From c0d352dd0f66b47ee91fb96eaf320f895fa78790 Mon Sep 17 00:00:00 2001 From: Prefetch Date: Sun, 14 Nov 2021 17:54:04 +0100 Subject: Expand knowledge base --- .../know/concept/kolmogorov-equations/index.pdc | 209 +++++++++++++++++++++ 1 file changed, 209 insertions(+) create mode 100644 content/know/concept/kolmogorov-equations/index.pdc (limited to 'content/know/concept/kolmogorov-equations/index.pdc') diff --git a/content/know/concept/kolmogorov-equations/index.pdc b/content/know/concept/kolmogorov-equations/index.pdc new file mode 100644 index 0000000..331d803 --- /dev/null +++ b/content/know/concept/kolmogorov-equations/index.pdc @@ -0,0 +1,209 @@ +--- +title: "Kolmogorov equations" +firstLetter: "K" +publishDate: 2021-11-14 +categories: +- Mathematics +- Statistics + +date: 2021-11-13T21:05:30+01:00 +draft: false +markup: pandoc +--- + +# Kolmogorov equations + +Consider the following general [Itō diffusion](/know/concept/ito-calculus/) +$X_t \in \mathbb{R}$, which is assumed to satisfy +the conditions for unique existence on the entire time axis: + +$$\begin{aligned} + \dd{X}_t + = f(X_t, t) \dd{t} + g(X_t, t) \dd{B_t} +\end{aligned}$$ + +Let $\mathcal{F}_t$ be the filtration to which $X_t$ is adapted, +then we define $Y_s$ as shown below, +namely as the [conditional expectation](/know/concept/conditional-expectation/) +of $h(X_t)$, for an arbitrary bounded function $h(x)$, +given the information $\mathcal{F}_s$ available at time $s \le t$. +Because $X_t$ is a [Markov process](/know/concept/markov-process/), +$Y_s$ must be $X_s$-measurable, +so it is a function $k$ of $X_s$ and $s$: + +$$\begin{aligned} + Y_s + \equiv \mathbf{E}[h(X_t) | \mathcal{F}_s] + = \mathbf{E}[h(X_t) | X_s] + = k(X_s, s) +\end{aligned}$$ + +Consequently, we can apply Itō's lemma to find $\dd{Y_s}$ +in terms of $k$, $f$ and $g$: + +$$\begin{aligned} + \dd{Y_s} + &= \bigg( \pdv{k}{s} + \pdv{k}{x} f + \frac{1}{2} \pdv[2]{k}{x} g^2 \bigg) \dd{s} + \pdv{k}{x} g \dd{B_s} + \\ + &= \bigg( \pdv{k}{s} + \hat{L} k \bigg) \dd{s} + \pdv{k}{x} g \dd{B_s} +\end{aligned}$$ + +Where we have defined the linear operator $\hat{L}$ +to have the following action on $k$: + +$$\begin{aligned} + \hat{L} k + \equiv \pdv{k}{x} f + \frac{1}{2} \pdv[2]{k}{x} g^2 +\end{aligned}$$ + +At this point, we need to realize that $Y_s$ is +a [martingale](/know/concept/martingale/) with respect to $\mathcal{F}_s$, +since $Y_s$ is $\mathcal{F}_s$-adapted and finite, +and it satisfies the martingale property, +for $r \le s \le t$: + +$$\begin{aligned} + \mathbf{E}[Y_s | \mathcal{F}_r] + = \mathbf{E}\Big[ \mathbf{E}[h(X_t) | \mathcal{F}_s] \Big| \mathcal{F}_r \Big] + = \mathbf{E}\big[ h(X_t) \big| \mathcal{F}_r \big] + = Y_r +\end{aligned}$$ + +Where we used the tower property of conditional expectations, +because $\mathcal{F}_r \subset \mathcal{F}_s$. +However, an Itō diffusion can only be a martingale +if its drift term (the one containing $\dd{s}$) vanishes, +so, looking at $\dd{Y_s}$, we must demand that: + +$$\begin{aligned} + \pdv{k}{s} + \hat{L} k + = 0 +\end{aligned}$$ + +Because $k(X_s, s)$ is a Markov process, +we can write it with a transition density $p(s, X_s; t, X_t)$, +where in this case $s$ and $X_s$ are given initial conditions, +$t$ is a parameter, and the terminal state $X_t$ is a random variable. +We thus have: + +$$\begin{aligned} + k(x, s) + = \int_{-\infty}^\infty p(s, x; t, y) \: h(y) \dd{y} +\end{aligned}$$ + +We insert this into the equation that we just derived for $k$, yielding: + +$$\begin{aligned} + 0 + = \int_{-\infty}^\infty \!