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

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?

• How is $h_s$ defined?
– ki3i
Commented Apr 2, 2015 at 23:45
• $h_s$ is a function depends on ds, for example, the drift part of gbm. Commented Apr 5, 2015 at 12:01
• 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.
– ki3i
Commented Apr 5, 2015 at 12:32
• Ok, so you mean h_s here has to be finite and non-negative to apply fubini-tonelli? Commented Apr 6, 2015 at 11:26
• Sort of. I mean it is sufficient that $h_s$ be a regular adapted process satisfying the finite expectations given in my answer below.
– ki3i
Commented Apr 6, 2015 at 15:00

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