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In Shreve's book, the value of cash flow for a future of discrete case is

$$\dfrac{1}{D(t)}E\Big[\sum\limits_{j=k}^{n-1}D(t_{j+1})(\textrm{Fut}_S(t_{j+1},T)-\textrm{Fut}_S(t_j,T))\Big|\mathcal{F}(t)\Big]$$ The continuous version is

$$\dfrac{1}{D(t)}E\Big[\int_t^T D(u)\textrm{d} \textrm{Fut}_S(u,T) \Big|\mathcal{F}(t)\Big]$$

But you know that Ito integral chooses the left-hand endpoint, we should replace $D(t_{j+1})$ by $D(t_j)$ in the first equation, the version in Shreve's book is actually the right-hand endpoint, but the author always regards as a Ito integral. So where is my misunderstanding? Here $D(t)$ is discounted factor.

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  • $\begingroup$ What is the exact definition of $D(t_j)$ ? $\endgroup$
    – Alex C
    Jan 3, 2017 at 4:15
  • $\begingroup$ here D(t) is the discounted factor from time t @ Alex C $\endgroup$ Jan 3, 2017 at 4:16

2 Answers 2

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Let $\mathcal {V} =\mathcal {V}(t,T)$ be the class of functions $$f(t,\omega):[0,\infty)\times\Omega\to\mathbb{R}$$ such that

  • $(t,\omega)\to f(t,\omega)$ is $\mathcal{B}\times\mathcal{F}$ where $\mathcal{B}$ denotes the Borel algebra on $[0,\infty)$.
  • $f(t,\omega)$ is $\mathcal{F}_t$ adapted.
  • $\mathbb{E}\left[\int_{t}^{T}f^2(s,\omega)ds\right]<\infty$

Suppose $f\in\mathcal V(t,T)$ and that $t\to f(t,\omega)$ is continuous . Let $I=\{u_i\}_{i=0}^{n}$ is a sequence of partitions of $[t,T]$, Indeed $t=u_0<u_1<\cdots<u_n=T$ .By definition of Ito integral, we have

$$\int\limits_t^T f(u,\omega)dW(u,\omega)=\lim_{\Delta u_j\to0}\sum_{j=0}^{n-1}f(u_j,\omega)(\,W(u_{j+1},\omega)-W(u_{j},\omega)\,)\qquad,\quad\text{ in }\, L^2(P).$$

Similarly we define the Stratonovich integral of $f$ by

$$\int\limits_t^T f(s,\omega)\circ dW(u,\omega)=\lim_{\Delta u_j\to0}\sum_{j=0}^{n-1}f(u_j^*,\omega)(\,W(u_{j+1},\omega)-W(u_{j},\omega)\,)$$

where $u_j^*=\frac12(u_j+u_{j+1}),$ whenever the limit exists in $L^2(P)$. In general these integrals are different.


I think the value of cash flow for a future of discrete case is

$$\dfrac{1}{D(t)}\mathbb{E}\left[\sum\limits_{j=k}^{n-1}D(t_{j})(\,\textrm{Fut}_S(t_{j+1},T)-\textrm{Fut}_S(t_j,T)\,)\Big|\mathcal{F}_t\right]$$

thus the continuous version is

$$\dfrac{1}{D(t)}\mathbb{E}\Big[\int_t^T D(u)\textrm{d} \textrm{Fut}_S(u,T) \Big|\mathcal{F}(t)\Big]$$

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  • $\begingroup$ but you surely receive the margin of future $\textrm{Fut}_S(t_{j+1},T)-\textrm{Fut}_S(t_j,T)$ at time $t_{j+1},$ thus the discounted factor should be $D(t_{j+1}).$ @Behrouz Maleki $\endgroup$ Jan 3, 2017 at 10:09
  • $\begingroup$ Yes I am sure , like Ito's integral. $\endgroup$
    – user16651
    Jan 3, 2017 at 10:11
  • $\begingroup$ sorry, I'd better use we...it is the definition of future. $\endgroup$ Jan 3, 2017 at 10:13
  • $\begingroup$ This is not about Stratonovich integral $\endgroup$
    – user16651
    Jan 3, 2017 at 10:13
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    $\begingroup$ If it is not an Ito integral then What can be it? $\endgroup$
    – user16651
    Jan 3, 2017 at 10:17
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I believe the confusion is due to indexes. $D(t_{j+1})$ is $\mathcal{F}(t_j)$ measurable, so it is known on the left-hand endpoint. If you substitute the formula (6.2.2) for $D(t_{j+1})$:

$$D(t_{j+1}) = \frac{1}{(1+R(t_0))(1+R(t_1)+ \cdots + (1+R(t_j))}$$

you will recognize an Ito integral:

$$\dfrac{1}{D(t)}E\Big[\sum\limits_{j=k}^{n-1}\frac{1}{(1+R(t_0))+ \cdots + (1+R(t_j))}(\textrm{Fut}_S(t_{j+1},T)-\textrm{Fut}_S(t_j,T))\Big|\mathcal{F}(t)\Big]$$

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