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If $M_t$ is a martingale, for $0<t<T$, prove $\Bbb E \left[ M_T\int_t^T h_s ds |F_t\right] = \Bbb E \left[ \int_t^T M_s h_s ds |F_t\right]$.

I can think of $LHS=M_T \Bbb E \left[ \int_t^T h_s ds |F_t\right] $

RHS look similar to Ito isometry but not quite. Anyone could give me a hint?

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  • $\begingroup$ How is $h_s$ defined? $\endgroup$
    – ki3i
    Commented Apr 2, 2015 at 23:45
  • $\begingroup$ $h_s $ is a function depends on ds, for example, the drift part of gbm. $\endgroup$
    – HLD25
    Commented Apr 5, 2015 at 12:01
  • $\begingroup$ Be careful,...$h_s$ does not depend on $\text ds$, although I think I know what you are trying to say. The reason I asked how $h_s$ is defined was to encourage you to see that this result does not just work for any process $h_s$: when you state the result you have to state the properties of the processes for which the result applies. $\endgroup$
    – ki3i
    Commented Apr 5, 2015 at 12:32
  • $\begingroup$ Ok, so you mean h_s here has to be finite and non-negative to apply fubini-tonelli? $\endgroup$
    – HLD25
    Commented Apr 6, 2015 at 11:26
  • $\begingroup$ Sort of. I mean it is sufficient that $h_s$ be a regular adapted process satisfying the finite expectations given in my answer below. $\endgroup$
    – ki3i
    Commented Apr 6, 2015 at 15:00

1 Answer 1

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Hints: If $\ \Bbb E[\int_{t}^{T}|h_s|\,\text ds],\,\Bbb E[\int_{t}^{T}|M_sh_s|\,\text ds]<\infty\ $ then, in succession, use the Fubini-Tonelli theorem, the tower property of conditional expectation, the martingale property, and Fubini-Tonelli one more time, as follows: $$ \Bbb E \left[ M_T\int_t^T h_s \text ds \, \vert\, \mathcal F_t\right] = \int_t^T \Bbb E \left[\, \ldots\, |\,\mathcal F_t\right] \text ds = \int_t^T \Bbb E \left[\, \Bbb E [\,\ldots\, \vert\, \ldots]\,|\,\mathcal F_t\right] \text ds \\= \int_t^T \Bbb E \left[\, \ldots\Bbb E [\ldots\, \vert\, \mathcal \ldots]\,|\,\mathcal F_t\right] \text ds = \int_t^T \Bbb E \left[\, \ldots\,|\,\mathcal F_t\right] \text ds = \Bbb E \left[\,\int_t^T M_sh_s \text ds\,|\,\mathcal F_t\right] $$

:) No peeking now :

$${\bf \text{The solution:}}\\\ \Bbb E \left[ M_T\int_t^T h_s \text ds \, \vert\, \mathcal F_t\right] = \int_t^T \Bbb E \left[\, M_T h_s\, |\,\mathcal F_t\right] \text ds = \int_t^T \Bbb E \left[\, \Bbb E [\,M_T h_s\, \vert\, \mathcal F_s]\,|\,\mathcal F_t\right] \text ds \\= \int_t^T \Bbb E \left[\, h_s\Bbb E [M_T\, \vert\, \mathcal F_s]\,|\,\mathcal F_t\right] \text ds = \int_t^T \Bbb E \left[\, M_sh_s\,|\,\mathcal F_t\right] \text ds = \Bbb E \left[\,\int_t^T M_sh_s \text ds\,|\,\mathcal F_t\right] $$

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  • $\begingroup$ Thank you. It is really helpful. BTW, could you recommend some textbooks on stochastic calculus? I am working on the Shreve's. I would like to have some practice problem set. $\endgroup$
    – HLD25
    Commented Apr 5, 2015 at 12:09
  • $\begingroup$ @HLD25, You are welcome (I hope you didn't peek at the answer before trying it out). If you are satisfied with the answer you can accept it by clicking on the "tick" to accept. Concerning books, I think Shreve's books (Volumes 1 and 2) are good. But, to be honest, there are many books out there and it really depends on what your needs are. You can get more advice from posts like this math.stackexchange.com/questions/842300/… and this math.stackexchange.com/questions/231712/… $\endgroup$
    – ki3i
    Commented Apr 5, 2015 at 12:23

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