updated montecarlo and dynamical systems

Mathematics/5th/Advanced_dynamical_systems/Advanced_dynamical_systems.tex

@@ -495,7 +495,7 @@
     $$
   \end{remark}
   \begin{definition}
-    Let $F\in \Homeo(\TT^1)$ and $X\subseteq \TT^1$ be closed and invariant. We say that $X$ is \emph{minimal} if $\forall x\in X$, $\overline{\mathcal{O}(x)}=X$. If $X=\TT^1$, we say that $F$ is \emph{minimal}.
+    Let $F\in \Homeo(\TT^1)$ and $X\subseteq \TT^1$ be nonempty closed invariant set. We say that $X$ is \emph{minimal} if $\forall x\in X$, $\overline{\mathcal{O}(x)}=X$. If $X=\TT^1$, we say that $F$ is \emph{minimal}.
   \end{definition}
   \begin{proposition}
     Let $F\in \Homeo(\TT^1)$ and $X\subseteq \TT^1$ be a closed and invariant. Then, $X$ is minimal $\iff$ $\forall Y\subseteq X$ closed, invariant and non-empty, $Y=X$.
@@ -512,5 +512,66 @@
 
     Now we do the general case. Assume $\rho(f)=\frac{p}{q}$. Then, again by \mcref{ADS:characterisation_rot_number}, we have that $\exists x\in \RR$ with $f^q(x)=x+p$. Assume we have $x'\in\RR$ and $p',q'\in\ZZ$ with $q'\geq 1$ such that $f^{q'}(x')=x'+p'$. By \mcref{ADS:characterisation_rot_number}, we have that $\frac{p}{q}=\frac{p'}{q'}$ and so $\exists k\in\NN$ such that $q'=kq$ and $p'=kp'$ because $\frac{p}{q}$ is irreducible. Now let $g=f^q-p$. Then, an easy calculation shows that $g^k(x')=x'$. But $\rho(g)=0$ and in the previous case we have seen that the periodic points are only fixed points, so $k=1$. For the second part, we proceed as in the previous case with the function $g=f^q-p$.
   \end{proof}
+  \subsubsection{Irrational rotation number}
+  \begin{definition}
+    Given $\mu\in \mathcal{M}(\TT^1)$ and $U\subseteq \TT^1$ open, we define the \emph{measure of $U$} as:
+    $$
+      \mu(U):=\sup\{\mu(\varphi):\varphi\in\mathcal{C}(\TT^1),\varphi\leq \indi{U}\}
+    $$
+    Let $A\subset \mathcal{B}(\TT^1)$ be a Borel measurable set. We define the \emph{measure of $A$} as:
+    $$
+      \mu(A):=\inf\{\mu(U):A\subseteq U, U\text{ open}\}
+    $$
+  \end{definition}
+  \begin{remark}
+    With this definition we have the usual properties of measure defined on subsets of $\TT^1$. In particular, $\text{Leb}([a,b])=b-a$ and $\delta_x(A)=\indi{x\in A}$ $\forall x\in \TT^1$ and $A\subseteq \TT^1$.
+  \end{remark}
+  \begin{definition}
+    Let $\mu\in \mathcal{M}(\TT^1)$. We define the \emph{support of $\mu$} as:
+    $$
+      \supp\mu:=\{x\in\TT^1:\forall U\subseteq \TT^1\text{ open with } x\in U,\mu(U)>0\}
+    $$
+  \end{definition}
+  \begin{remark}
+    Note that $\supp\mu $ is a closed set.
+  \end{remark}
+  \begin{remark}
+    $\supp\text{Leb}=\TT^1$ and $\supp\delta_{x}=\{x\}$.
+  \end{remark}
+  \begin{proposition}
+    Let $\mu\in \mathcal{M}(\TT^1)$ and $F\in\Homeo(\TT^1)$. $\mu$ is invariant by $F$ if and only if $\forall A\subseteq \TT^1$ Borel set, $\mu(A)=\mu(F^{-1}(A))$.
+  \end{proposition}
+  \begin{lemma}
+    Let $\mu\in \mathcal{M}(\TT^1)$. We have a lift to a measure $\mu$ on $\RR$ invariant by integer translations: $\mu(A+k)=\mu(A)$ $\forall k\in\ZZ$ and $A\subseteq \mathcal{B}(\RR)$.
+  \end{lemma}
+  \begin{definition}
