(X_n=j\mid X_m=j,X_{m-1}\ne j,\ldots, X_1\ne j)\cdot \\ &\hspace{7cm}\cdot f_{ij}^{(m)} \end{split} \\ - & =\sum_{m=1}^n \Prob_i(X_{n-m}=j)f_{ij}^{(m)} \\ + & =\sum_{m=1}^n \Prob_j(X_{n-m}=j)f_{ij}^{(m)} \\ & =\sum_{m=1}^nf_{ij}^{(m)}p_{jj}^{(n-m)} \end{align*} \end{proof} @@ -739,9 +766,10 @@ \begin{sproof} We will proof only the implication to the left, in order to keep the proof short. Note that we have: $$ + % p_{ii}^{(2n)}=\sum_{m=0}^{n}\binom{2n}{2m}\binom{2m}{m}\binom{2n-2m}{n-m}\frac{1}{4^{2n}}=\frac{1}{4^{2n}}\binom{2n}{n}\sum_{m=0}^n \binom{n}{m}^2 p_{ii}^{(2n)}=\sum_{m=0}^{n}\frac{(2n)!}{{(m!)}^2{(n-m)!}^2}\frac{1}{4^{2n}}=\frac{1}{4^{2n}}\binom{2n}{n}\sum_{m=0}^n \binom{n}{m}^2 $$ - In the formula $m$ denotes the number of steps rightwards, $n-m$, the number of steps upwards, $n-m$ the number of steps leftwards, and $m$, the number of steps downwards. Now using \mcref{SP:2n-n_convinatoria,SP:stirling_polya1} we have: + In the formula $m$ denotes the number of steps rightwards and leftwards, and $n-m$, the number of steps upwards and downwards. Now using \mcref{SP:2n-n_convinatoria,SP:stirling_polya1} we have: $$ p_{ii}^{(2n)}=\frac{1}{4^{2n}}\binom{2n}{n}^2\sim \frac{1}{n} $$ @@ -750,7 +778,7 @@ The simple random walk on $\ZZ^3$ is always transient. \end{theorem} \begin{corollary} - Let $(X_n)$ be a time-homogeneous Markov chain and $i,j\in I$. Then, if $j$ is recurrent $\Exp_i(N_i)<\infty$. + Let $(X_n)$ be a time-homogeneous Markov chain and $i,j\in I$. Then, if $j$ is transient, we have $\Exp_i(N_j)<\infty$. \end{corollary} \begin{proof} $$ @@ -765,19 +793,19 @@ \begin{theorem}[Ergotic theorem] Let $(X_n)$ be a time-homogeneous Markov chain and $i\in I$ be positively recurrent. Then: $$ - \lim_{n\to\infty}\frac{1}{n}\sum_{n=1}^\infty p_{ii}^{(n)}=\frac{1}{\mu_i} + \lim_{n\to\infty}\frac{1}{n}\sum_{m=1}^n p_{ii}^{(m)}=\frac{1}{\mu_i} $$ \end{theorem} \begin{proof} - Note that $T_i^k$ has finite expectation and so by the \mcref{P:stronglawKolmo} we have: - $$\frac{1}{k}\sum_{j=1}^{k}T_i^j=\frac{\tau_i^k}{k}\almoste{\longrightarrow}\mu_i$$ - Let $N_i^n=\sum_{j=1}^n\indi{X_j=i}\leq n$, which counts the number of visits of the state $i$ in the first $n$ steps. Note that if $N_i^n=k\leq n$, $\tau_i^k\leq n<\tau_i^{k+1}$ and so: + By hypothesis $T_i^k$ has finite expectation and so by the \mcref{P:stronglawKolmo} we have: + $$\frac{1}{k}\sum_{m=1}^{k}T_i^m=\frac{\tau_i^k}{k}\almoste{\longrightarrow}\mu_i$$ + Let $N_i^n=\sum_{m=1}^n\indi{\{X_m=i\}}\leq n$, which counts the number of visits of the state $i$ in the first $n$ steps. Note that if $N_i^n=k\leq n$, $\tau_i^k\leq n<\tau_i^{k+1}$ and so: $$ \frac{k}{k+1}\frac{k+1}{\tau_i^{k+1}}<\frac{N_i^n}{n}=\frac{k}{n}\leq \frac{k}{\tau_i^k} $$ Hence, taking the limit $k\to\infty$ we have: $$\displaystyle\lim_{n\to\infty}\frac{N_i^n}{n}=\lim_{k\to\infty}\frac{k}{\tau_i^k}\almoste{\longrightarrow}\frac{1}{\mu_i}$$ Moreover note that $\frac{N_i^n}{n}\leq 1$. Thus, by the \mcref{P:dominated} we have that: $$ - \Exp_i\left(\frac{N_i^n}{n}\right)=\frac{\sum_{j=1}^n \Prob_i(X_j=i)}{n}=\frac{\sum_{j=1}^n p_{ii}^{(j)}}{n}\almoste{\longrightarrow}\frac{1}{\mu_i} + \Exp_i\left(\frac{N_i^n}{n}\right)=\frac{\sum_{m=1}^n \Prob_i(X_m=i)}{n}=\frac{\sum_{m=1}^n p_{ii}^{(m)}}{n}\almoste{\longrightarrow}\frac{1}{\mu_i} $$ \end{proof} \begin{corollary} @@ -789,11 +817,12 @@ \begin{sproof} Recall \mcref{MA:cesaro}. \end{sproof} - \begin{theorem}[Erogotic theorem]\label{SP:ergotic2} + \begin{theorem}[Ergotic theorem]\label{SP:ergotic2} Let $(X_n)$ be a time-homogeneous Markov chain and $i\in I$ be recurrent and aperiodic. Then, the limit $\displaystyle\lim_{n\to\infty}p_{ii}^{(n)}$ exists and: $$ - \lim_{n\to\infty}\frac{1}{n}\sum_{n=1}^\infty p_{ii}^{(n)}=\frac{1}{\mu_i} + \lim_{n\to\infty}p_{ii}^{(n)}=\lim_{n\to\infty}\frac{1}{n}\sum_{n=1}^\infty p_{ii}^{(n)}=\frac{1}{\mu_i} $$ + In particular, if $i$ is positive recurrent, then $\displaystyle\lim_{n\to\infty}p_{ii}^{(n)}>0$ and if $i$ is null recurrent, then $\displaystyle\lim_{n\to\infty}p_{ii}^{(n)}=0$. \end{theorem} \begin{proposition} Let $(X_n)$ be a time-homogeneous Markov chain, $i\in I$ be recurrent and aperiodic and $j\in I$ be such that $i\leftrightarrow j$. Then: @@ -803,7 +832,7 @@ \end{enumerate} \end{proposition} \begin{proof} - By \mcref{SP:thmRec,SP:period_classes} we have that $j$ is recurrent and aperiodic. Thus, by \mnameref{SP:ergotic2}, the limits $\displaystyle \lim_{n\to\infty}p_{ij}^{(n)}$ and $\displaystyle \lim_{n\to\infty}p_{ji}^{(n)}$ exist. Moreover, since $i\leftrightarrow j$ $\exists r,s\in\NN$ such that $p_{ij}^{(r)}, p_{ji}^{(s)}>0$. By \mcref{SP:corolariChapKolmo} we have that $p_{jj}^{(n+r+s)}\geq C p_{ii}^{(n)}$. If $i$ is positive recurrent then: + By \mcref{SP:thmRec,SP:period_classes} we have that $j$ is recurrent and aperiodic. Thus, by \mnameref{SP:ergotic2}, the limits $\displaystyle \lim_{n\to\infty}p_{ii}^{(n)}$ and $\displaystyle \lim_{n\to\infty}p_{jj}^{(n)}$ exist. Moreover, since $i\leftrightarrow j$ $\exists r,s\in\NN$ such that $p_{ij}^{(r)}, p_{ji}^{(s)}>0$. By \mcref{SP:corolariChapKolmo} we have that $p_{jj}^{(n+r+s)}\geq C p_{ii}^{(n)}$. If $i$ is positive recurrent then: $$ \lim_{n\to\infty}p_{jj}^{(n+r+s)}\geq C\lim_{n\to\infty}p_{ii}^{(n)}>0 $$ @@ -812,17 +841,17 @@ \begin{theorem} Let $(X_n)$ be a time-homogeneous Markov chain and $i\in I$ be recurrent and periodic of period $d$. Then: $$ - \lim_{n\to\infty} p_{ii}^{(n)}=\frac{d}{\mu_i} + \lim_{n\to\infty} p_{ii}^{(nd)}=\frac{d}{\mu_i} $$ \end{theorem} \begin{proof} - $(Y_n)=(X_{nd})$ is a time-homogeneous Markov chain and $i\in I$ is recurrent and aperiodic. Thus, by \mnameref{SP:ergotic2} we have that $\displaystyle\lim_{n\to\infty}p_{ii}^{(nd)}=\frac{1}{\Exp_i(\tau_i^Y)}$. But: + $(Y_n):=(X_{nd})$ is a time-homogeneous Markov chain and $i\in I$ is recurrent and aperiodic. Thus, by \mnameref{SP:ergotic2} we have that $\displaystyle\lim_{n\to\infty}p_{ii}^{(nd)}=\frac{1}{\Exp_i(\tau_i^Y)}$. But: $$ \tau_i^Y=\inf\{ n\geq 1: Y_n=i\}=\frac{1}{d}\inf\{n\geq 1: X_{n}=i\}=\frac{\tau_i}{d} $$ \end{proof} \begin{theorem} - Let $(X_n)$ be a time-homogeneous irreducible and aperiodic Markov chain. Then we have exactly one of the following results: + Let $(X_n)$ be a time-homogeneous irreducible and aperiodic Markov chain. Then, we have exactly one of the following results: \begin{enumerate} \item All the states are transient and $\forall i,j\in I$: $$\lim_{n\to\infty} p_{ij}^{(n)}=\lim_{n\to\infty} \pi_j^{(n)}=0$$ Moreover $\sum_{n=1}^\infty p_{ij}^{(n)}<\infty$. \item All the states are null recurrent and $\forall i,j\in I$: $$\lim_{n\to\infty} p_{ij}^{(n)}=\lim_{n\to\infty} \pi_j^{(n)}=0$$ Moreover $\sum_{n=1}^\infty p_{ij}^{(n)}=\infty$. @@ -832,12 +861,12 @@ \begin{proof} It can be