Why a Bounded Itô Integral Has Zero Expectation

Summary

The note shows how to establish that the Itô integral of the process 1/(1 + W_s²) over a finite interval has expectation zero. It applies the martingale property of the Itô integral: an adapted integrand with finite expected time-integrated square is square-integrable, and its integral against standard Wiener motion has mean zero.

For this specific integrand, its squared value is bounded above by one, so its expected integral over any finite time interval is finite. This verifies the condition needed for the result. The argument is a general stochastic-calculus technique, rather than a trading strategy or market study; it provides no empirical evidence or pricing application.

Key ideas

• An Itô integral has zero expectation when its adapted integrand is square-integrable over the time interval.
• The squared integrand 1/(1 + W_s²)² is bounded above by one.
• The bound ensures finite expected integrated square on every finite interval.
• The result follows from the martingale property of the Itô integral.

Tags

# Expectation on a function of Wiener Process


# Expectation on a function of Wiener Process












If $W_t$ is a standard Wiener Process, then how should I prove that $E \left[ \int\limits_{0}^{t} \frac{1}{1+W_s^2} dW_s \right] = 0$?

## Answer by fes (score 2, accepted)

https://quant.stackexchange.com/a/57084

The proof uses the martingale property of the Ito integral. For an adapted stochastic process $X_t$ such that

$$\mathbb{E}\int_0^{t}|X_s|^2ds <\infty$$

we have

$$\mathbb{E}\int_0^{t}X_sdW_s =0$$

Now your result follows by setting

$$X_t=\frac{1}{W_t^2+1}.$$

To see that the square integrability condition is satisfied note

$$\mathbb{E}\int_0^{t}\frac{1}{(W_s^2+1)^2}ds <\int_0^{t}\frac{1}{(0+1)^2}ds=\int_0^{t}1ds<\infty$$

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