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How does one show that:

$$ \mathbb{E}\left[ \int f(W_s)dWs \right] = 0 $$

For all $f()$ that are powers of $W(s)$?? I assume that one would have to go via the definition of Ito integral and express the integral as a sum over martingale differences?

I tried doing that, but it didn't work for me: Taking $f(W(s))=W(s)$ and splitting the intergal into finite "constant" parts:

$$ \mathbb{E}\left[ \int f(W_s)dWs \right] = \mathbb{E} \sum_iW_i(W_i-W_{i-1}) = \sum_i\mathbb{E}[W_i^2-W_iW_{i-1}]=\sum_i\mathbb{E}W_i^2\neq0 $$

Obviously, the above is not the way to do show it. Any hints pls?

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3 Answers 3

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That the expectation is zero is often called the martingale property of Ito integral (see e.g. Oksendal Theorem 3.2.1). The formal proof consists of showing this for "simple" integrand functions and then generalising this by taking limits. This requires that the integrand process is adapted (i.e. not forward looking) and square integrable. Square integrability is important because in general the expectation of an Ito integral can take any value as explained here: https://math.stackexchange.com/questions/232932/it%C5%8D-integral-has-expectation-zero. However, these technical conditions are usually satisfied in practical applications. In your case it follows from the fact that the Wiener process has finite moments.

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An Ito integral is a martingale, and thus its expectation at anytime is it's value at t=0 - which is trivially 0; because the lower and upper limit of the integral would be 0.

For proof of martingality, you can refer to Shreve. It uses the definition of the ito integral by looking at it as the sum of many random variables generated from slicing the time axis. From martingality of Brownian motion, the proof follows.

Intuitively, you can then see the Ito integral then as the cummulative result of randomly allocating 'weights' (the Brownian increments) to the integrand. These weights are allocated independently of each other, and independent of their respective integrands ( you can't assign systematically higher/lower weights to a time point with a higher/lower integrand) . You would thus expect the sum to not be biased positively or negatively - since the assignment is at random and can not use the knowledge of the integrand so as to bias the sum. This is the martingale property.

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  • $\begingroup$ Thank you: how does one prove that an Ito Integral is a martingale? So basically, the expected value of an Ito integral over ANY integrand is zero? $\endgroup$ Aug 12, 2020 at 10:48
  • $\begingroup$ Just added it to the answer. You're right, with some technical condition. $\endgroup$
    – Arshdeep
    Aug 12, 2020 at 21:15
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A couple of things are required to make this work, the two key points are:

  1. The Ito Integral is a Martingale only when the integrand is not forward-looking

ie. when we DEFINE the summation to be this: \begin{align} \int^t_0 W_t dW_t = \sum^N_{i=1} W_{i-1}\bigl( W_i - W_{i-1}\bigr) \end{align}

As pointed out in the comments, this wouldn't matter in the Rienmann world, but in Ito calculus summing $W_i$ instead of $W_{i-1}$ gives us a different result.

Note the $i$ and $i-1$ terms, they will be important at the next step.

Some more formal proofs of this can be found here (page 17) and here (page 15)

  1. Your Expectation misses the correlation of $W_i$ and $W_{i-1}$

\begin{align} {\mathbb E} \Bigl[ W_iW_{i-1} - W_{i-1}^2\Bigr] &= {\mathbb E} [\Bigl(W_i - W_{i-1} + W_{i-1}\Bigr)W_{i-1} - W_{i-1}^2]\\ &= {\mathbb E} [ \Bigl(W_i - W_{i-1}\Bigr)W_{i-1} ] + {\mathbb E} [ W_{i-1}^2 ] - {\mathbb E} [ W_{i-1}^2 ]\\ &= {\mathbb E} [ \Bigl(W_i - W_{i-1}\Bigr)W_{i-1} ]\\ &= 0 \end{align}

Where ${\mathbb E} [ \Bigl(W_i - W_{i-1}\Bigr)W_{i-1} ] = 0$ because of independent increments in the Weiner process

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  • $\begingroup$ In fact, I might have made an additional mistake: it should be $\mathbb{E}[W_{i-1}(W_i-W_{i-1})]$, rather than $\mathbb{E}[W_{i}(W_i-W_{i-1})]$, isn't that right? $\endgroup$ Aug 12, 2020 at 12:03
  • $\begingroup$ Yup, needs to be forward diff too, great spot (I missed that earlier) $\endgroup$
    – StackG
    Aug 12, 2020 at 12:10
  • $\begingroup$ Why does it NEED to be forward diff? In fact, most definitions go with the integrator being defined as $W_{i+1}-W_i$ for each $i$: but if the $W(t)$ is adapted to $t$, how can we compute $W_{i+1}$ at each $i$? I am not sure I comprehend that. $\endgroup$ Aug 12, 2020 at 12:13
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    $\begingroup$ Another mystery of Ito calculus!! Just has to be forward... $\endgroup$
    – StackG
    Aug 12, 2020 at 12:32
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    $\begingroup$ @StackG In a Riemann world, once you build blocks of rectangles, it doesn't matter the height you build it at, the function is smooth enough that in the limit, the results converge to a single number called the definite integral. Here, BM path is not smooth, no matter how small you make your time slice, you can never approximate it by a linear step. This is because of the order of standard deviation is square root of time slice. BM vibrates a lot in any time slice. Thus it starts to matter which height you choose -Ito integral is the one where you choose the left hand height. $\endgroup$
    – Arshdeep
    Aug 12, 2020 at 22:18

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