Definition of Stratonovich Integral

Question

I have a doubt in definition of the Stratonovich integral. In "Stochastic Calculus for Finance" by Steven Shreve, he defines it using the midpoint $\frac {(t_i+t_{i+1})}{2}$ of the subinterval $[t_i,t_{i+1}]$, and at many places like google and calculating Stratonovich integration of Brownian motion with respect to itself, they use the average of the value of the process at these two points $t_i$ and $t_{i+1}$.

Are these two forms equivalent. If yes, then how (I tried this but found no way) and if not then why there is two different forms of definition. Any suggestion is highly appreciable.

Answer 1

This answer changed multiple times. I am very indebted to user Chaos who finally convinced me that both definitions of the Stratonovich integral are equivalent.

In his book Stochastic Calculus for Finance II Steven Shreve defines in Exercise 4.4 on p. 191, the Stratonovich integral as $$\tag{0} \int_0^TW_t\circ\,dW_t=\lim_{||\Pi||\to0}\sum_{j=0}^{n-1}W_{t^*_j}(W_{t_{j+1}}-W_{t_j})\,,\quad\quad t^*_j=\frac{t_{j+1}+t_j}{2}. $$ This is not in conflict with the definitions one can see elsewhere. For example in Karatzas & Shreve, Brownian Motion and Stochastic Calculus, Problem 2.29 on p. 148: $$\tag{1} S_T(\Pi):=\sum_{i=0}^{m-1}[(1-\epsilon)W_{t_i}+\epsilon W_{t_{i+1}}](W_{t_{i+1}}-W_{t_i}) $$ and $$ \int_0^TW_s\circ dW_s=\lim_{||\Pi||\to0}S_T(\Pi)\quad\text{ for }\epsilon=\frac{1}{2}. $$

• The OP in this post lists further references with the two seemingly different definitions which can be very disturbing.

• Thanks to Chaos I was made aware of the book H.H. Kuo, Introduction to Stochastic Integration. Kuo's Theorem 8.3.7. on p. 123 states that for a Brownian motion $B$ and a continuous function $f(t,x)$ with continuous partial derivatives $\frac{\partial f}{\partial t},\frac{\partial f}{\partial x},\frac{\partial f^2}{\partial x^2}$ one has \begin{align} &\int_a^bf(t,B(t))\circ dB(t)\\ &=\lim_{||\Delta_n||\to0}\sum_{i=1}^nf\Big(t^*_i,\frac{1}{2}(B(t_{i-1})+B(t_i))\Big)\Big(B(t_i)-B(t_{i-1})\Big)\\ &=\lim_{||\Delta_n||\to0}\sum_{i=1}^nf\Big(t^*_i,B\Big(\frac{t_{i-1}+t_i}{2}\Big)\Big)\Big(B(t_i)-B(t_{i-1})\Big) \end{align} where (in Kuo's notation) $t_{i-1}\le t_i^*\le t_i, \Delta_n=\{t_0,t_1,...,t_n\}$ is a partition of the finite interval $[a,b]$ and $||\Delta_n||=\max_{1\le i\le n}(t_i-t_{i-1})$.

Answer 2

Let $0=t_0<t_1<\cdots <t_N=T$ be an arbitrary partition of the interval $[0,T]$ with $\|\pi\|:=\max_{i} |t_{i+1}-t_i|$.

Define $t^*_i:=\frac{t_{i+1}+t_i}{2}$ and consider

\begin{align*} &\sum_{i=0}^{N-1}W\left(t_i^*\right)[W(t_{i+1})-W(t_i)]\\ &=\sum_{i=0}^{N-1}[W(t_i^*)-W(t_i)][W(t_{i+1})-W(t_i)]+\sum_{i=0}^{N-1} W(t_i)[W(t_{i+1})-W(t_i)]=\mathcal I_1+\mathcal I_2 \end{align*} We know that as $\|\pi\|\to 0$ the term $\mathcal I_2$ converges to $\int_0^T W(t)dW(t)$ in $L^2(\Omega)$. In order to prove the desired result it suffices to show that $\mathcal I_1$ converges in $L^2(\Omega)$ to $T/2$.

