Let $(S,\mathcal{S})$ and $(T,\mathcal{T})$ be measurable spaces. A transition probability from $S$ to $T$ is a function $\pi:S\times\mathcal{T}\to [0,1]$ such that $\pi(s,\cdot)$ is a probability measure for all $s\in S$ and $\pi(\cdot,B)$ is measurable for all $B\in\mathcal{T}$.

Now let $(\Omega,\Sigma,\mu)$ be a probability space and let $f:\Omega\times S\to T$ be a jointly measurable function.

Under what conditions is $\pi_f:S\times\mathcal{T}\to [0,1]$ given by $\pi_f(s,B)=\mu > \{\omega:f(s,\omega)\in B\}$ a transition probability?

Clearly, the issue is measurability.

Motivation: If $(T,\mathcal{T})$ is standard Borel, there always exists a function such that $\pi=\pi_f$ when $\Omega$ is atomless, this is Proposition 10.7.6. in Bogachev. Moreover, such random functions have been used as a definition of mixed strategies in game theory and I think it would be interesting to understand them in terms of transition probabilities in a general setting.


1 Answer 1


I've found a rather simple solution. No additional assumptions are necessary. There is a natural method of composing transition probabilities that gives rise to a new transition probability (see for example (2) here).

We can identify probability measures with transition probabilities constant in the first argument. Also, we can identify $f$ with the transition probability $\kappa:\Omega\times S\times\mathcal{T}\to[0,1]$ given by $\kappa(\omega,s,B)=1$ if $f(\omega,s)\in B$ and $\kappa(\omega,s,B)=0$ if $f(\omega,s)\notin B$.

Also, the pointwise product of transition probabilities is a transition probability. So let $\kappa_\mu:S\times\Sigma\to[0,1]$ the representation of $\mu$ as a tranition probability. Also, let $\kappa_1:S\times\mathcal{S}\to[0,1]$ be the identity transition probability on $S$. Let $\pi:S\times\mathcal{S}\otimes\mathcal{T}\to[0,1]$ be the pointwise product of $\kappa_\mu$ and $\kappa_1$.

Then $\pi_f=\kappa\circ\pi$ and is therefore a transition probability as the composition of transition probabilities.


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