Why Midpoint Sampling Cancels the Itô Correction

# Ito vs. Stratonovich: Why is it the exact midpoint that renders Ito-correction zero?


# Ito vs. Stratonovich: Why is it the exact midpoint that renders Ito-correction zero?












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?

Edit 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?

## Answer by vonjd (score 2, accepted)

https://quant.stackexchange.com/a/34322

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

## Answer by Gordon (score 2)

https://quant.stackexchange.com/a/34274

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.