Given two stochastic processes on a probability space, will their compound process be a valid stochastic process on the same probability space?

Let the stochastic process $M=(M_t, t\ge 0)$ and the stochastic pathwise continuous increasing process $Y=(Y_t,t\ge 0)$ be defined on the probability space $(\Omega, \mathcal F, P)$. Will the compound process $M_Y=(M_{Y_t},t\ge 0)$ also be valid (measureable on the same sigma algebra which $M$ and $Y$ maps from) on this probability space?

If it is not valid in general, what if $M$ and $v$ are independent of each other? Will it then be 'valid'?

Clarification:

Does

$\{\omega\in\Omega\colon M(\omega)\in B\}\in\mathcal{F}$, $\forall B\in \mathcal{B}(\mathbb{R})$ and $\{\omega\in\Omega\colon Y(\omega)\in B\}\in\mathcal{F}$, $\forall B\in \mathcal{B}(\mathbb{R})$ $\implies$ $\{\omega\in\Omega\colon M_Y(\omega)\in B\}\in\mathcal{F}$, $\forall B\in \mathcal{B}(\mathbb{R})$

hold? $\mathcal{B}(\mathbb{R})$ being the generated Borel $\sigma$-algebra.

• You're essentially asking if a composition of measurable maps is measurable - this depends on your definition of measurability. See math.stackexchange.com/questions/283443/… – Bowditch Aug 11 '17 at 11:56
• @Bowditch I have done some clarifications to the question. – noidea Aug 11 '17 at 12:50
• @Bowditch My question seem to differ slightly, even though they are related. – noidea Aug 11 '17 at 14:23

Your "compound" process $M_Y$ at a fixed $t\ge 0$ is the composition of two mappings $\psi_t(\omega):=(\omega,Y_t(\omega))$ and $\varphi(\omega,u):=M_u(\omega)$. The former is an $\mathcal F / \mathcal F\otimes\mathcal B$ measurable mapping of $\Omega$ to $\Omega\times[0,\infty)$. (Where $\mathcal B$ denotes the Borel subsets of $[0,\infty)$.) This is because $\mathcal F\otimes\mathcal B$ is generated by rectangles of the form $F\times B$, $F\in\mathcal F, B\in\mathcal B$, and $\psi_t^{-1}(F\times B)=F\cap Y_t^{-1}(B)$. To finish you need to know that the latter mapping $\varphi$ is $\mathcal F\otimes\mathcal B / \mathcal R$ measurable. (Here I use $\mathcal R$ to denote the Borel subsets of $\Bbb R$.) This situation is referred to as $M$ being a "measurable process" and is an additional hypothesis. For example, if each random variable $M_u$ is $\mathcal F$ measurable and $u\mapsto M_u(\omega)$ is right continuous for each $\omega$, then $M$ is a measurable process.

If both $\psi_t$ and $\varphi$ are measurable as indicated, then the composite function $M_{Y_t}=\varphi\circ\psi$ is $\mathcal F/\mathcal R$ measurable.

• Thank you for this detailed and well written answer. I have just two smaller questions and these are: why right continuity would be needed? Also, would requiring $M$ being a cadlag process be enough to satisfy the hypothesis? – noidea Aug 11 '17 at 16:20
• And if we know that $M$ maps to positive real line with the Borel subsets $\mathcal B$ using your definition? – noidea Aug 11 '17 at 16:34
• Also, you seem to have shown that $\psi_t^{-1}(F\times B)=F\cap Y_t^{-1}(B)$ is true only for the $\pi$-system $\{F\times B\colon F\in\mathcal F, B\in\mathcal B\}$. Why would this hold for all events in $\mathcal F\otimes \mathcal B$? Is the set of all events satisfying the same property as shown for the $\pi$-system a Dynkin system? – noidea Aug 11 '17 at 19:30
• Right continuity is a sufficient, though not necessary, condition. It guarantees that $M$ is the pointwise limit of $\mathcal F\otimes\mathcal B$ simple processes: $M_u(\omega) =\lim_n\sum_{k=0}^{n2^n}1_{[k2^{-n},(k+1)2^{-n})}M_{(k+1)2^{-n}}(\omega)$. – John Dawkins Aug 13 '17 at 16:09
• The set of rectangles of the form $F\times B$ is a $\pi$-system generating $\mathcal F\otimes\mathcal B$. The set $\mathcal G$ of $G\in\mathcal F\otimes\mathcal B$ such that $\psi_t^{-1}(G)\in \mathcal F$ is a $\sigma$-field containing that $\pi$-system. Therefore $\mathcal G\supset\mathcal F\otimes\mathcal B$, and finally $\mathcal G=\mathcal F\otimes\mathcal B$. – John Dawkins Aug 13 '17 at 16:17

I guess that by "same sigma algebra which $M$ and $Y$ maps onto" you mean the sigma algebra generated by the two processes.

However, this is not the case. For instance take $Y$ to be the deterministic process defined by $Y_t=2t$ and $M_t$ any stochastic process such that its filtration satisfies that $\mathcal{F}_t\subsetneq\mathcal{F}_s$ for $t,s$.

At time $t$ the sigma algebra generated by $M$ and $Y$ is the same one as the one generated by $M$.

However, even $M_{Y_t}=M_{2t}$ is not measurable at time $t$.

• I think you are confused. Or at least you need further explanation on your question as I suggested, this is no reason to downgrade my answer. If you give no further explanation the only way is to assume the "natural filtration". For instance, what exactly do you mean by $M_Y(\omega)$? $M_Y$ is not a r.v. $M_{Y_t}$ is. What do you mean by $\mathcal{F}$? – Diego F Medina Aug 11 '17 at 13:51
• It is just a shorthand writing for it to be true for each fix t. I did not write the filtered probability space but the probability space and therefore it is clear what $\mathcal F$ is. I have less than 150 rep so the downvote which is shown is from someone else. But yes I downvoted your answer also even though it isn't shown because, given the explanation you did not answer my question. – noidea Aug 11 '17 at 14:15