Jointly measurable and progressively measurable stochastic processes

1 Jointly measurable stochastic processes

Let $E={ℝ}^d$ with Borel $ℰ$, let $I={ℝ}_{\ge 0}$, which is a topological space with the subspace topology inherited from $ℝ$, and let $(\mathrm{\Omega },ℱ,P)$ be a probability space. For a stochastic process ${(X_t)}_{t\in I}$ with state space $E$, we say that $X$ is jointly measurable if the map $(t,\omega )\mapsto X_t(\omega )$ is measurable ${ℬ}_I\otimes ℱ\to ℰ$.

For $\omega \in \mathrm{\Omega }$, the path $t\mapsto X_t(\omega )$ is called left-continuous if for each $t\in I$,

$$X_s(\omega )\to X_t(\omega ),s\uparrow t.$$

We prove that if the paths of a stochastic process are left-continuous then the stochastic process is jointly measurable.

Theorem 1.

If $X$ is a stochastic process with state space $E$ and all the paths of $X$ are left-continuous, then $X$ is jointly measurable.

Proof.

For $n\ge 1$, $t\in I$, and $\omega \in \mathrm{\Omega }$, let

$$X_t^n(\omega )=\sum _{k=0}^{\mathrm{\infty }}{1}_{[k{2}^{-n},(k+1)2^{-n})}(t)X_{k{2}^{-n}}(\omega ).$$

Each $X^n$ is measurable ${ℬ}_I\otimes ℱ\to ℰ$: for $B\in ℰ$,

$$\{(t,\omega )\in I\times \mathrm{\Omega }:X_t^n(\omega )\in B\}=\bigcup _{k=0}^{\mathrm{\infty }}[k{2}^{-n},(k+1)2^{-n})\times \{X_{k{2}^{-n}}\in B\}.$$

Let $t\in I$. For each $n$ there is a unique $k_n$ for which $t\in [k_n{2}^{-n},(k_n+1)2^{-n})$, and thus $X_t^n(\omega )=X_{k_n{2}^{-n}}(\omega )$. Furthermore, $k_n{2}^{-n}\uparrow t$, and because $s\mapsto X_s(\omega )$ is left-continuous, $X_{k_n{2}^{-n}}(\omega )\to X_t(\omega )$. That is, $X^n\to X$ pointwise on $I\times \mathrm{\Omega }$, and because each $X_n$ is measurable ${ℬ}_I\otimes ℱ\to ℰ$ this implies that $X$ is measurable ${ℬ}_I\otimes ℱ\to ℰ$. Namely, the stochastic process ${(X_t)}_{t\in I}$ is jointly measurable, proving the claim. ∎

2 Adapted stochastic processes

Let ${ℱ}_I={({ℱ}_t)}_{t\in I}$ be a filtration of $ℱ$. A stochastic process $X$ is said to be adapted to the filtration ${\mathrm{F}}_I$ if for each $t\in I$ the map

$$\omega \mapsto X_t(\omega ),\mathrm{\Omega }\to E,$$

is measurable ${ℱ}_t\to ℰ$, in other words, for each $t\in I$,

$$\sigma (X_t)\subset {ℱ}_t.$$

For a stochastic process ${(X_t)}_{t\in I}$, the natural filtration of $X$ is

$$\sigma (X_s:s\le t).$$

It is immediate that this is a filtration and that $X$ is adapted to it.

3 Progressively measurable stochastic processes

Let ${ℱ}_I={({ℱ}_t)}_{t\in I}$ be a filtration of $ℱ$. A function $X:I\times \mathrm{\Omega }\to E$ is called progressively measurable with respect to the filtration ${\mathrm{F}}_I$ if for each $t\in I$, the map

$$(s,\omega )\mapsto X(s,\omega ),[0,t]\times \mathrm{\Omega }\to E,$$

is measurable ${ℬ}_{[0,t]}\otimes {ℱ}_t\to ℰ$. We denote by ${ℳ}^0({ℱ}_I)$ the set of functions $I\times \mathrm{\Omega }\to E$ that are progressively measurable with respect to the filtration ${ℱ}_I$. We shall speak about a stochastic process ${(X_t)}_{t\in I}$ being progressively measurable, by which we mean that the map $(t,\omega )\mapsto X_t(\omega )$ is progressively measurable.

We denote by $\mathrm{Prog}({ℱ}_I)$ the collection of those subsets $A$ of $I\times \mathrm{\Omega }$ such that for each $t\in I$,

$$([0,t]\times \mathrm{\Omega })\cap A\in {ℬ}_{[0,t]}\otimes {ℱ}_t.$$

We prove in the following that this is a $\sigma$-subalgebra of ${ℬ}_I\otimes ℱ$ and that it is the coarsest $\sigma$-algebra with which all progressively measurable functions are measurable.

