I am asked to prove that the following processes are predictable. I am used to looking at stochastic processes as sequences of random variables (by fixing time) or as a collection of paths (by fixing $\omega$) but I find it hard to picture the two varying together. For this reason the notions of predictability and progressive measurability somewhat escape me. Anyway, here are the processes I am interested in.

(a) Suppose $h : \mathbf{R}_+ \rightarrow \mathbf{R}$ is a Borel measurable function. Show that $X(t,\omega) = h(t)$ is predictable.

Here is my attempt. First I fix an arbitrary Borel set $B\in\mathbf{R}$. Then I say $\{(t,\omega): X(t,\omega)\in B\} = \{(t,\Omega): h(t)\in B\} = (\bar{B},\Omega)$ for some $\bar{B}\in \mathcal{B}_{\mathbf{R}_+}$. The first equality is due to the deterministic nature of $h$ and the second one is due to the fact it is Borel measurable. Since I can generate $(\bar{B},\Omega)$ by using sets of the form $((a,b],\Omega)$, which are all in the predictable $\sigma$-algebra, $(\bar{B},\Omega)$ is also in the predictable $\sigma$-algebra. Does this sound right?

(b): $X$ is a predictable process. $g :\mathbf{R} \rightarrow \mathbf{R}$ is a Borel function. Then, $Z_t = g(X_t)$ is also predictable. The intuitive answer that I would give is that Borel functions preserve measurability. Hence, if $X$ is measurable with respect to a sigma algebra, then $g(X)$ is also measurable with respect to that sigma algebra. But this doesn't constitute a rigorous answer so I write the following.

$$\{(t,\omega): g(X(t,\omega)) \in B\} = \{(t,\omega): X(t,\omega) \in \bar{B}\}$$ The RHS is in the predictable sigma algebra since $X$ is predictable so this is it I guess.

(c): $X$ is a predictable process. $g :\mathbf{R}_+\times\mathbf{R} \rightarrow \mathbf{R}$ is a Borel function. $Z_t = g(t,X_t)$ is predictable. There is a hint which suggests that I start with functions of the form $h(t)g(X_t)$ and apply the $\pi-\lambda$ lemma. I know that the product of two measurable functions is again measurable. So $h(t)g(X_t)$ is predictable. But I don't see how one could apply the $\pi-\lambda$ lemma to extend the result to arbitrary functions of the form $g(t,X_t)$?


1 Answer 1


(a) Well, the idea of your proof is correct, but there a several things which I would like to point out:

  • There is some abuse of notation in it. So, for example, every probabilist will know what you mean by $$\{(t,\Omega); h(t) \in B\},$$ but (at least from my point of view) that's not a nice way to put it. Just write $$\{(t,\omega); \omega \in \Omega, h(t) \in B\},$$ it's not really more effort to do so and it is rigorous.
  • You write "... for some $\bar{B} \in \mathcal{B}_{\mathbb{R}_+}$"; well, why not simply given an explicit expression for the set? What about $\bar{B} = h^{-1}(B)$?
  • You write "Since I can generate $(\bar{B},\Omega)$ by using [...]"; how can you generate it? Please be a bit more precise... (e.g. "Since the Borel $\sigma$-algebra $\mathcal{B}_{\mathbb{R}_+}$ is generated by $\{(a,b]; 0<a<b\}$ ...").

I would phrase it like that: Since $h$ is Borel-measurable, there exists a sequence of simple functions $(s_n)_{n \in \mathbb{N}}$ of the form, $$s(t) = \sum_{j=1}^n c_j 1_{(a_j,b_j]}(t),$$ such that $s_n \to f$ as $n \to \infty$. Since each $s_n$ is obviously predictable (it is left-continuous), $h$ is predictable as a pointwise limit of predictable functions.

(b) Actually, you show that the composition of two measurable functions $S: (\Omega_1,\mathcal{A}_1) \to (\Omega_2,\mathcal{A}_2)$ and $T: (\Omega_2,\mathcal{A}_2) \to (\Omega_3,\mathcal{A}_3)$ is measurable. Since this is well-known, there is no need to prove this, but nevertheless your proof is correct.

(c) Define $$\Sigma := \{A \in \mathcal{B}(\mathbb{R}_+) \otimes \mathcal{B}(\mathbb{R}); (t,\omega) \mapsto 1_A(t,X_t(\omega)) \, \text{is predictable}\}.$$ Show that $\Sigma$ is a $\sigma$-algebra. Conclude from $\mathcal{B}(\mathbb{R}_+) \times \mathcal{B}(\mathbb{R}) \subseteq \Sigma$ that $$\mathcal{B}(\mathbb{R}_+) \otimes \mathcal{B}(\mathbb{R}) = \Sigma.$$ Finally, approximate (similar as in (a)) any Borel-measurable function $g: \mathbb{R}_+ \times \mathbb{R} \to \mathbb{R}$ by simple functions and conclude that $Z_t$ it is predictable as the pointwise limit of predictable functions.

  • $\begingroup$ Thanks! I am having a bit of a trouble understanding your answer to (c), though. Why is it true that $\mathcal{B}(\mathbb{R}_+) \times \mathcal{B}(\mathbb{R}) \subseteq \Sigma$? And how does $\Sigma$ play a role in the last approximation argument? $\endgroup$
    – Calculon
    Commented May 17, 2015 at 14:18
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    $\begingroup$ @Calculon For $A \in \mathcal{B}(\mathbb{R}+) \times \mathcal{B}(\mathbb{R})$, we have $A = B \times C$ (for suitable Borel sets $B$, $C$). Since -as you already noted - the product of msble. functions is measurable, it follows easily that $1_A(t,X_t) = 1_B(t) 1_C(X_t)$ is predictable. Hence, $A \in \Sigma$. Concerning $\Sigma$: Well, if we approximate $g$ by simple functions, then the simple functions are of the form $$s(t,x) = \sum_{j=1}^n c_j 1_{A_j}(t,x)$$ where $A_j \in \mathcal{B}(\mathbb{R}_+) \otimes \mathcal{B}(\mathbb{R})$. $\endgroup$
    – saz
    Commented May 17, 2015 at 15:37
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    $\begingroup$ So in order to conclude that $Z_t$ is predictable we need so know that $s(t,X_t)$ is predictable - and this follows from the fact that $\Sigma = \mathcal{B}(\mathbb{R}_+) \otimes \mathcal{B}(\mathbb{R})$. $\endgroup$
    – saz
    Commented May 17, 2015 at 15:38

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