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.

  • $\begingroup$ 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


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


thus the continuous version is

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

  • $\begingroup$ 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
  • 1
    $\begingroup$ If it is not an Ito integral then What can be it? $\endgroup$
    – user16651
    Jan 3, 2017 at 10:17

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:

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.