April 25, 2024
\pi\colon M(X)\to M(X), positive, isometric, w-star continuous.
A measure \mu on X\times Y determines a linear transformaton M\colon B(X)\to B(Y) by Mf(y) = \int_X f\,d\mu_y where \mu_Y(E) = \mu(X,E)
Mf(y) = sum_x f(x) m_{x,y}
B(X\times Y)^* = M(X\times Y)
C_y\colon B(X\times Y) \to B(X) by f\mapsto (x \to f(x,y)).
C_y^*\colon ba(X)\to ba(X\times Y). $(X) =
A Markov chain on a set X is a positive measure \pi on X\times X where \pi_x(E) = \pi(X,E), E\subseteq X, is a probability measure for all x\in X.
Define \Pi\colon B(X)\to B(X) by \Pi(f)(x) = \int_X f\,d\pi_x.
A stochastic process (X_t)_{t\in T} satisfies the Markov property if given t_0 < \cdots < t_n, t_j\in T then P(X_t \le x \mid X_{t_0} = x_0, \ldots, X_{t_n} = x_n) = P(X_t \le x \mid X_{t_n} = x_n) whenever t \ge t_n. Informally, the process forgets its history.
Exercise. If (X_t) has independent increments then it satisfies the Markov property.
If the sample space \Omega = \{\omega_j\} is finite and T = \{t_0,t_1,\ldots\} is discrete then …
Let M\colon ba(\Omega)\to ba(\Omega) be a positive unitary linear transformation.
M\delta_\omega = \sum_{\omega'} \pi_{\omega,\omega'} \delta_{\omega'}. \pi_{\omega,\omega'} \ge 0, \sum_{\omega'} \pi_{\omega,\omega'} = 1, \omega\in\Omega.
\mu = \sum \mu_\omega\delta_\omega.
M\mu = M(\sum \mu_\omega\delta_\omega) = \sum \mu_\omega \sum_{\omega'} \pi_{\omega,\omega'} \delta_{\omega'}.
(M\mu)_{\omega'} = \sum \mu_\omega \sum_{\omega'} \pi_{\omega,\omega'}.
M_\star\colon B(\Omega)\to B(\Omega).
\begin{aligned} \langle M_\star X,\mu\rangle &= \langle X, M\mu\rangle \\ &= \langle X, M\sum_{\omega\in\Omega}\mu_\omega\delta_\omega\rangle \\ &= \langle X, \sum_{\omega\in\Omega}\mu_\omega M\delta_\omega\rangle \\ &= \langle X, \sum_{\omega}\mu_\omega \sum_{\omega'} \pi_{\omega,\omega'} \delta_{\omega'}\rangle \\ &= \langle X, \sum_{\omega'}\sum_{\omega}\mu_\omega \pi_{\omega,\omega'} \delta_{\omega'}\rangle \\ &= \langle \sum_{\omega'} X(\omega'), \sum_{\omega}\mu_\omega \pi_{\omega,\omega'} \rangle \\ \end{aligned}