+    Let $\mu\in\mathcal{M}(\TT^1)$. We define $h_\mu:[0,1]\to [0,1]$ as the function with $h_\mu(0)=0$ and $h_\mu(x)=\mu([0,x))$ for $0<x\leq 1$. This definition extends to a non-decreasing function $h_\mu:\RR\to\RR$ such that $h_\mu(x+k)=h_\mu(x)+k$ $\forall k\in\ZZ$
+  \end{definition}
+  \begin{lemma}
+    Let $\mu\in\mathcal{M}(\TT^1)$. We say that $\mu$ has atoms if $\exists x\in\TT^1$ such that $\mu(\{x\})>0$
+  \end{lemma}
+  \begin{lemma}\label{ADS:lemma_atom}
+    Let $\mu\in\mathcal{M}(\TT^1)$. $h_\mu$ is continuous if and only if $\forall x\in\RR$, $\mu(\{x\})=0$, that is $\mu$ has no atoms.
+  \end{lemma}
+  \begin{definition}
+    A subset $C\subseteq \RR$ is a \emph{Cantor set} if it is closed, it has no isolated points and it has empty interior.
+  \end{definition}
+  \begin{theorem}
+    Let $F\in\Homeo(\TT^1)$ with $\rho(F)\notin\quot{\QQ}{\ZZ}$. Then, there exists a surjective continuous map $H:\TT^1\to \TT^1$ such that $H\circ F=R_{\rho(F)}\circ H$. Moreover, we have exactly one of the following two properties:
+    \begin{enumerate}
+      \item $F$ is conjugated to $R_{\rho(F)}$ and in that case $F$ is minimal.
+      \item $\exists X\subsetneq \TT^1$ minimal which is a Cantor set and $X=\Omega(F)$.
+    \end{enumerate}
+  \end{theorem}
+  \begin{proof}
+    Let $\mu\in\mathcal{M}_F(\TT^1)$ and consider $h_\mu:\RR\to\RR$ as defined above. Now assume $x\in \TT^1$ is such that $\mu(\{x\})=c>0$, then by invariance $\mu(A_n)=c>0$, where $A_n:=\{F^n(x)\}$. Note that since $\mu\leq 1$, $(A_n)$ cannot be disjoint. So $\exists n,m\in\NN$ with $n<m$ such that $F^n(x)=F^m(x)$. But then $F^{m-n}(x)=x$ and so $x$ is periodic, which is not possible since $\rho(F)\in\quot{\QQ}{\ZZ}$ by \mcref{ADS:characterisation_rot_number}. Thus, $\mu$ has no atoms and so $h_\mu$ is continuous by \mcref{ADS:lemma_atom}. Now, define $H:\TT^1\to \TT^1$ as the projection of $h_\mu$ to $\TT^1$, which is continuous and surjective. Let $f\in\mathcal{D}^0(\TT^1)$ be a lift of $F$. Then:
+    $$
+      h(f(x))-h(f(0))=\mu([f(0),f(x)))=\mu([0,x))=h(x)
+    $$
+    where we have used the invariance of $\mu$. Thus, $h\circ f=R_{h(f(0))}\circ h$ and necessarily we need $h(f(0))=\rho(R_{h(f(0))})=\rho(f)$, by the invariance of the rotation number by semi-conjugacy. This gives $H\circ F=R_{\rho(F)}\circ H$. Now, we can express the dichotomy as follows: either $\supp\mu=\TT^1$ or $\supp\mu=:X\subsetneq \TT^1$. The first case is equivalent to $h$ being strictly increasing and so $h$ is a homeomorphism. Then, $H$ conjugates $F$ and $R_{\rho(F)}$ and so $F$ is minimal because $R_{\rho(F)}$ is minimal. In the second case, we have that $X$ is a nonempty closed invariant set that has no isolated points because $\mu$ has no atoms. To show that $X$ is minimal, let $\TT^1=X\sqcup U$ with $U$ open, and so it can be written as a countable union of open intervals. Let $D\subseteq X$ be the set containing the endpoints of those intervals and let $$
+      Y=\{y\in\TT^1:H^{-1}(\{y\})\text{ is a closed interval}\}
+    $$
+    $Y$ is countable. Now take $M\subseteq X$ be nonempty, closed and invariant. We want to prove that $M=X$. Then, $H(M)\subseteq \TT^1$ is nonempty, closed and invariant by $R_{\rho(F)}$. But $R_{\rho(F)}$ is minimal, so $H(M)=\TT^1$.
+  \end{proof}
 \end{multicols}
 \end{document}

