# Expectation on a function of Wiener Process

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

• I get the feeling we are doing your homework for you?! – StackG Aug 2 at 10:15
• Certainly not. I am picking this expression from another post - quant.stackexchange.com/questions/28009/… – Bogaso Aug 2 at 10:25
• I would say it is zero almost by definition... The $dW_s$ terms are increments of a Brownian motion which are independent and normally distributed. The integral is a weighted sum of these, and each of them has an expectation of zero. Expectation of sums = sum of expectations, so $0$ – StackG Aug 2 at 11:42
• So if I try to generalise this with $f \left( W_s \right)$ instead of $\frac{1}{1+W_s}$, then still the expectation will be zero? – Bogaso Aug 2 at 18:27
• Yes setting $f(W_s)$ instead works as long as the integral is well-defined. – fesman Aug 2 at 18:38

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

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

• But I failed to understand this solution. I needed to prove the expectation is $0$ – Bogaso Aug 2 at 18:29
• @Bogaso I clarified the answer a bit. The point is that essentially any Ito integral has expectation zero as long as it is well-defined. Usually to see it is well-defined we check for the above square integrability condition. – fesman Aug 2 at 18:37