Theorem 2.

Let ${ℱ}_I={({ℱ}_t)}_{t\in I}$ be a filtration of $ℱ$.

1. $\mathrm{Prog}({ℱ}_I)$ is a $\sigma$-subalgebra of ${ℬ}_I\otimes ℱ$, and is the $\sigma$-algebra generated by the collection of functions $I\times \mathrm{\Omega }\to E$ that are progressively measurable with respect to the filtration ${ℱ}_I$:

   $$\mathrm{Prog}({ℱ}_I)=\sigma ({ℳ}^0({ℱ}_I)).$$

2. If $X:I\times \mathrm{\Omega }\to E$ is progressively measurable with respect to the filtration ${ℱ}_I$, then the stochastic process ${(X_t)}_{t\in I}$ is jointly measurable and is adapted to the filtration.

Proof.

If $A_1,A_2,\mathrm{\dots }\in \mathrm{Prog}({ℱ}_I)$ and $t\in I$ then

$$([0,t]\times \mathrm{\Omega })\cap \bigcup _{n\ge 1}{A}_n=\bigcup _{n\ge 1}(([0,t]\times \mathrm{\Omega })\cap A_n),$$

which is a countable union of elements of the $\sigma$-algebra ${ℬ}_{[0,t]}\otimes {ℱ}_t$ and hence belongs to ${ℬ}_{[0,t]}\otimes {ℱ}_t$, showing that ${\bigcup }_{n\ge 1}{A}_n\in \mathrm{Prog}({ℱ}_I)$. If $A_1,A_2\in \mathrm{Prog}({ℱ}_I)$ and $t\in I$ then

$$([0,t]\times \mathrm{\Omega })\cap (A_1\cap A_2)=(([0,t]\times \mathrm{\Omega })\cap A_1)\cap (([0,t]\times \mathrm{\Omega })\cap A_2),$$

which is an intersection of two elements of ${ℬ}_{[0,t]}\otimes {ℱ}_t$ and hence belongs to ${ℬ}_{[0,t]}\otimes {ℱ}_t$, showing that $A_1\cap A_2\in \mathrm{Prog}({ℱ}_I)$. Thus $\mathrm{Prog}({({ℱ}_t)}_{t\in I})$ is a $\sigma$-algebra.

If $X:I\times \mathrm{\Omega }\to E$ is progressively measurable, $B\in ℰ$, and $t\in I$, then

$$([0,t]\times \mathrm{\Omega })\cap X^{-1}(B)=\{(s,\omega )\in [0,t]\times \mathrm{\Omega }:X(s,\omega )\in B\}.$$

Because $X$ is progressively measurable, this belongs to ${ℬ}_{[0,t]}\otimes {ℱ}_t$. This is true for all $t$, hence $X^{-1}(B)\in \mathrm{Prog}({ℱ}_I)$, which means that $X$ is measurable $\mathrm{Prog}({ℱ}_I)\to ℰ$.

If $X:I\times \mathrm{\Omega }\to E$ is measurable $\mathrm{Prog}({ℱ}_I)\to ℰ$, $t\in I$, and $B\in ℰ$, then because $X^{-1}(B)\in \mathrm{Prog}({ℱ}_I)$, we have $([0,t]\times \mathrm{\Omega })\cap X^{-1}(B)\in {ℬ}_{[0,t]}\otimes {ℱ}_t$. This is true for all $B\in ℰ$, which means that $(s,\omega )\mapsto X(s,\omega )$, $[0,t]\times \mathrm{\Omega }\to E$, is measurable ${ℬ}_{[0,t]}\otimes {ℱ}_t$, and because this is true for all $t$, $X$ is progressively measurable. Therefore a function $I\times \mathrm{\Omega }\to E$ is progressively measurable if and only if it is measurable $\mathrm{Prog}({ℱ}_I)\to ℰ$, which shows that $\mathrm{Prog}({ℱ}_I)$ is the coarsest $\sigma$-algebra with which all progressively measurable functions are measurable.

