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Let $b:[0,T]\times\Bbb R\to\Bbb R$ measurable.

Consider now $Y\in M^2[0,T]$, thus we have behind our shoulders a probability space $(\Omega,\mathcal A,P)$ and a filtration $\Bbb F=(\mathcal F_t)_{t\in[0,T]}$ on $\mathcal A$.

How can we prove that $$ [0,T]\times\Omega\to\Bbb R,\;\;\;(s,\omega)\mapsto b(s,Y_s(\omega)) $$ is progressively measurable?

I tried directly but I didn't get nothing. Can someone give me an hint?

EDIT: $M^2[0,T]$ consists of the processes $X:[0,T]\times\Omega\to\Bbb R$ progressively measurable and such that a.s. $\Bbb E[\int_0^TX_s^2(\omega)\,ds]<+\infty$.

EDIT2: $f:\Omega\times[0,T]\to\Bbb E$ is progressively measurable, where $(E,\mathcal E)$ is a measurable space, if $\forall t\in[0,T]$ the map $$ [0,t]\times\Omega\to E\;\;\;\;(s,\omega)\mapsto f(s,\omega) $$ is $\mathcal B[0,t]\otimes\mathcal F_t$ -measurable.

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  • $\begingroup$ What is $M^2[0,T]$? $\endgroup$ – John Dawkins Dec 19 '16 at 20:11
  • $\begingroup$ @JohnDawkins I edited $\endgroup$ – Joe Dec 19 '16 at 20:13
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Recall the following statement on the measurability with respect to a product $\sigma$-algebra:

Let $(E_i,\mathcal{A}_i)$, $i=1,2,3$, be measurable spaces. Then a mapping $$f: E_1 \to E_2 \times E_3, x \mapsto f(x) = (f_1(x),f_2(x)) \in E_2 \times E_3$$ is $\mathcal{A}_1$/$\mathcal{A}_2 \otimes \mathcal{A}_3$-measurable if and only if the projections $f_1: (E_1,\mathcal{A}_1) \to (E_2,\mathcal{A}_2)$ and $f_2: (E_1,\mathcal{A}_1) \to (E_3,\mathcal{A}_3)$ are measurable.

Applying this result, we find that the mapping

$$[0,t] \times \Omega \ni (s,\omega) \mapsto f(s,\omega):= (s,Y_s(\omega)) \in [0,t] \times \mathbb{R}$$

is measurable for any $t \leq T$ since both

$$[0,t] \times \Omega \ni (s,\omega) \mapsto s \in [0,t]$$

and

$$[0,t] \times \Omega \ni (s,\omega) \mapsto Y_s(\omega) \in \mathbb{R}$$

are measurable. Consequently, we find that the composition

$$[0,t] \times \Omega \ni (s,\omega) \mapsto (b \circ f)(s,\omega) = b(s,Y_s(\omega)) \in \mathbb{R}$$

is measurable as a composition of measurable functions. As $t \leq T$ is arbitrary, this proves that $(s,\omega) \mapsto b(s,Y_s(\omega))$ is progressively measurable.

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  • $\begingroup$ many thanks.. but I think you proved the measurability, not the progressively measurability (I edited my post and I wrote the definition) $\endgroup$ – Joe Dec 19 '16 at 22:23
  • $\begingroup$ @Joe My answer shows that $$[0,T] \times \Omega \ni (s,\omega) \mapsto (b \circ f)(s,\omega) = b(s,Y_s(\omega))$ is measurable, and since $T>0$ is arbitrary, this gives the progressive measurability. $\endgroup$ – saz Dec 20 '16 at 6:40

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