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Given measurable spaces $(\Omega_k,\mathcal {F_k })_{k=1 } ^n$, $P_1 $ a probability measure on $\Omega_1 $, $P_2 $ a transition kernel from $\mathcal {F_1}$ to $\mathcal {F_2 }$,$P_3$ a transition kernel from $\mathcal {F_1}\otimes \mathcal {F_2} $ to $\mathcal {F_3}$ ...$P_n $ a transition kernel from $\mathcal {F_1}\otimes \ldots\otimes \mathcal {F_{n-1 }} $ to $\mathcal {F_n } \ $. Here $ \ P_m $ is a transition kernel from $\mathcal {F_1}\otimes \ldots \otimes \mathcal {F_{m-1 }} $ to $\mathcal F_m $ means that for every $B \in \mathcal {F_m } , \ P_m( \ , B) $ is a measurable function from $\Omega_1 \times \ldots \times \Omega_{m-1 } $ to $[0,1]$ and for every $(\omega_1,\ldots,\omega_{m-1 } ) \in \Omega_1 \times \ldots \times \Omega_{m-1 } $ a probability measure on $\mathcal {F_m} $.

I would like to show that $$P(B)=\int P_1(d\omega_1) \int P_2(\omega_1,d \omega_2) \ldots \int 1_B(\omega_1,\ldots,\omega_n)P_n(\omega_1,\ldots,\omega_{n-1 } , d \omega_n)$$ defines a measure on $\Omega_1 \times \ldots\times \Omega_n $

This should be done by an argument using induction. That is one shows that $P$ is a countably additive set function on the set of measurable rectangles $B_0\times \ldots\times B_n$ (and then extend this set function by the usual arguments). To show countable additivity of $P $ on measurable rectangles we derive this for $n=1 $ and then extended the result to an arbitrary number $n$ with an argument using induction. And this inductive step is what I need help with.

For the base case $n=1 $ one do as follows:

First one shows that $\omega_1 \mapsto \int 1_B(\omega_1,\omega_2)P_2(\omega_1,d \omega _2)$ is $\mathcal F_1 $ measurable and this is done exactly as usually done in a first step in Fubini's theorem. Then noting that $$B \mapsto \int P_1(d \omega_1)\int 1_B(\omega_1,\omega_2)P_2(\omega_1,d \omega _2)$$ is nonnegative and monotone all we need to verify is the $\sigma $-additivity. This follows from the monotone convergence theorem:

\begin{align*}\int P_1(d \omega_1)\int 1_{\cup_{k=1}^{\infty }F_k} (\omega_1,\omega_2)P_2(\omega_1,d \omega _2) &=\int P_1(d \omega_1)\sum_{k=1 }^ {\infty } \int 1_{F_k} (\omega_1,\omega_2)P_2(\omega_1,d \omega _2) \\ &=\sum_{k=1 }^ {\infty }\int P_1(d \omega_1) \int 1_{F_k} (\omega_1,\omega_2)P_2(\omega_1,d \omega _2). \end{align*}

And the inductive step should be something like: assume that $$Q(B)=\int P_1(d\omega_1) \int P_2(\omega_1,d \omega_2)\ldots\int 1_B(\omega_1,\ldots,\omega_{n-1 } )P_{n-2 } (\omega_1,\ldots,\omega_{n-2 } , d \omega_{n-1 } )$$ defines a measure on $\Omega_1 \times \ldots\times \Omega_{n-1 }$ then

\begin{align*} & \int P_1(d\omega_1) \int P_2(\omega_1,d \omega_2)\ldots\int 1_{B_0 \times \ldots \times B_n}(\omega_1,\ldots,\omega_n)P_n(\omega_1,\ldots,\omega_{n-1 } , d \omega_n) \\ &=\int 1_{B_0 \times \ldots \times B_{n-1}} \left(\int 1_{B_n } P_n(\omega_1,\ldots,\omega_{n-1 } , d \omega_n) \right) Q(d \omega_1,\ldots,d \omega_{n-1 } ) \end{align*} And we would then be in a similar situation as in the base case.

My question is: If this is correct, how is the equality motivated? Or if not, how should the inductive step be?

Thanks in advance!

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Let's start with a technical lemma which we will use several times:

Lemma. Let $(\Omega_i,\mathcal{F}_i)$, $i=1,2$,be measurable spaces, and let $P$ a transition kernel from $\mathcal{F}_1$ to $\mathcal{F}_2$. If $$X:(\Omega_1 \times \Omega_2, \mathcal{F}_1 \otimes \mathcal{F}_2) \to (\mathbb{R},\mathcal{B}(\mathbb{R}))$$ is a bounded measurable mapping then $$(\Omega_1,\mathcal{F}_1) \ni \omega_1 \mapsto \int_{\Omega_2} X(\omega_1,\omega_2) \, P(\omega_1,d\omega_2) \in (\mathbb{R},\mathcal{B}(\mathbb{R}))$$ is measurable.

