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

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    $\begingroup$ I get the feeling we are doing your homework for you?! $\endgroup$ – StackG Aug 2 '20 at 10:15
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    $\begingroup$ Certainly not. I am picking this expression from another post - quant.stackexchange.com/questions/28009/… $\endgroup$ – Bogaso Aug 2 '20 at 10:25
  • $\begingroup$ 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 $0$ $\endgroup$ – StackG Aug 2 '20 at 11:42
  • $\begingroup$ 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? $\endgroup$ – Bogaso Aug 2 '20 at 18:27
  • $\begingroup$ Yes setting $f(W_s)$ instead works as long as the integral is well-defined. $\endgroup$ – fesman Aug 2 '20 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


To see that the square integrability condition is satisfied note

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

  • $\begingroup$ But I failed to understand this solution. I needed to prove the expectation is $0$ $\endgroup$ – Bogaso Aug 2 '20 at 18:29
  • $\begingroup$ @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. $\endgroup$ – fesman Aug 2 '20 at 18:37

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