seen that $\displaystyle\lim_{n\to\infty}p_{ij}^{(n)}=\lim_{n\to\infty} p_{jj}^{(n)}$ $\forall i, j\in I$. We will prove that $\displaystyle\lim_{n\to\infty}p_{ij}^{(n)}=\lim_{n\to\infty} \pi_j^{(n)}$ $\forall i, j\in I$. We have that: \begin{multline*} - \lim_{n\to\infty}\pi_j^{(n)}=\lim_{n\to\infty}\Prob(X_n=j)=\lim_{n\to\infty}\sum_{i\in I}p_{ij}^{(n)}\pi_i=\\=\sum_{i\in I}\lim_{n\to\infty}p_{ij}^{(n)}\pi_i=\sum_{i\in I}\frac{\pi_i}{\mu_i}=\mu_i + \lim_{n\to\infty}\pi_j^{(n)}=\lim_{n\to\infty}\Prob(X_n=j)=\lim_{n\to\infty}\sum_{i\in I}p_{ij}^{(n)}\pi_i=\\=\sum_{i\in I}\lim_{n\to\infty}p_{ij}^{(n)}\pi_i=\sum_{i\in I}\frac{\pi_i}{\mu_j}=\frac{1}{\mu_j} \end{multline*} where we have used the dominated convergence theorem for series. \end{proof} \begin{corollary}\label{SP:coroClassificationStates} - Let $(X_n)$ be a time-homogeneous irreducible and aperiodic Markov chain such that $I$ is finite. Then all the states are positive recurrent. + Let $(X_n)$ be a time-homogeneous irreducible and aperiodic Markov chain such that $I$ is finite. Then, all the states are positive recurrent. \end{corollary} \begin{sproof} Note that we must have $\sum_{j\in I}\pi_j^{(n)}=1$ $\forall n\in\NN$. @@ -849,7 +878,7 @@ $$ \end{definition} \begin{remark} - In general we cannot guarantee neither existence nor uniqueness of stationary distributions. + In general, we cannot guarantee the existence or uniqueness of stationary distributions. \end{remark} \begin{lemma} Let $(X_n)$ be a time-homogeneous Markov chain, $\vf\nu$ be a stationary distribution and suppose $\vf\pi_0=\vf\nu$. Then, $\vf\pi_n=\vf\nu$ $\forall n\in\NN$. @@ -868,7 +897,7 @@ \sum_{j\in I}\nu_j & =\sum_{j\in I}\lim_{n\to\infty}p_{ij}^{(n)}=\lim_{n\to\infty}\sum_{j\in I}p_{ij}^{(n)}=1 \\ \sum_{i\in I}\nu_i p_{ij} & =\sum_{i\in I}\lim_{n\to\infty}p_{ki}^{(n)}p_{ij}=\lim_{n\to\infty}p_{kj}^{(n+1)}=\nu_j \end{align*} - where we have used \mnameref{SP:ChapKolmo}. Hence, $\vf\nu$ is a stationary distribution. Now, for the uniqueness, suppose $\vf\nu$ is an arbitrary stationary distribution. Then, $\nu_j=\sum_{i\in I}\nu_i p_{ij}^{(n)}$ $\forall n\in\NN$. Thus, taking $n\to\infty$ we get that $\nu_i = \frac{1}{\mu_i}$ $\forall i\in I$. + where we have used \mnameref{SP:ChapKolmo}. Hence, $\vf\nu$ is a stationary distribution. Now, for the uniqueness, suppose $\vf\nu$ is an arbitrary stationary distribution. Then, $\nu_j=\sum_{i\in I}\nu_i p_{ij}^{(n)}$ $\forall n\in\NN$. Thus, taking $n\to\infty$ we get that $\nu_j = \frac{1}{\mu_j}$ $\forall j\in I$. \end{proof} \subsection{Continuous-time Markov chains} \subsubsection{Introduction} @@ -906,18 +935,11 @@ \begin{proposition} Let ${(X_t)}_{t\geq 0}$ be a CTHMC. Then, for all $0\leq t_1 < \cdots < t_n$ and all $i_1,\ldots,i_{n}\in I$ we have that: \begin{multline*} - \Prob(X_{t_n}=i_n\mid X_{t_{n-1}}=i_{n-1},\ldots,X_{t_1}=i_1)=\\ + \Prob(X_{t_n}=i_n,X_{t_{n-1}}=i_{n-1},\ldots,X_{t_1}=i_1)=\\ =p_{i_1}(t_1)p_{i_1i_2}(t_2-t_1)\cdots p_{i_{n-1}i_n}(t_n-t_{n-1}) \end{multline*} where $p_{i}(t):= \Prob(X_t=i)$. \end{proposition} - \begin{sproof} - First note that: - $$ - p_j(t)=\sum_{i\in I}p_i(0)\Prob(X_t=j\mid X_0=i)=\sum_{i\in I}p_i(0)p_{ij}(t) - $$ - The formula is a consequence of the \mnameref{P:compound}. - \end{sproof} \subsubsection{Poisson process} \begin{definition} Let $\lambda>0$. A stochastic process ${(N_t)}_{t\geq 0}$ is called a \emph{Poisson