Idea of proof: For $X=1_A 1_B$ with $A \in \mathcal{F}_1$ and $B \in \mathcal{F}_2$ the assertion is immediate and then we can extend it using a classical monotone class argument.


For measurable spaces $(\Omega_i,\mathcal{F}_i)$, $i \in \{0,\ldots,n\}$, and transition kernels $P_i$ from $\mathcal{F}_0 \otimes \ldots \otimes \mathcal{F}_{i-1}$ to $\mathcal{F}_i$ set

$$Q_n(\omega_0,B) := \int P_1(\omega_0,d\omega_1) \int P_2(\omega_0,\omega_1,d\omega_2) \ldots \int 1_B(\omega_1,\ldots,\omega_n) \, P_n(\omega_0,\ldots,\omega_{n-1},d\omega_n)$$

Claim: For each $n \in \mathbb{N}$ it holds for any choice of measurable spaces and transition kernels that $Q_n$ is a transition kernel from $\mathcal{F}_0$ to $\mathcal{F}_1 \otimes \ldots \otimes \mathcal{F}_n$.

Base: Since $Q_1(\omega_0,B) = P_1(\omega_0,B)$, the assertion is obvious for $n=1$.

Inductive step: Assume that the assertion holds for some $n \in \mathbb{N}$. Let $(\Omega_k,\mathcal{F}_k)$, $k \leq n+1$, be measurable spaces, and $P_i$ transition kernels from $\mathcal{F}_0 \otimes \ldots \otimes \mathcal{F}_{i-1}$ to $\mathcal{F}_i$. For fixed $(\omega_0,\omega_1) \in \Omega_0 \times \Omega_1$ consider

\begin{align*} &U((\omega_0,\omega_1),A) \\ &:= \int P_2(\omega_0,\omega_1,d\omega_2) \int P_3(\omega_0,\omega_1,\omega_2,d\omega_3) \ldots \int 1_{A}(\omega_2,\ldots,\omega_{n+1}) \, P_{n+1}(\omega_0,\ldots,\omega_n,d\omega_{n+1}) \end{align*}

for $A \in \mathcal{F}_2 \otimes \ldots \otimes \mathcal{F}_{n+1}$. Since $P_i(\omega_0,\omega_1,\cdot)$, $i \in \{2,\ldots,n+1\}$ are transition kernels from $\mathcal{F}_2 \otimes \ldots \otimes \mathcal{F}_{i-1}$ to $\mathcal{F}_i$, it follows from our induction hypothesis that $U$ is a transition kernel from $\mathcal{F}_0 \otimes \mathcal{F}_1$ to $\mathcal{F}_2 \otimes \ldots \times \mathcal{F}_{n+1}$. By the above lemma, this implies that $$(\Omega_0 \times \Omega_1, \mathcal{F}_0 \otimes \mathcal{F}_1) \ni (\omega_0,\omega_1) \mapsto \int_{\Omega_2 \times \ldots \times \Omega_{n+1}} 1_A(\omega_0,\omega_1,\tilde{\omega}) \, U((\omega_0,\omega_1),d\tilde{\omega}) \tag{1}$$ is measurable for any $A \in \mathcal{F}_0 \otimes \ldots \otimes \mathcal{F}_{n+1}$. In particular, $Q_{n+1}(\omega_0,B)$ is well defined and

\begin{align*} Q_{n+1}(\omega_0,B) &= \int_{\Omega_1} \int_{\Omega_2 \times \ldots \times \Omega_{n+1}} 1_B(\omega_1,\tilde{\omega}) \, U((\omega_0,\omega_1),d\tilde{\omega}) \, P_1(\omega_0,d\omega_1) \tag{2} \end{align*}

for any $B \in \mathcal{F}_1 \otimes \ldots \otimes \mathcal{F}_{n+1}$ and $\omega_0 \in \Omega_0$.

Now let's show that $Q_{n+1}$ has all the desired properties. Let $(B_j)_{j \in \mathbb{N}} \subseteq \mathcal{F}_1 \otimes \ldots \otimes \mathcal{F}_{n+1}$ be pairwise disjoint sets and set $B:= \bigcup_{j \geq 1} B_j$. Since $$1_B = \sum_{j \geq 1} 1_{B_j}$$ it follows from the monotone convergence theorem that

$$\int_{\Omega_2 \times \ldots \times \Omega_n} 1_B(\omega_1,\tilde{\omega}) \, U((\omega_0,\omega_1),d\tilde{\omega}) = \sum_{j \geq 1} \int_{\Omega_2 \times \ldots \times \Omega_{n+1}} 1_{B_j}(\omega_1,\tilde{\omega}) \, U((\omega_0,\omega_1),d\tilde{\omega})$$

