Why Adapted Itô Integrals Have Zero Expectation

# Ito Integral of functions of Brownian motion


# Ito Integral of functions of Brownian motion












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?

## Answer by fes (score 3)

https://quant.stackexchange.com/a/57284

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.

## Answer by Arshdeep (score 2)

https://quant.stackexchange.com/a/57283

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.

## Answer by StackG (score 1)

https://quant.stackexchange.com/a/57281

A couple of things are required to make this work, the two key points are:

- 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)

- 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