If $X:I\times \mathrm{\Omega }\to E$ is a progressively measurable function and $B\in ℰ$,

$$X^{-1}(B)=\bigcup _{k\ge 1}(([0,k]\times \mathrm{\Omega })\cap X^{-1}(B)).$$

Because $X$ is progressively measurable,

$$([0,k]\times \mathrm{\Omega })\cap X^{-1}(B)\in {ℬ}_{[0,k]}\otimes {ℱ}_k\subset {ℬ}_I\otimes ℱ,$$

thus $X^{-1}(B)$ is equal to a countable union of elements of ${ℬ}_I\otimes ℱ$ and so itself belongs to ${ℬ}_I\otimes ℱ$. Therefore $X$ is measurable ${ℬ}_I\otimes ℱ\to ℰ$, namely $X$ is jointly measurable.

Because $\mathrm{Prog}({ℱ}_I)$ is the $\sigma$-algebra generated by the collection of progressively measurable functions and each progressively measurable function is measurable ${ℬ}_I\otimes ℱ$,

$$\mathrm{Prog}({ℱ}_I)\subset {ℬ}_I\otimes ℱ,$$

and so $\mathrm{Prog}({ℱ}_I)$ is indeed a $\sigma$-subalgebra of ${ℬ}_I\otimes ℱ$.

Let $t\in I$. That $X$ is progressively measurable means that

$$(s,\omega )\mapsto X(s,\omega ),[0,t]\times \mathrm{\Omega }$$

is measurable ${ℬ}_{[0,t]}\otimes {ℱ}_t\to ℰ$. This implies that for each $s\in [0,t]$ the map $\omega \mapsto X(s,\omega )$ is measurable ${ℱ}_t\to ℰ$. (Generally, if a function is jointly measurable then it is separately measurable in each argument.) In particular, $\omega \mapsto X(t,\omega )$ is measurable ${ℱ}_t\to ℰ$, which means that the stochastic process ${(X_t)}_{t\in I}$ is adapted to the filtration, completing the proof. ∎

We now prove that if a stochastic process is adapted and left-continuous then it is progressively measurable.

Theorem 3.

Let ${({ℱ}_t)}_{t\in I}$ be a filtration of $ℱ$. If ${(X_t)}_{t\in I}$ is a stochastic process that is adapted to this filtration and all its paths are left-continuous, then $X$ is progressively measurable with respect to this filtration.

Proof.

Write $X(t,\omega )=X_t(\omega )$. For $t\in I$, let $Y$ be the restriction of $X$ to $[0,t]\times \mathrm{\Omega }$. We wish to prove that $Y$ is measurable ${ℬ}_{[0,t]}\otimes {ℱ}_t\to ℰ$. For $n\ge 1$, define

$$Y_n(s,\omega )=\sum _{k=0}^{2^n-1}{1}_{[kt{2}^{-n},(k+1)t{2}^{-n})}(s)Y(kt{2}^{-n},\omega )+1_{\{t\}}(s)Y(t,\omega ).$$

Because $X$ is adapted to the filtration, each $Y_n$ is measurable ${ℬ}_{[0,t]}\otimes {ℱ}_t\to ℰ$. Because $X$ has left-continuous paths, for $(s,\omega )\in [0,t]\times \mathrm{\Omega }$,

$$Y_n(s,\omega )\to Y(s,\omega ).$$

Since $Y$ is the pointwise limit of $Y_n$, it follows that $Y$ is measurable ${ℬ}_{[0,t]}\otimes {ℱ}_t\to ℰ$, and so $X$ is progressively measurable. ∎

4 Stopping times

Let ${ℱ}_I={({ℱ}_t)}_{t\in I}$ be a filtration of $ℱ$. A function $T:\mathrm{\Omega }\to [0,\mathrm{\infty }]$ is called a stopping time with respect to the filtration ${\mathrm{F}}_I$ if

$$\{T\le t\}\in {ℱ}_t,t\in I.$$

It is straightforward to prove that a stopping time is measurable $ℱ\to {ℬ}_{[0,\mathrm{\infty }]}$. Let

$${ℱ}_{\mathrm{\infty }}=\sigma ({ℱ}_t:t\in I).$$

We define

$${ℱ}_T=\{A\in {ℱ}_{\mathrm{\infty }}:\text{if }t\in I\text{ then }A\cap \{T\le t\}\in {ℱ}_t\}.$$

It is straightforward to check that $T$ is measurable ${ℱ}_T\to {ℬ}_{[0,\mathrm{\infty }]}$, and in particular .

For a stochastic process ${(X_t)}_{t\in I}$ with state space $E$, we define $X_T:\mathrm{\Omega }\to E$ by

We prove that if $X$ is progressively measurable then $X_T$ is measurable ${ℱ}_T\to ℰ$.

Theorem 4.