process} with parameter $\lambda$ if: @@ -937,10 +959,10 @@ \end{proposition} \begin{proof} \begin{align*} - \Prob(X_{t_{n+1}} & = j\mid X_{t_n}=i, X_{t_{n-1}}=i_{n-1}\ldots,X_{t_1}=i_1) \\ - & =\Prob(X_{t_{n+1}}-X_{t_n}=j-i\mid X_{t_n}-X_{t_{n-1}}=i- \\ - & \hspace{0.45cm}-i_{n-1},\ldots,\ldots,X_{t_2}-X_{t_1}=i_2-i_1,X_{t_1}=i_1) \\ - & =\Prob(X_{t_{n+1}}-X_{t_n}=j-i) \\ + \Prob(X_{t_{n+1}} & = j\mid X_{t_n}=i, X_{t_{n-1}}=i_{n-1}\ldots,X_{t_1}=i_1) \\ + & =\Prob(X_{t_{n+1}}-X_{t_n}=j-i\mid X_{t_n}-X_{t_{n-1}}=i- \\ + & \hspace{1cm}-i_{n-1},\ldots,X_{t_2}-X_{t_1}=i_2-i_1,X_{t_1}=i_1) \\ + & =\Prob(X_{t_{n+1}}-X_{t_n}=j-i) \\ & =\Prob(X_{t_{n+1}}=j\mid X_{t_n}=i) \end{align*} \end{proof} @@ -950,14 +972,28 @@ p_{ij}(t)=\Prob(N_t=j\mid N_0=i)=\frac{(\lambda t)^j}{j!}e^{-\lambda t} $$ \end{corollary} + \begin{proposition} + Let ${(N_t^1)}_{t\geq 0}$ and ${(N_t^2)}_{t\geq 0}$ be two independent Poisson processes with parameters $\lambda_1$ and $\lambda_2$ respectively. Then, ${(N_t^1+N_t^2)}_{t\geq 0}$ is a Poisson process with parameter $\lambda_1+\lambda_2$. + \end{proposition} + \begin{proof} + Let $N_t:=(N_t^1+N_t^2)$. We only check the independent increment, the other properties are easier. We need to check that for all $0\leq t_1< \cdots < t_n$ and all $n\in\NN$ the random variables $X_{t_1}:=N_{t_1},X_{t_2}:= N_{t_2}-N_{t_1},\ldots,X_{t_n}:=N_{t_n}-N_{t_{n-1}}$ are independent. We have that + $$ + X_{t_\ell}=N_{t_\ell}^1-N_{t_{\ell-1}}^1+N_{t_\ell}^2-N_{t_{\ell-1}}^2=:Y_\ell^1+Y_\ell^2 + $$ + By hypothesis the variables $Y_\ell^1$ and $Y_\ell^2$ are independent. Moreover, since $N_t^i$ are Poisson processes, we have that $\{(Y_k^i)\}_{k=1,\ldots,n}$ pairwise independent, for $i=1,2$. Now using the characterization of independence with the characteristic function, we have: + \begin{align*} + \varphi_{X_{t_1},\ldots,X_{t_n}}(u_1,\ldots,u_n) & =\Exp\left(\exp{\ii\sum_{j=1}^nu_j X_j}\right) \\ + & =\Exp\left(\exp{\ii\sum_{j=1}^nu_jY_j^1}\exp{\ii\sum_{j=1}^nu_jY_j^2}\right) \\ + & =\prod_{j=1}^n\Exp\left(\exp{\ii u_jY_j^1}\right)\Exp\left(\exp{\ii u_jY_j^2}\right) \\ + & = \prod_{j=1}^n\varphi_{X_j}(u_j) + \end{align*} + \end{proof} \begin{lemma} Let ${(N_t)}_{t\geq 0}$ be a Poisson process with parameter $\lambda$. Then: $$\Prob(N_h\geq 2)=\o{h}$$ \end{lemma} \begin{proof} - \begin{multline*} - \Prob(N_h\geq 2)=1-\Prob(N_h=0)-\Prob(N_h=1)=\\ - =1-e^{-\lambda h}-\lambda h e^{-\lambda h}=\o{h} - \end{multline*} + $\displaystyle + \Prob(N_h\geq 2)=1-e^{-\lambda h}-\lambda h e^{-\lambda h}=\o{h}$ \end{proof} \begin{proposition} Let ${(N_t)}_{t\geq 0}$ be a Poisson process with parameter $\lambda$. Then, the trajectories are almost surely non-decreasing and have jumps of size at most $1$. @@ -969,15 +1005,16 @@ $$ because the trajectories are càd. Finally, since $\Prob(N_s\leq N_t)=\Prob(N_t-N_s\geq 0)=1$, the intersection has probability $1$. Now, let: \begin{align*} - A & :=\{\omega\in\Omega:N(\omega)\text{ has jumps of size }\geq 2\} \\ - A_R & :=\{\omega\in\Omega:N(\omega)\text{ has jumps of size }\geq 2\text{ in $[0,R]$}\} \\ + A & :=\{\omega\in\Omega:N(\omega)\text{ has jumps of size}\geq 2\} \\ + A_R & :=\{\omega\in\Omega:N(\omega)\text{ has jumps of size}\geq 2\text{ in $[0,R]$}\} \\ B_R^n & :=\{\exists k\in\{1,\ldots,n\}: N_{\frac{kR}{n}}-N_{\frac{(k-1)R}{n}}\geq 2\} \end{align*} - Note that $A=\bigcup_{R=1}^\infty A_R$ and $A_R\subseteq B_R^n$ $\forall n\geq 1$ because the trajectories are càd. Thus: + Note that $A=\bigcup_{R=1}^\infty A_R$ and $A_R\subseteq B_R^n$ $\forall n\geq 1$ because the trajectories are càd. Thus, $\forall R>0$: \begin{multline*} \Prob(A_R)\leq \Prob(B_R^n)=\Prob\left(\bigcup_{k=1}^n\left\{N_{\frac{kR}{n}}-N_{\frac{(k-1)R}{n}}\geq 2\right\}\right)\leq\\ \leq\sum_{k=1}^n\Prob\left(N_{\frac{kR}{n}}-N_{\frac{(k-1)R}{n}}\geq 2\right)=n\Prob(N_{\frac{R}{n}}\geq 2)=\\=n\o{\frac{R}{n}}\overset{n\to\infty}{\longrightarrow}0 \end{multline*} + Hence, $\Prob(A)=0$. \end{proof} \begin{definition} Let ${(N_t)}_{t\geq 0}$ be a Poisson process with parameter $\lambda$. We define the \emph{holding times} as: @@ -997,9 +1034,10 @@ \begin{multline*} \Prob(T_k=\infty)=\Prob(\forall t\in\RR: N_t\leq k-1)\leq \Prob(N_1=0,\\N_2-N_1 =0,\ldots,N_n-N_{n-1}=0)=\exp{-\lambda n} \overset{n\to\infty}{\longrightarrow}0 \end{multline*} + because the inequality is true for all $n\in\NN$. \end{proof} \begin{theorem} - Let ${(N_t)}_{t\geq 0}$ be a Poisson process with parameter $\lambda$. Then, the inter-arrival times $(S_k)$ are \iid variables distributed as $\Exp(\lambda)$. + Let ${(N_t)}_{t\geq 0}$ be a Poisson process with parameter $\lambda$. Then, the inter-arrival times $(S_k)$ are \iid random variables distributed as $\text{Exp}(\lambda)$. \end{theorem} \begin{proof} Let $\vf{T}:=(T_1,\ldots,T_n)$. Recall that: @@ -1018,20 +1056,20 @@ $$ \function{\vf{g}}{\{00$ such that $p_{ij}(t_1)>0$ and $p_{ji}(t_2)>0$. + Let ${(X_t)}_{t\geq 0}$ be a CTHMC. We say that the chain is \emph{irreducible} if $\forall i,j\in I$, $\exists t_1,t_2>0$ such that $p_{ij}(t_1)>0$ and $p_{ji}(t_2)>0$. \end{definition} \begin{theorem}\label{SP:limit_distribution} Let $X$ be an irreducible CTHMC and a regular jump process. Then, we have exactly one of the following: \begin{enumerate} - \item The balance equation has a unique solution $\vf{\overline{p}}$ and in that case $\displaystyle \lim_{t\to\infty}p_{ij}(t)=\overline{p}_j$ $\forall i,j\in I$. In that case, $\vf{\overline{p}}$ is called a \emph{limit distribution}. + \item The balance equation has a unique solution $\vf{\overline{p}}$ (which must be the stationary distribution) and $\displaystyle \lim_{t\to\infty}p_{ij}(t)=\overline{p}_j$ $\forall i,j\in I$. In that case, $\vf{\overline{p}}$ is called a \emph{limit distribution}. \item The balance equation has no solution and in that case $\displaystyle \lim_{t\to\infty}p_{ij}(t)=0$ $\forall i,j\in I$. \end{enumerate} \end{theorem} @@ -1176,7 +1214,7 @@ \end{remark} \subsubsection{Birth and death processes} \begin{definition} - Let $I=\NN\cup\{0\}$. The following infinitesimal transition scheme describe the \emph{birth and death processes}: + Let $I=\NN\cup\{0\}$. A \emph{birth and death process} is a CTHMC and a regular jump process with the following infinitesimal transition scheme: $$ \begin{cases} p_{i,i+1}(h)=\lambda_i h+\o{h} & i\geq 0 \\ @@ -1186,7 +1224,7 @@ p_{ij} = \o{h} & \text{otherwise} \end{cases} $$ - This model describes a population of individuals, each of whom having $\lambda_i h + \o{h}$ probability of giving birth to a new individual in the time interval $[t,t+h)$ and $\mu_i h + \o{h}$ probability of dying in the same time interval. The probability of having more than one birth or death in that interval is $\o{h}$. In this case infinitesimal generator is: + This model