I don't quite understand the definition of a probability kernel (or Markov kernel).

Is this correct, the reason to introduce this a transition kernel is, if we have a source $(X,\mathcal{A})$ and a target $(Y,\mathcal{B})$, both measurable spaces, we want to have a new measurable space $(X,\mathcal{B})$? There is this example on Wikipedia for a random walk:

Take $X=Y=\mathbb{Z}$ and $\mathcal A = \mathcal B = \mathcal P(\mathbb{Z})$, then the Markov kernel $\kappa$ with $$\kappa(x,B)=\frac{1}{2}\mathbf{1}_{B}(x-1)+\frac{1}{2}\mathbf{1}_{B}(x+1), \quad \forall x \in \mathbb{Z}, \quad \forall B \in \mathcal P(\mathbb{Z})$$ describes the transition rule

But I don't understand it. Why are we using here $Y$ and $\mathcal{A}$? Is the measurable space $(Y,\mathcal{B})$ the next position on the random walk with new event from $\mathcal{B}$?


2 Answers 2


For two measurable space $(E,\mathcal{E})$ and $(F,\mathcal{F})$, we call a mapping $\kappa:E\times\mathcal{F}\rightarrow \mathbb{R}$ a kernel, if $\kappa(x,.):\mathcal{F}\rightarrow\mathbb{R}$ is a measure for all $x\in E$ and $\kappa(.,B):E\rightarrow\mathbb{R}$ is $\mathcal{E}-\mathcal{B}(\mathbb{R})$-measurable for all $B\in\mathcal{F}$. It is called a Markov kernel, if in addition $\kappa(x,.):\mathcal{F}\rightarrow[0,1]$ is a probability measure.

Your example just says that if you are in state $x\in\mathbb{Z}$, then you can jump to state $x+1$ or $x-1$ and you do this with probability $1/2$ resp (since $\kappa(x,\{x+i\})=0$ for $i\in\mathbb{Z}\backslash \{1;-1\}$).

  • $\begingroup$ Hi, thank you for the answer, but this doesn't help me a lot. I know the definition but I have no intuition for it, it seems quite abstract. Is my intuition that we make a new measurable space $(E,\mathcal{F})$ wrong? $\endgroup$
    – MarcE
    Mar 20, 2017 at 11:21
  • $\begingroup$ What confuses me is that it is a map from $E \times \mathcal{F}$ but the transition says $(x,x+i)$ which seems to me like a map from $E\times F$ $\endgroup$
    – MarcE
    Mar 20, 2017 at 11:34
  • $\begingroup$ @MarcE Sorry, i forgot to add the brackets in my post. But in general, as your second argument you want to have an event, that is an element of the sigma-algebra $\mathcal{F}$. Note that in general, $(X,\mathcal{B})$ is not a measurable space, since $\mathcal{B}\subseteq \mathcal{P}(X)$ is not true in general, so $\mathcal{B}$ is not a sigma-algebra on $X$. $\endgroup$
    – peer
    Mar 20, 2017 at 11:53
  • $\begingroup$ @MarcE Answering your last question in your post, i would say that the values in Y are the possible states of your random walk in the next step. $\endgroup$
    – peer
    Mar 20, 2017 at 12:10
  • 1
    $\begingroup$ @MarcE Yes, you want for example write $\kappa(X_s,B)$ which is a mapping from $\Omega$ into $[0,1]$ for fixed $B\in\mathcal{F}$ defined by $\omega\mapsto\kappa(X_s(\omega),B)$ and $X_s(\omega)$ is in $E$ (or in your case in $\mathbb{Z}$) (here $X=(X_n)_{n\in\mathbb{N}}$ is a stochastic process; for example your random walk). Does that help you? $\endgroup$
    – peer
    Mar 20, 2017 at 16:39

To explain the wikipedia definition:

Let $(X, \mathcal{A})$ and $(Y,\mathcal{B})$ be two measurable spaces. A markov kernel with source $(X, \mathcal{A})$ and target $(Y,\mathcal{B})$ is a map $\kappa: \mathcal{B} \times X \rightarrow [0,1]$ such that

  1. for any fixed $B \in \mathcal{B}$, the map $x\mapsto \kappa(B,x)$ is $\mathcal{A}$-measurable.
  2. for any fixed $x\in X$, the map $B\mapsto \kappa(B,x)$ is a probability measure on $(Y,\mathcal{B})$.

Point 2. gives some intuition on what exactly the probability kernel is doing:

  1. tells us that $\kappa$ sends any element $x\in X$ to a probability measure on $(Y,\mathcal{B})$. Another way to put this is "for any $x \in X$, $\kappa_x$ is a measure on $(Y,\mathcal{B})$.

So let's use the notation $\kappa_x(\cdot) = \kappa(\cdot,x)$. Point 1. requires some sense of injectivity:

  1. tells us that for any fixed $B\in\mathcal{B}$, every measure $\kappa_x(\cdot)$ corresponds to some element of $\mathcal{A}$ (see definition of $\mathcal{A}$-measurable).

In your example:

$\kappa(x,B)=\frac{1}{2}\mathbf{1}_{B}(x-1)+\frac{1}{2}\mathbf{1}_{B}(x+1), \quad \forall x \in \mathbb{Z}, \quad \forall B \in \mathcal P(\mathbb{Z})$

This means, for any state $x \in \mathbb{Z}$ there is an associated probability measure $\kappa(x, \cdot)$ (i.e. $\kappa_x$) on the state space $\mathbb{Z}$, and that measure represents the probability of where the particle will go. It places $\frac12$ mass at the $x\pm 1$ and $0$ everywhere else.


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