Stopping Time Integral

Theorem (Itô Stochastic Integral on E\mathcal{E} is a R.C. Martingale)

Let XEX\in\mathcal{E}, let Yt=1[0,t]XdMY_{t}=\int\limits \mathbb{1}_{[0,t]}X \, dMwhere M=(Mt)t0M=(M_{t})_{t\ge 0} is a right continuous L2L^{2}-martingale. Then (Yt)t0(Y_{t})_{t\ge 0} is a right continuous L2L^{2}-martingale.

Theorem (Itô Stochastic Integral on Λ\Lambda is a R.C. Martingale)

Let MM be a right continuous L2L^{2}-martingale. Let XΛ2(P,M)X\in\Lambda^{2}(\mathscr{P},M). t0\forall t\ge 0 let Yt=1[0,t]XdMY_{t}=\int\limits \mathbb{1}_{[0,t]}X \, dM Then (Yt)t0(Y_{t})_{t\ge 0} is a L2L^{2}-martingale and hence admits a right continuous version.

Theorem (Independence of Bounded RV)

Let XΛ2(P,M)X\in\Lambda^{2}(\mathscr{P},M), let 0s<t0\le s<t, let ZZ be Fs\mathcal{F}_{s}-measurable and bounded. Then:

  1. 1(s,t]Z\mathbb{1}_{(s,t]}Z is P\mathscr{P}-measurable (i.e. predictable)
  2. 1(s,t]ZXL2(R+×Ω,P,μM)\mathbb{1}_{(s,t]}ZX\in L^{2}(\mathbb{R}^{+}\times\Omega,\mathscr{P},\mu_{M})
  3. 1(s,t]ZXdM=Z1(s,t]XdM a.s.\int\limits \mathbb{1}_{(s,t]}ZX \, dM =Z\int\limits \mathbb{1}_{(s,t]}X \, dM \text{ a.s.}

Proposition (Predictable Stochastic Intervals)

Let τ\tau be a stopping time. Then, the stochastic interval <spanclass="wikilinkunresolved"title="Notenotpublished">0,τ</span>P<span class="wikilink-unresolved" title="Note not published">0,\tau</span>\in\mathscr{P} is predictable.

Theorem (Stopping Time Integral)

Let MM be a right continuous L2L^{2}-martingale, let XΛ2(P,M)X\in\Lambda^{2}(\mathscr{P},M), and let (Yt)t0(Y_{t})_{t\ge 0} be the stochastic integral process Yt=1[0,t]XdMY_{t}=\int\limits \mathbb{1}_{[0,t]}X \, dM Let τ\tau be a bounded stopping time. Then Yτ=1<spanclass="wikilinkunresolved"title="Notenotpublished">0,τ</span>XdM a.s.Y_{\tau}=\int\limits \mathbb{1}_{<span class="wikilink-unresolved" title="Note not published">0,\tau</span>}X \, dM\text{ a.s.}

Proof

The proof uses: