Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I read a theorem, stating

Let $X_t$ be a Markov process w.r.t. to its natural filtration $(\mathcal F_t)$ on the space of cadlag functions on $\mathbb R_{\geq 0}$ and $(Z_t)_{t\geq 0}$ be a non-negative martingale s.t. $\mathsf E_xZ_t = 1$ for all $x,t$. Let $ \mathsf{\tilde{P}}$ be such that $$ \mathsf{\tilde{P}}_x(A):=\mathsf E_x[1_A\cdot Z_t] $$ for any $A\in \mathcal F_t$, then the family $(\mathsf{\tilde{P}}_x)$ defines a time-homogeneous Markov process iff $Z_t$ is multiplicative functional, i.e. $$ Z_{t+s} = Z_t\cdot(Z_s\circ \theta_t) $$ where $\theta_t$ is the shift operator.

I wonder which form can $Z$ take for a discrete time Markov process. The motivation is that for random walks $Y$ there is closed form for $Z$ given by $$ Z_n = h(Y_1)\cdot h(Y_2)\cdot\dots\cdot h(Y_n) $$ where $\mathsf E h(Y_1) = 1$.

share|cite|improve this question

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.