Why the Stratonovich Integral of Brownian Motion Is Not a Martingale

Summary

The document evaluates whether the Stratonovich integral of Brownian motion with respect to itself is a martingale. It derives the integral from symmetric partition sums: each term becomes a difference of squared Brownian values, and the sum telescopes. Continuity of the Brownian path then gives one half of the square of the process at the integration time, since its starting value is zero.

Applying Itô's lemma to that squared process shows that the result contains a time-dependent drift term, so it is not a martingale. The partition argument gives the Stratonovich expression, while Itô's lemma supplies the comparison needed to identify the drift. This is a focused stochastic calculus example rather than a trading method. The explanation relies on standard Brownian motion starting at zero and does not discuss broader integrability conditions or applications to asset models.

Key ideas

• The Stratonovich integral of Brownian motion against itself equals one half of its squared value when the process starts at zero.
• Symmetric partition sums reduce to a telescoping sum of differences in squared Brownian values.
• Itô's lemma expresses the squared Brownian motion with a time drift term.
• That drift means the Stratonovich integral is not a martingale.
• The derivation is a stochastic calculus result, not a trading strategy.

Tags

# Stratonovich Integral and Ito's lemma


# Stratonovich Integral and Ito's lemma












Let $(\Omega, \mathcal{F},\mathbb{P},\{\mathcal{F}\}_t)$ be a filtered- probability space and $W_t$ be standard Wiener process. I want to show stratonovich integral of $W_t$, i.e $\int_{0}^{t} W_s ○ dW_s$ , is not a martingale by Ito lemma.

Thanks.

## Answer by user16651 (score 5, accepted)

https://quant.stackexchange.com/a/28301

It is well-known, $W_t$ is a continues, adapted and locally bounded process.Let $I=\{t_i\}_{i=0}^{n}$ is a sequence of partitions of $[0,t]$, Indeed $0=t_0<t_1<\cdots<t_n=t$ .By definition of stratonovich integral, we have $$\int_{0}^{t} W_s\circ dW_s=\frac{1}{2}\underset{n\to \infty }{\mathop{\lim }}\sum_{i=0}^{n-1}(W(t_{i+1})+W(t_i))(W(t_{i+1})-W(t_i))$$ $$\int_{0}^{t} W_s\circ dW_s=\frac{1}{2}\underset{n\to \infty }{\mathop{\lim }}\sum_{i=0}^{n-1}(W^2(t_{i+1})-W^2(t_i))\quad(\operatorname{Telescoping series })$$ $$\int_{0}^{t} W_s\circ dW_s=\frac{1}{2}\underset{n\to \infty }{\mathop{\lim }}(W^2(t_n)-W^2(t_0))$$ Therefore $$\int_{0}^{t} W_s\circ dW_s=\frac{1}{2}(\,W^2(t)-W^2(0)\,)=\frac{1}{2}W_t^2\quad (\underset{n\to \infty }{\mathop{\lim }}W(t_n)=W(t))$$ Hence

> $$\int_{0}^{t} W_s\circ dW_s=\frac{1}{2}W_t^2$$

Now use Ito lemma.

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