Why a Bounded Itô Integrand Gives a Zero-Mean Integral

Summary

The document shows why the expectation of an Itô integral with integrand equal to the reciprocal of one plus squared Brownian motion is zero. Its argument is that an Itô integral is a martingale when the integrand’s expected squared value is integrable over the time interval. A martingale starting at zero has expectation zero at each later time.

For this integrand, the squared value is bounded above by one, so its expected integral over a finite interval is at most the interval length and is finite. That verifies the condition needed to use the martingale property. The result depends on the integrability condition and the standard Itô integral setup; the explanation does not address other integrands for which that condition may fail. The topic is a stochastic-calculus result rather than a trading strategy or empirical market analysis.

Key ideas

• An Itô integral is a martingale when the integrand is square-integrable over the time interval.
• A martingale that starts at zero has zero expectation at later times.
• The squared integrand in this example is bounded by one.
• Over a finite interval, that bound ensures the required integrability condition.

Tags

# Expectation of $\int_0^t \frac{1}{1+W_s^2} \text dW_s$


# Expectation of $\int_0^t \frac{1}{1+W_s^2} \text dW_s$












I am trying to calculate the expectation of

$$\int\limits_0^t \frac{1}{1+W_s^2} \text dW_s,$$

where $(W_t)$ is a Wiener process.

I was told that the value of this expectation is zero. Can someone please provide any clue why it would be zero?

## Answer by Kevin (score 11, accepted)

https://quant.stackexchange.com/a/59831

By construction, the Itô integral, $I_t=\int_0^t X_s\text{d}W_s$, is a martingale if $\int_0^t \mathbb{E}[X_s^2]\text{d}s<\infty$.

The martingale property, $\mathbb{E}_s[I_t]=I_s$ implies $\mathbb{E}[I_t]=I_0=0$.

Because $W_s\overset{d}{=}\sqrt{s}Z$, where $Z\sim N(0,1)$, we indeed have \begin{align*} \int_0^t\mathbb{E}\left[\frac{1}{(1+W_s^2)^2}\right]\text{d}s &= \int_0^t\frac{1}{\sqrt{2\pi}}\int_\mathbb{R}\frac{1}{(1+sz^2)^2}e^{-\frac{1}{2}z^2}\text{d}z\text{d}s \\ &\leq \int_0^t\frac{1}{\sqrt{2\pi}}\int_\mathbb{R}e^{-\frac{1}{2}z^2}\text{d}z\text{d}s\\ &=\int_0^t1\text{d}s \\ &=t<\infty. \end{align*}

@NHN suggests using the above argument, $\frac{1}{(1+x^2)^2}\leq1$ for all $x\in\mathbb{R}$, to directly get \begin{align*} \int_0^t\mathbb{E}\left[\frac{1}{(1+W_s^2)^2}\right]\text{d}s &\leq \int_0^t\mathbb{E}\left[1\right]=t<\infty. \end{align*}

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.