Perhaps I am approaching this from the wrong direction but I was just thinking about the relationship between Ito and Stratonovich integrals:

It is a well known result that to convert one into the other you need an extra Ito-correction term which basically is a convexity correction.

My question
I understand that you need this extra term in the Ito case. My problem is that I don't have a good intuition why you lose this extra term exactly at the midpoint when you construct the Stratonovich integral. Why not more to the right or more to the left? Or put another way: Why is the situation always that symmetric and thus independent of the integrated function?

One hunch I have is that it has to do with bounded quadratic variation: Because the quadratic function is symmetric the midpoint automatically balances the left and the right hand side's "distortions". Is this idea correct and if yes, how can you show this behaviour in the definition of the Stratonovich (and the Ito) integral?

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    $\begingroup$ Fix a number $u \in [0,1]$. For a given partition $\pi = \{0 = t_0 < t_1 < \ldots < t_{m{\pi}}=t\}$, let $s_i = (1-u)t_i + ut_{i+1}$ and define $$S(\pi) = \sum_{i=0}^{m(\pi)-1}B_{s_i}(B_{t_{i+1}}-B_{t_i})$$ Then $S(\pi) \to \frac{1}{2}B_t^2 + (u-\frac{1}{2})t$ in $L^2$ as $m(\pi) \to 0$. You can try to prove this and see where the second term comes from. Maybe that helps. $\endgroup$
    – Calculon
    May 5, 2017 at 8:04
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    $\begingroup$ You are welcome. I will add more details as soon as I find a good moment. $\endgroup$
    – Calculon
    May 5, 2017 at 14:13
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    $\begingroup$ You can read the fourth chapter here for the proof of the result in my first comment: math.wisc.edu/~seppalai/courses/735/notes.pdf $\endgroup$
    – Calculon
    May 9, 2017 at 13:45
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    $\begingroup$ Interesting question and good insights, but keep in mind that Stratonovich Integral is not much used in Finance. $\endgroup$
    – nbbo2
    May 15, 2017 at 19:23
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    $\begingroup$ I don't see why this was migrated to quant.stackexchange. It's a math question not a finance question. $\endgroup$
    – Mark Joshi
    May 15, 2017 at 21:14

2 Answers 2


Adding to @Gordon's mathematical explanation I want to give some intuition:

The reason that the mid-point renders the correction term $0$ is that when you take the right-hand point you get a result like the Ito integral but with a negative correction term (because now you kind of look “from the future into the past” where you have the same effect from convexity – but “in the other direction”!). The Stratonovich integral is just the arithmetic mean of the two and therefore loses the correction term!

I updated the following paper accordingly, the details can be found on page 14:
von Jouanne-Diedrich, Holger, Ito, Stratonovich and Friends (May 18, 2017). Available at SSRN: https://ssrn.com/abstract=2956257


There is a good discussion in Chapter 5 of the book Stochastic Integration and Differential Equations by Protter. Note that, generally, the Stratonovich integral is defined by \begin{align*} \int_0^t Y_{s-} \circ dX_s = \lim_{\pi \rightarrow 0}\sum_{i=1}^n\frac{1}{2}(Y_{t_i}+Y_{t_{i-1}})(X_{t_i}-X_{t_{i-1}}). \end{align*} Here, $\pi$ is the module of the (can be random) partition $0=t_0 <t_1 < \cdots < t_n=t$. However, for a standard Brownian motion $B=\{B_t, t \ge0\}$, we can define \begin{align*} \int_0^t f(B_s) \circ dB_s = \lim_{\pi \rightarrow 0}\sum_{i=1}^n f\Big(B_{\frac{t_i+t_{i-1}}{2}}\Big)(B_{t_i}-B_{t_{i-1}}). \end{align*}

From Theorem 30 in Chapter 5 of the above book, \begin{align*} \lim_{\pi \rightarrow 0}\sum_{i=1}^n B_{\lambda t_{i-1}+(1-\lambda)t_i}(B_{t_i}-B_{t_{i-1}}) &= \int_0^1 B_s dB_s + 1-\lambda\\ &=\frac{1}{2}B_1^2 + \frac{1}{2}-\lambda,\tag{1} \end{align*} for any $0<\lambda <1$. That is, only taking values at the interval mid-points, or $\lambda=\frac{1}{2}$, the correction term can disappear.

Theorem 30: Let X be a semimartingale and Y be a continuous semimartingale. Let $\mu$ be a probability measure on [0,1] and let $\alpha = \int_0^1 k \mu(dk)$. Further suppose that $[X, Y]_t$ is absolutely continuous. Then \begin{align*} &\ \lim_{\pi \rightarrow 0}\sum_{i=1}^n \int_0^1 f(Y_{t_{i-1}+k (t_i-t_{i-1})})\mu(dk)(X_{t_i}-X_{t_{i-1}})\\ =&\ \int_0^1 f(Y_s)dX_s + \alpha \int_0^1f'(Y_s)d[Y, X]_s. \end{align*}

Now, for $0<\lambda <1$, taking $\mu =\delta_{1-\lambda}(dk)$, that is, the point mass at $1-\lambda$, we obtain $(1)$ above.

  • $\begingroup$ The book Basics of Stochastic Analysis mentioned by @Calculon is also good. $\endgroup$
    – Gordon
    May 17, 2017 at 18:15
  • $\begingroup$ Thank you. I am looking at Theorem 30 in Chapter 5 on p. 288ff. in the 2nd edition and cannot find your last equation there? $\endgroup$
    – vonjd
    May 17, 2017 at 19:28
  • $\begingroup$ ...it seems that I have a different version but there is another version on google books (2nd ed., v2.1), here the theorem is on pages 294-294 and there isn't the equation either: books.google.de/… $\endgroup$
    – vonjd
    May 17, 2017 at 19:39
  • $\begingroup$ @vonjd: I outlined Theorem 30 now. $\endgroup$
    – Gordon
    May 17, 2017 at 20:49
  • $\begingroup$ Thank you again... so I see all those theorems and corollaries, yet I still don't see the big picture... could you add some intuition to all those symbols?!? Some effect from algebra where you multiply two terms of differences and one vanishes in certain cases or something connected to parabolas where because of symmetry considerations the slope gets $0$ in the middle or something else that could lead to a real understanding of this stuff ... do you know what I mean? $\endgroup$
    – vonjd
    May 18, 2017 at 9:06

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