Mathematics/5th/Montecarlo_methods/Montecarlo_methods.tex

@@ -459,5 +459,75 @@
     $$
   \end{lemma}
   \subsection{Computation of sensitivities}
+  In this chapter we aim to construct Montecarlo methods in order to compute sensitivities of the price of an option. From the PDE point of view, we aim to compute the derivatives of the solution ${(\vf{X}_t^{\vf{x}})}_{t\geq 0}$ of the SDE:
+  $$
+    \begin{cases}
+      \dd{\vf{X}_t}=\vf{b}(\vf{X}_t)\dd{t}+\vf{\sigma}(\vf{X}_t)\dd{\vf{B}_t} \\
+      \vf{X}_0=\vf{x}
+    \end{cases}
+  $$
+  \subsubsection{Finite difference method}
+  Here we will focus on the case $d=1$.
+  \begin{definition}
+    Let $u(0,x):=\Exp(g(X_T^x))$ and $\tilde{u}^n(0,x):=\frac{1}{n}\sum_{i=1}^n g(\tilde{X}_T^{x,(i)})$ where $\tilde{X}_T^{x,(i)}$ are \iid copies of $\tilde{X}_T^x$. We define the \emph{finite difference estimator} as:
+    \begin{align*}
+      \partial_xu(0,x) & \approx \frac{u(0,x+\varepsilon) - u(0,x-\varepsilon)}{2\varepsilon}                     \\
+                       & \approx \frac{\tilde{u}^n(0,x+\varepsilon) - \tilde{u}^n(0,x-\varepsilon)}{2\varepsilon}
+    \end{align*}
+  \end{definition}
+  \begin{proposition}
+    If $g$ is smooth enough, then:
+    $$
+      \Var\left( \frac{\tilde{u}^n(0,x+\varepsilon) - \tilde{u}^n(0,x-\varepsilon)}{2\varepsilon}\right)\approx \frac{1}{n}\Var(g'(X_T^x))
+    $$
+  \end{proposition}
+  \begin{remark}
+    When $g$ is irregular, the optimal choice of $\varepsilon$ is not clear, as the bias increases with $\varepsilon$ and the variance increases as $\varepsilon$ decreases.
+  \end{remark}
+  \subsubsection{Black-Scholes model}
+  \begin{proposition}
+    Recall the one-dimensional \emph{Black-Scholes model}:
+    $$
+      \dd{X}_t=r X_t\dd{t}+\sigma X_t\dd{B}_t
+    $$
+    Then: $$
+      \partial_x \Exp(g(X_T^x))=\frac{1}{\sigma x T}\Exp(g(X_T^x)B_T)
+    $$
+  \end{proposition}
+  \subsubsection{Pathwise differentiation}
+  \begin{theorem}
+    If $\vf{b},\vf\sigma\in \mathcal{C}^2$ with bounded derivatives, then the flow ${\vf{x}}\mapsto \vf{X}_t^{\vf{x}}$ is $\mathcal{C}^1$ a.s.\ and the \emph{tangent process} $\vf{DX}^{\vf{x}}$ satisfies:
+    $$
+      \vf{DX}^{\vf{x}}_t=\vf{I}_d+\!\int_0^t\! \vf{Db}(\vf{X}_s^{\vf{x}})\vf{DX}_s^{\vf{x}}\dd{s}+\sum_{j=1}^{d}\int_0^t\! \vf{D\sigma}_j(\vf{X}_s^{\vf{x}})\vf{DX}_s^{\vf{x}}\dd{\vf{B}_s^j}
+    $$
+    where $\vf{D\sigma}_j$ is the $j$-th column of $\vf{D\sigma}$.
+  \end{theorem}
+  \begin{remark}
+    In one dimension, we have:
+    $$
+      \partial_xX_t^x=1+\int_0^t b'(X_s^x)\grad X_s^x\dd{s}+\int_0^t \sigma'(X_s^x)\grad X_s^x\dd{B}_s
+    $$
+    which yields:
+    $$
+      \partial_xX_t^x\!=\!\expp{\int_0^t\!\left[b'(X_s^x)-\frac{1}{2}{\sigma'(X_s^x)}^2\right]\dd{s} \!+\!\! \int_0^t \sigma'(X_s^x)\dd{B}_s\!}
+    $$
+  \end{remark}
+  \begin{proposition}
+    If $\vf{b},\vf\sigma\in \mathcal{C}^2$ and $g\in\mathcal{C}^1$ with bounded derivatives, then:
+    $$
+      \grad \Exp(g(\vf{X}_T^{\vf{x}}))=\Exp\left(\grad g(\vf{X}_T^{\vf{x}})\grad \vf{X}_T^{\vf{x}}\right)
+    $$
+  \end{proposition}
+  \begin{remark}
+    In practice, to find the derivative of $\vf{X}_T^{\vf{x}}$ we proceed as in the deterministic case, i.e.\ solving the coupled variational equations.
+  \end{remark}
+  \subsubsection{Malliavin differentiation}
+  \begin{proposition}
+    If $\vf{b},\vf\sigma\in \mathcal{C}^2$ with bounded derivatives, $\vf\sigma$ is uniformly elliptic and $g$ is measurable with polynomial growth, then:
+    $$
+      \grad \Exp(g(\vf{X}_T^{\vf{x}}))=\Exp\left[g(\vf{X}_T^{\vf{x}})\frac{1}{T}\transpose{\left(\int_0^T\transpose{(\vf\sigma^{-1}(X_t^{\vf{x}})\vf{DX}_t^{\vf{x}})}\!\dd{\vf{B}_t}\!\right)\!\!}\right]
+    $$
+  \end{proposition}
+  \subsection{American options}
 \end{multicols}
 \end{document}

preamble_formulas.sty

@@ -252,6 +252,7 @@
 } % partial differential operator
 \newcommand{\ii}{\mathrm{i}} % imaginary unit
 \renewcommand{\exp}[1]{\mathrm{e}^{#1}} % exponential function
+\newcommand{\expp}[1]{\mathrm{exp}\!\left(#1\right)} % exponential function 2.0
 \newcommand{\upint}[2]{\overline{\int_{#1}^{#2}}} % upper integral
 \newcommand{\lowint}[2]{\underline{\int_{#1}^{#2}}} % lower integral
 \DeclareMathOperator{\domain}{dom} % domain of a function