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

Question

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 1

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*}