\! \Big( \pdv{s} p(s, x; t, y) + \hat{L} p(s, x; t, y) \Big) h(y) \dd{y} +\end{aligned}$$ + +Because $h$ is arbitrary, and this must be satisfied for all $h$, +the transition density $p$ fulfills: + +$$\begin{aligned} + 0 + = \pdv{s} p(s, x; t, y) + \hat{L} p(s, x; t, y) +\end{aligned}$$ + +Here, $t$ is a known parameter and $y$ is a "known" integration variable, +leaving only $s$ and $x$ as free variables for us to choose. +We therefore define the **likelihood function** $\psi(s, x)$, +which gives the likelihood of an initial condition $(s, x)$ +given that the terminal condition is $(t, y)$: + +$$\begin{aligned} + \boxed{ + \psi(s, x) + \equiv p(s, x; t, y) + } +\end{aligned}$$ + +And from the above derivation, +we conclude that $\psi$ satisfies the following PDE, +known as the **backward Kolmogorov equation**: + +$$\begin{aligned} + \boxed{ + - \pdv{\psi}{s} + = \hat{L} \psi + = f \pdv{\psi}{x} + \frac{1}{2} g^2 \pdv[2]{\psi}{x} + } +\end{aligned}$$ + +Moving on, we can define the traditional +**probability density function** $\phi(t, y)$ from the transition density $p$, +by fixing the initial $(s, x)$ +and leaving the terminal $(t, y)$ free: + +$$\begin{aligned} + \boxed{ + \phi(t, y) + \equiv p(s, x; t, y) + } +\end{aligned}$$ + +With this in mind, for $(s, x) = (0, X_0)$, +the unconditional expectation $\mathbf{E}[Y_t]$ +(i.e. the conditional expectation without information) +will be constant in time, because $Y_t$ is a martingale: + +$$\begin{aligned} + \mathbf{E}[Y_t] + = \mathbf{E}[k(X_t, t)] + = \int_{-\infty}^\infty k(y, t) \: \phi(t, y) \dd{y} + = \braket{k}{\phi} + = \mathrm{const} +\end{aligned}$$ + +This integral has the form of an inner product, +so we switch to [Dirac notation](/know/concept/dirac-notation/). +We differentiate with respect to $t$, +and use the backward equation $\pdv*{k}{t} + \hat{L} k = 0$: + +$$\begin{aligned} + 0 + = \pdv{t} \braket{k}{\phi} + = \braket{k}{\pdv{\phi}{t}} + \braket{\pdv{k}{t}}{\phi} + = \braket{k}{\pdv{\phi}{t}} - \braket{\hat{L} k}{\phi} + = \braket{k}{\pdv{\phi}{t} - \hat{L}{}^\dagger \phi} +\end{aligned}$$ + +Where $\hat{L}{}^\dagger$ is by definition the adjoint operator of $\hat{L}$, +which we calculate using partial integration, +where all boundary terms vanish thanks to the *existence* of $X_t$; +in other words, $X_t$ cannot reach infinity at any finite $t$, +so the integrand must decay to zero for $|y| \to \infty$: + +$$\begin{aligned} + \braket{\hat{L} k}{\phi} + &= \int_{-\infty}^\infty \pdv{k}{y} f \phi + \frac{1}{2} \pdv[2]{k}{y} g^2 \phi \dd{y} + \\ + &= \bigg[ k f \phi + \frac{1}{2} \pdv{k}{y} g^2 \phi \bigg]_{-\infty}^\infty + - \int_{-\infty}^\infty k \pdv{y}(f \phi) + \frac{1}{2} \pdv{k}{y} \pdv{y}(g^2 \phi) \dd{y} + \\ + &= \bigg[ -\frac{1}{2} k g^2 \phi \bigg]_{-\infty}^\infty + + \int_{-\infty}^\infty - k \pdv{y}(f \phi) + \frac{1}{2} k \pdv[2]{y}(g^2 \phi) \dd{y} + \\ + &= \int_{-\infty}^\infty k \: \big( \hat{L}{}^\dagger \phi \big) \dd{y} + = \braket{k}{\hat{L}{}^\dagger \phi} +\end{aligned}$$ + +Since $k$ is arbitrary, and $\pdv*{\braket{k}{\phi}}{t} = 0$ for all $k$, +we thus arrive at the **forward Kolmogorov equation**, +describing the evolution of the probability density $\phi(t, y)$: + +$$\begin{aligned} + \boxed{ + \pdv{\phi}{t} + = \hat{L}{}^\dagger \phi + = - \pdv{y}(f \phi) + \frac{1}{2} \pdv[2]{y}(g^2 \phi) + } +\end{aligned}$$ + + + +## References +1. U.H. Thygesen, + *Lecture notes on diffusions and stochastic differential equations*, + 2021, Polyteknisk Kompendie. -- cgit v1.2.3