7
$\begingroup$

Supposing $X_t$ is a Markov Process, can the transition kernel be defined by $K_t(x,A):= P(X_{t+1} \in A | X_t = x)?$ Assume that $X_t : \Omega \to \mathbb{R}^n$.

The issue is that under the normal definition of conditional probability, r.h.s is defined as $P(X_{t+1} \in A | X_t = x) =\frac{P( (X_{t+1} \in A) \cap (X_t = x))}{P(X_t = x)}$ and the denominator is zero for most random variables. Even if this is assumed to be $E[I_A(X_{t+1}) | \sigma(X_t = x)]$, the conditional expectation can take arbitrary values on the set $\{X_t =x\}$ if $P(X_t = x) =0$.

Another definition I could gather from the web is that $K_t(x,A)$ is called a transition kernel if $K_t(X_t(\omega),A) := E[I_A(X_{t+1})|X_t](\omega)~\forall \omega \in \Omega.$ Also, $K_t(x,\cdot)$ should be a probability measure so that the conditional expectations should be regular (if I am not wrong).

The book (page 18) I am reading uses the first definition given above.

Thanks for the help.

  • 0
    @did: Thanks for your answer! I was thinking about it but havent logged in for a long time.2012-08-27

1 Answers 1

6

The transition kernel $K_t$ is defined by some measurability conditions and by the fact that, for every measurable Borel set $A$ and every (bounded) measurable function $u$, $ \mathrm E(u(X_t):X_{t+1}\in A)=\mathrm E(u(X_t)K_t(X_t,A)). $ Hence, each $K_t(\cdot,A)$ is defined only up to sets of measure zero for the distribution of $X_t$, in the following sense: if $K_t$ is a transition kernel for $X_t$ and if, for every measurable Borel set $A$, $X_t$ is almost surely in $C_A$, where $ C_A=\{x\in\mathbb R^n\,\mid\,K_t(x,A)=\tilde K_t(x,A)\}, $ then $\tilde K_t$ is also a transition kernel for $X_t$.