for any $(\omega_0,\omega_1) \in \Omega_0 \times \Omega_1$; note that we can apply the monotone convergence theorem because we have already shown that $U((\omega_0,\omega_1),\cdot)$ is a measure. Integrating both sides with respect to $P_1$ we find from $(2)$ that

$$Q_{n+1}(\omega_0,B) = \int_{\Omega_1} \sum_{j \geq 1} \left( \int_{\Omega_2 \times \ldots \times \Omega_{n+1}} 1_{B_j}(\omega_1,\tilde{\omega}) \, U((\omega_0,\omega_1),d\tilde{\omega}) \right) P(\omega_0,d\omega_1).$$

Applying once more the monotone convergence theorem we get

\begin{align*} Q_{n+1}(\omega_0,B) & = \sum_{j \geq 1} \int_{\Omega_1}\left( \int_{\Omega_2 \times \ldots \times \Omega_{n+1}} 1_{B_j}(\omega_1,\tilde{\omega}) \, U((\omega_0,\omega_1),d\tilde{\omega}) \right) P(\omega_0,d\omega_1) \\ &= \sum_{j \geq 1} Q_{n+1}(\omega_0,B_j). \end{align*}

This proves the $\sigma$-addivity of $Q_{n+1}(\omega_0,\cdot)$. Finally, we note that it follows from $(1)$, $(2)$ and the above lemma that $$(\Omega_0,\mathcal{F}_0) \ni \omega \mapsto Q_{n+1}(\omega_0,B)$$ is measurable for any $B \in \mathcal{F}_1 \otimes \ldots \otimes \mathcal{F}_{n+1}$. Consequently, $Q_{n+1}$ is a transition kernel from $\mathcal{F}_0$ to $\mathcal{F}_1 \otimes \ldots \otimes \mathcal{F}_{n+1}$.


Choosing $\Omega_0$ as a trivial space (e.g. consisting of a single element), it follows immediately that

$$P_n(B) = \int P_1(d\omega_1) \int P_2(\omega_1,d\omega_2) \ldots \int 1_B(\omega_1,\ldots,\omega_n) \, P_n(\omega_1,\ldots,\omega_{n-1},d\omega_n)$$

is a measure on $(\Omega_1 \times \ldots \times \Omega_n,\mathcal{F}_1 \otimes \ldots \otimes \mathcal{F}_n)$ for any probability measure $P_1$ on $(\Omega_1,\mathcal{F}_1)$ and any transition kernels $P_i$.

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  • $\begingroup$ Thank you! Clear and detailed as ever! I have a small question for the countable additivity, how do you get $Q_{n+1}(\omega_0,\cup _{k=1 } ^{\infty} B^{(k) } ) = \int \sum _{k=1 } ^{\infty } 1_{B^{(k)} _1}(\omega_1) U((\omega_0,\omega_1),B_2 ^{(k) } \times \ldots \times B_{n+1}^{(k)} ) \, P_1(\omega_0,d\omega_1).$? On which we then could apply the monotone convergence theorem. $\endgroup$
    – MrFranzén
    May 17, 2019 at 13:07
  • $\begingroup$ On the other hand when I think about it, could we instead apply the monotone convergence theorem $n+1 $ times to "pull" the limit out from the $n+1 $ integrals. Here using that for any $k $ and for $(\omega_1,...,\omega_k)$ the map $(\omega_{k+1 } ,...,\omega_{n+1 } )\mapsto \int P_{k+1 } ...\int \sum_{k=1 } ^m 1_{B_1^{(k)} \times ... \times B_{n+1 }^{(k)} } P_{n+1 } (\omega_0,...\omega_n, d \omega_{n+1 } ) $ is increasing with $m $. (Or maybe this last claim would need some motivation??) $\endgroup$
    – MrFranzén
    May 17, 2019 at 13:08
  • $\begingroup$ @MrFranzén See my edited answer, I hope it's fine now. Let me know if something seems out of place (... there are still some typos, I guess... sorry already in advance) $\endgroup$
    – saz
    May 17, 2019 at 15:35
  • $\begingroup$ Excellent! There is one minor thing. I believe that we get $(2) $ first for sets of the form $B_1\times C $ for sets $B_1 \in \mathcal {F_1 } $ and $C \in \mathcal {F_2} \times ... \times \mathcal {F_{n+1 } } $, since we have to use $1_{B_1 \times C } = 1_{B_1 } 1 _C $ and then we extend the class of sets for which the equation holds using a standard $\pi - \lambda $ argument. $\endgroup$
    – MrFranzén
    May 18, 2019 at 6:35
  • $\begingroup$ And lots of thanks again! $\endgroup$
    – MrFranzén
    May 18, 2019 at 6:37

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