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Fix a continuous local martingale $M$ starting at $0$. Suppose $X \in \mathscr{P}^*(M)$, i.e. $X$ is progressively measurable and $\int_0^tX_u^2 d\langle M \rangle_u<\infty$ a.s. Then suppose $Z$ is $\mathscr{F}_0$-measurable (we are NOT assuming $\mathscr{F}_0$ is trivial, so $Z$ does not necessarily have to be constant). How do I show that $$ \int_0^t Z X_udM_u = Z \int_0^t X_u dM_u. $$

What I have tried so far: If $X$ simple and $M$ is a martingale, then the result is immediate by definition using sums (all we need to note is that $ZX_u$ will still be progressive since $Z$ is known at time $0$). If $Z$ is bounded then taking simple $X^{(n)} \to X$ (in the sense of $E \int_0^t|X_u^{(n)}-X_u|^2d\langle M \rangle_u \to 0$) we also have $Z X^{(n)} \to Z X$, which gives the result for $Z$ bounded and $M$ a martingale. By taking a localizing sequence we may easily extend this to $Z$ bounded and $M$ a local martingale.

Both sides of the proposed equality make sense without assuming any extra assumptions on $Z$, but if $Z$ blows up in $\omega$, a localizing sequence $\tau_n$ for $X$ may not be a localizing sequence for $ZX$, and we do not necessarily have $ZX^{(n)} \to ZX$ in the sense $L^2(P,\langle M\rangle)$ above. Intuitively this doesn't matter, but I don't know how to show it.

How can I get rid of this requirement that $Z$ be bounded? Maybe there is a more direct way to do this without resorting to simple functions?

nullUser
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1 Answers1

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Suppose that $M$ is a martingale and $Z$ satisfies

$$\mathbb{E}\left( \int_0^t |Z \cdot X_u|^2 \, d\langle M \rangle_u \right) < \infty \tag{1}$$

Define cut-off functions

$$Z_k := (-k) \vee Z \wedge k, \quad (k \in \mathbb{N})$$

By applying the dominated convergence theorem, we see

$$\|Z_k \cdot X-Z \cdot X\|_{L^2(\langle M \rangle)}^2 = \mathbb{E} \bigg( \int_0^t \underbrace{|Z_k \cdot X_u-Z \cdot X_u|^2}_{\leq 2|X_u \cdot Z|^2} \, d\langle M \rangle_u \bigg) \stackrel{k \to \infty}{\to} 0$$

Using the well-known isometry, we thus get

$$\int_0^t Z_k \cdot X_u \, dM_u \stackrel{k \to \infty}{\to} \int_0^t Z \cdot X_u \, dM_u \tag{2}$$

in $L^2(\mathbb{P})$. Consequently, there exists a subsequence which converges almost surely. Since $Z_k$ is bounded,

$$\int_0^t Z_k \cdot X_u \, dM_u = Z_k \cdot \int_0^t X_u \, dM_u.$$

From $Z_k \to Z$ a.s. we conclude

$$Z_k \cdot \int_0^t X_u \, dM_u \stackrel{k \to \infty}{\to} Z \cdot \int_0^t X_u \, dM_u \quad \text{a.s.} \tag{3}$$

Combining $(2)$ and $(3)$ finishs the proof.

saz
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  • If $Z_k \to Z$ in $L^2(P)$ this would imply $Z \in L^2$ which I am not assuming. (Though this certainly is a much better result than what I was stuck with). – nullUser Oct 15 '13 at 15:28
  • @nullUser So what do you assume? Local integrability or ...? ($Z \in L^1$ should work similarly, by the way.) – saz Oct 15 '13 at 15:33
  • All I'm assuming is $Z$ is $\mathscr{F}_0$ measurable. It does not have to have expectation. – nullUser Oct 15 '13 at 15:34
  • Since for every $\omega \in \Omega$ we have $\int_0^t |Z(\omega)X_u|^2d\langle M \rangle(\omega) = Z(\omega)\int_0^t |X_u|^2d\langle M \rangle(\omega)$ the random variable $Z$ will not affect the set on which $\int_0^t|X|^2d\langle M \rangle$ is finite. Thus $\int ZX dM$ is defined for $M$ as a local martingale (even if $M$ is in fact a martingale.) – nullUser Oct 15 '13 at 15:37
  • Actually I think your proof might still work. For dominated convergence all we need is a.s. convergence, not $L^2$ convergence. Let me think about it for a bit. – nullUser Oct 15 '13 at 15:39
  • @nullUser I edited my answer. Note that $Z(\omega)^2 \cdot \int_0^t |X_u|^2 , d\langle M \rangle_u(\omega)<\infty$ does in general not imply that $Z \cdot X_u$ is locally $L^2$-integrable. (And stochastic integrals are only defined for locally integrable functions, as far as I know!) – saz Oct 15 '13 at 15:48
  • As long as $\int_0^t |Y|^2 d\langle M \rangle$ is finite a.s. for every $t>0$, there is a sequence of stopping times $\tau_n \to \infty$ such that $E\int_0^t |Y^{\tau_n}|^2 d\langle M^{\tau_n} \rangle<\infty$ so that $\int_0^t Y^{\tau_n} dM^{\tau_n}$ makes sense. We then define $\int_0^t Y dM = \lim \int_0^t Y^{\tau_n}dM^{\tau_n}$. It can be shown that the limit is independent of the choice of the $\tau_n$. – nullUser Oct 15 '13 at 19:49
  • @nullUser I came across this same question and couldn't solve it. Any ideas on how to prove this for the general case of $M$ being only a continuous local martingale? – nomadicmathematician Sep 07 '20 at 19:08