describes a population of individuals, each of whom having $\lambda_i h + \o{h}$ probability of giving birth to a new individual in the time interval $[t,t+h)$ and $\mu_i h + \o{h}$ probability of dying in the same time interval. The probability of having more than one birth or death in that interval is $\o{h}$. In this case the infinitesimal generator is: $$ \vf{Q}=\begin{pmatrix} -\lambda_0 & \lambda_0 & 0 & 0 & \cdots \\ @@ -1218,8 +1256,19 @@ (\lambda_j+\mu_j) \frac{\lambda_0\cdots\lambda_{j-1}}{\mu_1\cdots\mu_j}\overline{p}_0 & = \lambda_{j-1}\frac{\lambda_0\cdots\lambda_{j-2}}{\mu_1\cdots\mu_{j-1}}\overline{p}_0+\mu_{j+1}\overline{p}_{j+1} \\ \frac{\lambda_0\cdots\lambda_{j-1}\lambda_j}{\mu_1\cdots\mu_j}\overline{p}_0 & = \mu_{j+1}\overline{p}_{j+1} \end{align*} - The first argument is determined from the condition $1=\sum_{i\in I} \overline{p}_i=\overline{p}_0+\overline{p}_0\sum_{i\in I}\frac{\lambda_0\lambda_1\cdots\lambda_{i-1}}{\mu_1\mu_2\cdots\mu_i}$. Now if we see that for $\lambda_i>0$ and $\mu_i>0$ $\forall i\in I$ the chain is irreducible, then the theorem will be proved by \mcref{SP:limit_distribution}. + The first argument is determined from the condition $1=\sum_{i\in I} \overline{p}_i=\overline{p}_0+\overline{p}_0\sum_{i\in I}\frac{\lambda_0\lambda_1\cdots\lambda_{i-1}}{\mu_1\mu_2\cdots\mu_i}$. Now if we see that for $\lambda_i>0$ and $\mu_i>0$ $\forall i\in I$ the chain is irreducible, then the theorem will be proved by \mcref{SP:limit_distribution}. But this is clear because, for example if $i 0 + \end{align*} + if $\lambda_i>0$ $\forall i\in I$ and for some $h$ small enough. The case $i>j$ is analogous. \end{proof} + \begin{theorem}[Reuter criterion] + Consider an infinitesimal generator for a birth and death process. Then, there is a CTHMC of regular jumps with this infinitesimal generator if and only if: + $$ + \sum_{n=1}^\infty \left[\frac{1}{\lambda_n}+\frac{\mu_n}{\lambda_n\lambda_{n-1}}+ \cdots+ \frac{\mu_n\cdots \mu_1}{\lambda_n\cdots\lambda_0}\right]=\infty + $$ + \end{theorem} \subsection{Brownian motion} \subsubsection{Gaussian processes} \begin{proposition}\label{SP:gaussian_vector} @@ -1232,7 +1281,7 @@ Let ${(X_t)}_{t\geq 0}$ be a gaussian process. Then, the \emph{mean function} is defined as: $$ - \function{\mu}{[0,\infty)}{\RR}{t}{\Exp{X_t}=:\mu_t} + \function{\mu}{[0,\infty)}{\RR}{t}{\Exp(X_t)=:\mu_t} $$ and the \emph{covariance function} is defined as: $$ @@ -1241,7 +1290,7 @@ \end{definition} \subsubsection{Brownian motion} \begin{definition} - A stochastic process ${(B_t)}_{t\geq 0}$ is called a \emph{Brownian motion} with parameter $\lambda$ if: + A stochastic process ${(B_t)}_{t\geq 0}$ is called a \emph{Brownian motion} (or a \emph{Wiener process}) with parameter $\lambda$ if: \begin{enumerate} \item $B_0=0$. \item $B_t$ has independent increments. @@ -1254,9 +1303,15 @@ Let $B:={(B_t)}_{t\geq 0}$ be a standard Brownian motion. Then, $B$ is a gaussian process with mean function $\mu_t=0$ and covariance function $C(s,t)=\min(s,t)$. \end{proposition} \begin{proof} - Let $0< t_1<\cdots0$ such that: $$ \Exp(\abs{X_t-X_s}^\alpha)\leq C\abs{t-s}^{1+\beta} $$ for all $t,s\geq 0$. Then, there exists a version of ${(X_t)}_{t\geq 0}$ with continuous trajectories. \end{theorem} - \begin{lemma} + \begin{lemma}\label{SP:post_kolmo} Let $X\sim N(0,\sigma^2)$. Then, $\Exp(\abs{X}^n)=C_n\sigma^n$ where: $$ C_n=\Exp(\abs{Z}^n)=\Gamma\left(\frac{n+1}{2}\right)\frac{2^{n/2}}{\sqrt{\pi}} $$ and $Z\sim N(0,1)$. \end{lemma} + \begin{corollary} + Let $B:={(B_t)}_{t\geq 0}$ be a standard Brownian motion. Then, there exists a version of $B$ with continuous trajectories. + \end{corollary} + \begin{proof} + We use \cref{SP:kolmogorov_continuity,SP:post_kolmo} with $\alpha=3$ and $\beta=1/2$. + \end{proof} \begin{proposition} Let $B:= {(B_t)}_{t\geq 0}$ be a standard Brownian motion. Then, for any interval $[a,b]\subset \RR$: $$ @@ -1365,11 +1426,11 @@ $$ \end{proof} \begin{proposition} - Let $B$ be a standard Brownian motion. Then, $\forall t\geq 0$ the set + Let $B:= {(B_t)}_{t\geq 0}$ be a standard Brownian motion. Then, $\forall t\geq 0$ the set $$ A:=\left\{\omega\in\Omega:\limsup_{h\to 0}\frac{\abs{B_{t+h}(\omega)-B_t(\omega)}}{h}=+\infty\right\} $$ - which could not be in the $\sigma$-algebra, contains an event of probability $1$. + which may not belong in the $\sigma$-algebra, contains an event of probability $1$. \end{proposition} \begin{proof} Note that $A\supseteq \{\omega\in\Omega:\sup_{n\in\NN}\frac{\abs{B_{t+1/n}(\omega)-B_t(\omega)}}{1/n}=+\infty\}=\bigcap_{M\geq 1} A_{M}$ where: @@ -1380,10 +1441,17 @@ \begin{multline*} \Prob(A_M)\geq \Prob \left(\frac{\abs{B_{t+1/n}(\omega)-B_t(\omega)}}{1/n}\geq M\right) = \\ = \Prob\left(\abs{Z}\geq \frac{M}{\sqrt{n}}\right)=2 \left(1-\Phi\left(\frac{M}{\sqrt{n}}\right)\right)\overset{n\to\infty}{\longrightarrow} 1 \end{multline*} - where $Z\sim N(0,1)$ and $\Phi$ is the cumulative distribution function of the standard normal distribution. + where the first inequality holds $\forall n\in\NN$, $Z\sim N(0,1)$ and $\Phi$ is the cumulative distribution function of the standard normal distribution. \end{proof} \begin{theorem}[Paley-Wiener-Zygmund theorem] - The Brownian trajectories are almost surely nowhere differentiable. + The Brownian trajectories are almost surely nowhere differentiable. Namely, the set + \begin{multline*} + \left\{ + \omega\in\Omega:\forall t\geq 0, \limsup_{h\to 0^+}\frac{{B_{t+h}(\omega)-B_t(\omega)}}{h}=+\infty\text{ or }\right.\\ + \left.\liminf_{h\to 0^+}\frac{{B_{t+h}(\omega)-B_t(\omega)}}{h}=-\infty + \right\} + \end{multline*} + contains an event of probability $1$. And the same occurs for the left limit $h\to 0^-$ (in this case we need to exclude $t=0$). \end{theorem} \begin{figure}[H] \centering @@ -1412,7 +1480,7 @@ $$ \end{lemma} \begin{corollary} - Let $B$ be a standard Brownian motion and $a>0$. Then, $\Prob(\tau_a<\infty)=1$. + Let $B$ be a standard Brownian motion and $a\in\RR^*$. Then, $\Prob(\tau_a<\infty)=1$. \end{corollary} \begin{proof} Assume $a>0$, the other case is similar. Then: @@ -1441,7 +1509,7 @@ The Brownian trajectories have infinite zeros almost surely, and they tend to infinity. \end{corollary} \begin{proof} - Let $A=\{\omega\in\Omega:B_{\cdot}(\omega)\text{ has finite zeros}\}\subseteq \bigcup_{n=1}^\infty\{\omega\in\Omega:B_{\cdot}(\omega)\text{ doesn't vanish in $[n,\infty)$}\}$. Let's see that all the events $A_n:=\{B_{\cdot}\text{ doesn't vanish in $[n,\infty)$}\}$ in the union have probability 0. + Let $B:=\{B_t:t\geq 0\}$ be a standard Brownian motion and let $A=\{\omega\in\Omega:B_{\cdot}(\omega)\text{ has finite zeros}\}\subseteq \bigcup_{n=1}^\infty\{\omega\in\Omega:B_{\cdot}(\omega)\text{ doesn't vanish in $[n,\infty)$}\}$. Let's see that all the events $A_n:=\{B_{\cdot}\text{ doesn't vanish in $[n,\infty)$}\}$ in the union have probability 0. \begin{multline*} \Prob(A_n)=\Prob(B_{\cdot}>0\text{ in $[n,\infty)$})+\Prob(B_{\cdot}<0\text{ in $[n,\infty)$})\leq\\\leq \Prob\left(\inf_{t\geq n}B_t\ne-\infty\right)+\Prob\left(\sup_{t\geq n}B_t\ne+\infty\right)=0 \end{multline*} @@ -1476,13 +1544,13 @@ because it diverges at $\infty$. \end{proof} \begin{definition} - An $n$-dimensional Brownian motion is a $d$-dimensional stochastic process $\vf{B}=(B^1,\dots,B^d)$ such that $\forall i\in\{1,\dots,d\}$, $B^i$ is a standard Brownian motion, and it is independent of the other components. + A \emph{$d$-dimensional standard Brownian motion} is a $d$-dimensional stochastic process $\vf{B}=(B^1,\dots,B^d)$ such that $\forall i\in\{1,\dots,d\}$, $B^i$ is a standard Brownian motion, and it is independent of the other components. \end{definition} \begin{theorem} Let $\vf{B}$ be a $d$-dimensional Brownian motion. Then: \begin{enumerate} - \item If $d=2$, then $B$ is recurrent, that is $\forall \vf{x}\in\RR^2$ and $\forall \delta>0$ $\exists (\tau_n)$ such that $\Prob(\tau_n)$ - \item If $d\geq 3$, then $B$ is transient, that is $\forall T>0$ + \item If $d=2$, then $B$ is recurrent, that is $\forall \vf{x}\in\RR^2$ and $\forall \delta>0$ $\exists (\tau_n)\in\RR$, with $\displaystyle\lim_{n\to\infty} \tau_n=+\infty$, such that: $$\Prob(B_{\tau_n}\in B_\delta(\vf{x})\ \forall n\in\NN)=1$$ Here $B_\delta(\vf{x})$ denotes the open ball of radius $\delta$ centered at $\vf{x}$. + \item If $d\geq 3$, then $B$ is transient, that is $\forall M>0$, $\exists T>0$ such that: $$\Prob(B_t\in B_M(\vf{0})\ \forall t\geq T)=0$$ \end{enumerate} \end{theorem} \begin{theorem}[Law of the iterated logarithm] @@ -1507,6 +1575,7 @@ $$ Then, $Y_t^n\overset{\mathrm{d}}{\longrightarrow} B_t$. \end{proposition} + \subsubsection{Existence of Brownian motion} \begin{definition}[Finite-dimensional distributions] Let $(\Omega, \mathcal{A}, \Prob)$ be a probability space and $X: I\times\Omega\rightarrow (E,\mathcal{E})$ be a stochastic process. The \emph{finite-dimensional distributions} of $X$ are the probability measures $\Prob_{t_1,\dots,t_n}$ defined on $(E^n,\mathcal{E}^n)$ by: $$ @@ -1518,13 +1587,13 @@ Let $(\Omega, \mathcal{A}, \Prob)$ be a probability space and $X: I\times\Omega\rightarrow (E,\mathcal{E})$ be a stochastic process. Then, the finite-dimensional distributions satisfy the following \emph{consistency condition}: \begin{enumerate} \item For all $n\in\NN$, $t_1,\dots,t_n\in I$, $B_1,\dots,B_n\in\mathcal{E}$ and $\sigma\in\S_n$, we have: - $$ - \Prob_{t_1,\dots,t_n}(B_1\times\dots\times B_n)=\Prob_{t_{\sigma(1)},\dots,t_{\sigma(n)}}(B_{\sigma(1)}\times\dots\times B_{\sigma(n)}) - $$ + \begin{multline*} + \Prob_{t_1,\dots,t_n}(B_1\times\dots\times B_n)=\\=\Prob_{t_{\sigma(1)},\dots,t_{\sigma(n)}}(B_{\sigma(1)}\times\dots\times B_{\sigma(n)}) + \end{multline*} \item For all $n\in\NN$, $t_1,\dots,t_n\in I$ and $B_1,\dots,B_{n-1}\in\mathcal{E}$, we have: - $$ - \Prob_{t_1,\dots,t_n}(B_1\times\dots\times B_{n-1}\times E)=\Prob_{t_1,\dots,t_{n-1}}(B_1\times\dots\times B_{n-1}) - $$ + \begin{multline*} + \Prob_{t_1,\dots,t_n}(B_1\times\dots\times B_{n-1}\times E)=\\=\Prob_{t_1,\dots,t_{n-1}}(B_1\times\dots\times B_{n-1}) + \end{multline*} \end{enumerate} \end{lemma} \begin{theorem}[Kolmogorov extension theorem]