If ${ℱ}_I={({ℱ}_t)}_{t\in I}$ is a filtration of $ℱ$, ${(X_t)}_{t\in I}$ is a stochastic process that is progressively measurable with respect to ${ℱ}_I$, and $T$ is a stopping time with respect to ${ℱ}_I$, then $X_T$ is measurable ${ℱ}_T\to ℰ$.

Proof.

For $t\in I$, using that $T$ is a stopping time we check that $\omega \mapsto T(\omega )\wedge t$ is measurable ${ℱ}_t\to {ℬ}_{[0,t]}$, and then $\omega \mapsto (T(\omega )\wedge t,\omega )$, $\mathrm{\Omega }\to [0,t]\times \mathrm{\Omega }$, is measurable ${ℱ}_t\to {ℬ}_{[0,t]}\otimes {ℱ}_t$. Because $X$ is progressively measurable, $(s,\omega )\mapsto X_s(\omega )$ is measurable ${ℬ}_{[0,t]}\otimes {ℱ}_t\to ℰ$. Therefore the composition

$$\omega \mapsto X_{T(\omega )\wedge t}(\omega ),\mathrm{\Omega }\to E,$$

is measurable ${ℱ}_t\to ℰ$, and a fortiori it is measurable ${ℱ}_{\mathrm{\infty }}\to ℰ$. We have

$$X_T(\omega )=\underset{n\to \mathrm{\infty }}{lim}{1}_{\{T\le n\}}(\omega )X_{T(\omega )\wedge n}(\omega ),$$

and because $\omega \mapsto 1_{\{T\le n\}}(\omega )X_{T(\omega )\wedge n}(\omega )$ is measurable ${ℱ}_{\mathrm{\infty }}\to ℰ$, it follows that $\omega \mapsto X_T(\omega )$ is measurable ${ℱ}_{\mathrm{\infty }}\to ℰ$. For $B\in ℰ$,

$$\{X_T\in B\}\cap \{T\le t\}=\{\omega \in \mathrm{\Omega }:X_{T(\omega )\wedge t}(\omega )\in B\}\cap \{T\le t\}\in {ℱ}_t,$$

therefore $\{X_T\in B\}\in {ℱ}_T$. This means that $X_T$ is measurable ${ℱ}_T\to ℰ$. ∎

For a stochastic process ${(X_t)}_{t\in I}$, a filtration ${ℱ}_I={({ℱ}_t)}_{t\in I}$, and a stopping time $T$ with respect to the filtration, we define

$$X_t^T(\omega )=X_{T(\omega )\wedge t}(\omega ),$$

and ${(X_t^T)}_{t\in I}$ is a stochastic process. We prove that if $X$ is progressively measurable with respect to ${ℱ}_I$ then the stochastic proces $X^T$ is progressively measurable with respect to ${ℱ}_I$.

Theorem 5.

If ${(X_t)}_{t\in I}$ is a stochastic process that is progressively measurable with respect to a filtration ${ℱ}_I={({ℱ}_t)}_{t\in I}$ and $T$ is a stopping time with respect to ${ℱ}_I$, then $X^T$ is progressively measurable with respect to ${ℱ}_I$.

Proof.

Let $t\in I$. Because $T$ is a stopping time, for each $s\in [0,t]$ the map $\omega \mapsto T(\omega )\wedge s$ is measurable ${ℱ}_s\to {ℬ}_{[0,t]}$ and a fortiori is measurable ${ℱ}_t\to {ℬ}_{[0,t]}$. Therefore $(s,\omega )\mapsto T(\omega )\wedge s$ is measurable ${ℬ}_{[0,t]}\otimes {ℱ}_t\to {ℬ}_{[0,t]}$, and $(s,\omega )\mapsto \omega$ is measurable ${ℬ}_{[0,t]}\otimes {ℱ}_t\to {ℱ}_t$. This implies that

$$(s,\omega )\mapsto (T(\omega )\wedge s,\omega ),[0,t]\times \mathrm{\Omega }\to [0,t]\times \mathrm{\Omega },$$

is measurable ${ℬ}_{[0,t]}\otimes {ℱ}_t\to {ℬ}_{[0,t]}\otimes {ℱ}_t$. Because $X$ is progressively measurable,

$$(s,\omega )\mapsto X_s(\omega ),[0,t]\times \mathrm{\Omega }\to E,$$

is measurable ${ℬ}_{[0,t]}\otimes {ℱ}_t\to ℰ$. Therefore the composition

$$(s,\omega )\mapsto X_{T(\omega )\wedge s}(\omega ),[0,t]\times \mathrm{\Omega }\to E,$$

is measurable ${ℬ}_{[0,t]}\otimes {ℱ}_t\to ℰ$, which shows that $X^T$ is progressively measurable. ∎