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

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?

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*}
• Thanks. One quick question, what is $X_s$? Can it be any function, including a function of the Wiener process $W_s$ itself? Dec 8, 2020 at 17:44
• @Daniel Yes, it's any adapted (and square-integrable) process, so it may be a function of $W_s$ or indeed, $W_s$ itself. Dec 8, 2020 at 17:45