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First of all, sry for the title. I just couldn't figure out any better description for this weird problem:

Let $X$ be a bounded real r.v. and $(A_t)_{t\geq0}$ an increasing bounded process (hence $A_{\infty}\leq M$ a.s for some real $M$.). EDIT: And assume additionally that $A(0)=0$ a.s.

Show that $\mathbb{E}[XA_{\infty}]=\mathbb{E}[\int_{0}^{\infty}\mathbb{E}[X|\mathcal{F}_t]\mathbb{d}A_t]$

Since this exercise is really different from others I have done and differs from the stuff I already worked with I have to admit that I got no clue what to do. Especially the integral with respect to $A_t$ bothers me...

Nevertheless I tried to figure out a simple case. $X=5$ and $A_t=t-1$, $t\in[0,1]$, $A_t=0$ if $t\geq1$. Here the LHS becomes $0$ and the RHS (not sure if that's true) $5\int_0^1\mathbb{d}A_t=5\neq0$??

I'm really confused...maybe we additionally need positivity of $A$?

Mr. Barrrington
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    What does $\ast$ denote? – Nate Eldredge Nov 08 '13 at 20:59
  • sry, for the bad style. * was just ment to meat to say $X$ times $A_{\infty}$ – Mr. Barrrington Nov 09 '13 at 09:47
  • What's $\mathcal{F}t$? An arbritrary filtration? The canonical filtration of $(A_t){t \geq 0}$...? And yes, you have to assume $A(0)=0$. – saz Nov 09 '13 at 10:20
  • There is no futher specification about the filtration, so I guess it should be arbitrary? (but for sure satisfying the "usual" conditions of right continuity and completeness). And thanks for your your comment. The fact that $A(0)=0$ is necessary is indeed obvious using $A=\pm C$ for any constant $C\neq0$ and $X=1$, since the RHS is always $0$ while the LHS is $C\neq0$. – Mr. Barrrington Nov 09 '13 at 12:38
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    I doubt that the statement is correct for any filtration. For example, if $X$ is independent of $(\mathcal{F}t){t \geq 0}$, then the RHS equals $\mathbb{E}(X) \cdot \mathbb{E}(A_{\infty})$ whereas the LHS is $\mathbb{E}(X \cdot A_{\infty})$. And as long as $X$ and $(A_t)_{t \geq 0}$ are not independent, these expectations do not necessarily coincide. (Note that this counterexample doesn't work if $A_t$ is $\mathcal{F}_t$-adapted.) – saz Nov 09 '13 at 13:15
  • Hmm you are right. So I guess we should assume that $A$ is in fact adapted to the filtration, but I'm not sure if the assumption that$\mathcal{F}_t$ is even the canonical filtration of $A$ is valid. – Mr. Barrrington Nov 09 '13 at 14:11

1 Answers1

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Suppose that $A_t$ is $\mathcal{F}_t$-measurable. Otherwise the statement is in general not correct; for example if $X$ is independent of $(\mathcal{F}_t)_{t \geq 0}$, but not independent of $(A_t)_{t \geq 0}$.

Hints Since $X$ as well as the process $A$ are bounded it suffices to show

$$\mathbb{E}(X \cdot A_T) = \mathbb{E} \left( \int_0^T \mathbb{E}(X \mid \mathcal{F}_t) \, dA_t \right)$$

for any $T \geq 0$. As $A$ is increasing, we have

$$X \cdot A_T = \sum_{j=1}^n X \cdot (A_{t_j}-A_{t_{j-1}})$$

for any partition $0=t_0<\ldots<t_n = T$. Conclude that

$$\mathbb{E}(X \cdot A_T) = \mathbb{E} \left( \sum_{j=1}^n \mathbb{E}(X \mid \mathcal{F}_{t_{j}}) \cdot (A_{t_j}-A_{t_{j-1}}) \right)$$

Finally, use the definition of the Riemann-Stieltjes integral.

saz
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