We start by noticing that

\begin{align*} \mathbb E\left[\mathcal I_1\right]=\sum_{i=0}^{N-1}\mathbb E\left([W(t_i^*)-W(t_i)][W(t_{i+1})-W(t_i)]\right)&=\sum_{i=0}^{N-1} t_i^*\wedge t_{i+1}-t_i\wedge t_{i+1}-t_i^*\wedge t_{i}+t_i\\ &=\sum_{i=0}^{N-1}t_i^*-t_i=\sum_{i=0}^{N-1}\frac{t_{i+1}-t_i}{2}=T/2 \end{align*} Then \begin{align*} \|\mathcal I_1-T/2\|_{L^2(\Omega)}^2= \mathbb V\left(\mathcal I_1\right)=\mathbb V\left(\sum_{i=0}^{N-1}[W(t_i^*)-W(t_i)][W(t_{i+1})-W(t_i)]\right), \end{align*} due to the disjointness of the intervals in each term of the sum we can write the latter as \begin{align*} \|\mathcal I_1-T/2\|_{L^2(\Omega)}^2= \mathbb V\left(\mathcal I_1\right)&=\sum_{i=0}^{N-1}\mathbb V\left([W(t_i^*)-W(t_i)][W(t_{i+1})-W(t_i)]\right)\\ &=\sum_{i=0}^{N-1}\mathbb V\left([W(t_i^*)-W(t_i)][(W(t_{i+1})-W(t_i^*))+(W(t_i^*)-W(t_i))]\right) \end{align*}

Let $\Delta_*(i):=[W(t^*_i)-W(t_i)]$ and $\Delta^*(i):=[W(t_{i+1})-W(t^*_i)]$ \begin{align*} &\sum_{i=0}^{N-1}\mathbb V\left(\Delta_*(i)[\Delta^*(i)+\Delta_*(i)]\right)\\ &=\sum_{i=0}^{N-1}\mathbb E\left(\Delta_*(i)^2[\Delta^*(i)+\Delta_*(i)]^2\right)- (t_i^*-t_i)^2\\ &=\sum_{i=0}^{N-1}\mathbb E\left(\Delta_*(i)^2[\Delta^*(i)^2+2\Delta^*(i)\Delta_*(i)+\Delta_*(i)^2]\right)- (t_i^*-t_i)^2\\ &=\sum_{i=0}^{N-1}\mathbb E\left(\Delta_*(i)^2\Delta^*(i)^2\right)+2E\left(\Delta^*(i)\Delta_*(i)^3\right)+\mathbb E\left(\Delta_*(i)^4\right)- (t_i^*-t_i)^2\\ &=\sum_{i=0}^{N-1}\mathbb E\left(\Delta_*(i)^2\Delta^*(i)^2\right)+\mathbb E\left(\Delta_*(i)^4\right)- (t_i^*-t_i)^2\\ &=\sum_{i=0}^{N-1} (t_i^*-t_i)(t_{i+1}-t_i^*)+3(t_i^*-t_i)^2- (t_i^*-t_i)^2\\ &=\sum_{i=0}^{N-1} (t_i^*-t_i)(t_{i+1}-t_i^*)+2(t_i^*-t_i)^2 \end{align*}

Now notice that \begin{align*} (t_{i+1}-t_i^*)(t_i^*-t_i)= \left(t_{i+1}-\frac{t_i+t_{i+1}}{2}\right)\left(\frac{t_i+t_{i+1}}{2}-t_i\right)=\frac{(t_{i+1}-t_i)^2}{4}, \end{align*} and \begin{align*} (t_i^*-t_i)^2=\frac{(t_{i+1}-t_i)^2}{4} \end{align*}

and thus the latter equals

\begin{align*} \frac{3}{4}\sum_{i=0}^{N-1} (t_{i+1}-t_i)^2\leq \frac{3}{4}\|\pi\|\sum_{i=0}^{N-1} (t_{i+1}-t_i)=\frac{3}{4}\|\pi\|T \end{align*} and the last term on the right vanished as $\|\pi\|\to 0$.

EDIT:

An interesting property is that if we replace the standard product $\times$ in $$\sum_{i=0}^{N-1}W\left(t_i^*\right)\times [W(t_{i+1})-W(t_i)],$$ with the so-called Wick product "$\diamond$", then the choice of the evaluation point is irrelevant in fact $$\sum_{i=0}^{N-1}W\left(t_i^{\alpha}\right)\diamond [W(t_{i+1})-W(t_i)]\to \int_0^T W(t)dW(t)$$ where $t_i^{\alpha}:=[1-\alpha]t_{i}+\alpha t_{i+1}$ for any choice of $\alpha\in [0,1]$.

This is due to the fact that the Wick product is somehow implicit in the Itô integration via the formula

$$\int_0^T f(W(t))dW(t)=\int_0^T f(W(t))\diamond \dot W(t)dt$$ where $\dot W(t)$ denotes the distributional derivative of the Brownian motion (i.e. a white noise process).