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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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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? $\endgroup$ – Novice555 Aug 12 '20 at 10:48
  • $\begingroup$ Just added it to the answer. You're right, with some technical condition. $\endgroup$ – Arshdeep Singh Duggal Aug 12 '20 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? $\endgroup$ – Novice555 Aug 12 '20 at 12:03
  • $\begingroup$ Yup, needs to be forward diff too, great spot (I missed that earlier) $\endgroup$ – StackG Aug 12 '20 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. $\endgroup$ – Novice555 Aug 12 '20 at 12:13
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    $\begingroup$ Another mystery of Ito calculus!! Just has to be forward... $\endgroup$ – StackG Aug 12 '20 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 Singh Duggal Aug 12 '20 at 22:18
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My take on this would be via the intuitive understanding of an Ito Integral. I feel it's best to interpret the Ito Integral via relating it to a gambling game: the integrator (i.e. the Brownian motion with respect to which we are integrating) is the (random) outcome of the gambling game, whilst the integrand (the function we are integrating) is the betting strategy. The betting strategy can be deterministic or random.

By design, at each point in time when the betting strategy is placed, the (random) outcome of the gambling game is not yet known, similarly to playing a roulette in a casino (hence why the integrator has to be forward-looking: by design, when the bet is placed (i.e. $f()$ becomes known), the game outcome (i.e. the integrator $W(t)$) is not yet known.

I believe that we can construct the Ito Integral both: (a) from the better's time point of view as well as (b) from the casino's time point of view:

(a) Ito Integral from the better's time point: let $f(\omega_{t_i},t_i)$ be the (possibly random) bet at time $t_i$, with $f(t_0)$ being the initial bet and $\omega_{t_i}$ denoting some random outcome at time $t_i$ ($\omega_t$ is adapted to the same filtration as $W_t$).

Indeed, the bet could be deterministic and even constant, in which case $f(\omega_{t_i},t_i)=k$, or it could be related to the outcomes that gradually become known as the game goes on, i.e. $f(\omega_{t_i},t_i)=f(W_{t_i})$

In general:

$$ I(f(\omega_h,h))_{h=t_0}^{h=t_n}:=\int_{h=t_0}^{t_n}f(\omega_h,h)dW(h)=\lim_{n \to\infty}\sum_{i=0}^{i=n-1}f(\omega_{t_i},t_i)\left(W(t_{i+1})-W(t_i) \right)$$

Above, at each time point, the better places a bet but does not yet know the random outcome of the game at the next time point.

(b) Ito Integral from the casino's time point: let $f(\omega_{t_i},t_i)$ be the (possibly random) bet at time $t_i$, with $f(t_0)$ being the initial bet. Then:

$$ I(f(\omega_h,h))_{h=t_0}^{h=t_n}=\lim_{n \to\infty} \sum_{i=1}^{i=n}f(\omega_{t_{i-1}},t_{i-1})\left(W(t_{i})-W(t_{i-1}) \right)$$

Above, at each time point, the casino knows the outcome of the random game, but it had known the better's bet before the random game had commenced.

Bottom line: intuitively, the expected value of the Ito integral is zero, because the integrator (i.e. the random game) is (by design) independent of the betting strategy. Since the integrator is a sum of independent Brownian motion increments, the expected value of Ito integral has to be zero, i.e.:

$$\mathbb{E}[I(f(\omega_h,h))_{h=t_0}^{h=t_n}]\approx \mathbb{E} \left[ \sum_{i=0}^{i=n-1}f(\omega_{t_i},t_i)\left(W(t_{i+1})-W(t_i) \right) \right]=\\=\sum_{i=1}^{i=n}\mathbb{E}[f(\omega_{t_i},t_{i-1})] \mathbb{E}[\left(W(t_{i})-W(t_{i-1}) \right)]=\\=\sum_{i=1}^{i=n}\mathbb{E}[f(\omega_{t_i},t_{i-1})] *0=0$$

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  • $\begingroup$ This assumes that $f$ (the betting strategy) is independent of the Brownian motion (gamble) and hence deterministic? But this logic wouldn't apply to OP's problem of $\int f(W_s)dW_s$, where the betting strategy explicitly depends on the gamble? In particular, $E\left[ f(W_{t_i})(W_{t_{i+1}}-W_{t_i})\right] \neq E[f(W_{t_i})]E[(W_{t_{i+1}}-W_{t_i})] $ due to potential covariance terms? $\endgroup$ – Alex Aug 18 '20 at 10:36
  • $\begingroup$ if the betting strategy is $W_t$, the game outcome is still independent of the betting strategy, because the game is "one step ahead": so the better sets his bet to whatever the outcome of the game had just materialized, but the outcome of the next round of the game only happens at the next time step. Coincidentally, $E\left[ W_{t_i}(W_{t_{i+1}}-W_{t_i})\right] = E[W_{t_i}]E[(W_{t_{i+1}}-W_{t_i})]=0$, but as you rightly point out, it is because the covariance term cancels out the $\mathbb{E}W_{t_i}^2$ term. $\endgroup$ – Jan Stuller Aug 18 '20 at 10:44
  • $\begingroup$ But yeh, in general, $E\left[ f(W_{t_i})(W_{t_{i+1}}-W_{t_i})\right] \neq E[f(W_{t_i})]E[(W_{t_{i+1}}-W_{t_i})]$ as $f()$ is a function of the integrator. $\endgroup$ – Jan Stuller Aug 18 '20 at 10:46
  • $\begingroup$ So we cannot use this gamble analogy to answer OP's question that $E[\int f(W_s)dW_s]=0$ because it only works if $f(x)=x$ (and some other special cases)? The equation in your first reply just corresponds to the independent increments of Brownian motion, I think. $\endgroup$ – Alex Aug 18 '20 at 10:55
  • $\begingroup$ I amended my answer to reflect that the gambling strategy can be random, and not just $f(t)$, i.e. deterministic function of time. Let me think a bit more about the specific case where $f()=f(W_t)$, but I think the gambling analogy still definitely works: because in the case where $f()=f(W_t)$, the better randomly places a bet that is equal to the last observed game outcome (with the game being the integrator $W_t$). $\endgroup$ – Jan Stuller Aug 18 '20 at 11:44

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