Why an Itô Integral Has Zero Expected Future Change
Summary
The document explains why an Itô integral driven by Brownian motion is a martingale. It splits the integral at the current time: the part up to that time is already known, while the later part has conditional expectation zero under the usual assumptions. Thus the conditional expected future value equals the current value, and the expected change from the present is zero. A trader can interpret this as the absence of predictable drift in the accumulated gains from an adapted strategy in the model.
The explanation is conceptual rather than a full proof of the conditions. The integrand must meet suitable integrability and adaptation requirements, and the Brownian motion must be Brownian relative to the filtration in use. In particular, the claim that the future integral has conditional expectation zero is a property of the Itô integral under these assumptions, not merely a consequence of each Brownian increment having mean zero when the integrand may depend on information over the interval. The frictionless-market interpretation is illustrative and does not account for trading costs or other market constraints.
Key ideas
- An Itô integral can be split into a known past component and a future component.
- The past component is measurable at the current time, so its conditional expectation is its current value.
- Under suitable adaptation and integrability conditions, the future Itô integral has conditional expectation zero.
- The expected future change is zero even when the current accumulated integral is nonzero.
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Full text
# Prove that an Itô integral is a martingale
# Prove that an Itô integral is a martingale
### Background
> Q. Prove that an Itô integral with respect to Brownian motion is a martingale
### My Attempt
To prove that an Itô integral with respect to Brownian motion is a martingale (representing a "fair game" with no predictable drift), we look at the properties of the Itô integral stochastic process:$$I_t = \int_0^t H_s \, dW_s$$ where $W_s$ is a standard Brownian motion and $H_s$ is a suitably integrable, adapted process (meaning the trader cannot look into the future to pick their strategy $H_s$).
For $I_t$ to be a martingale, it must satisfy two main conditions for any future time $t$ and present time $s$ (where $s \le t$):
Proof. Let's split the integral into past and future segments. We can split the integral from $0$ to $t$ into two pieces: $\color{blue}{\text{from}\ 0 \ \text{to}\ s}$, and $\color{red}{\text{from}\ s \ \text{to}\ t}$:$$I_t = \int_0^t H_u \, dW_u = \color{blue}{\int_0^s H_u \, dW_u} + \color{red}{\int_s^t H_u \, dW_u}$$
Notice that the first part ($\int_0^s H_u \, dW_u$) is simply $I_s$, and it is already fully "known" (measurable) at time $s$.
Take the conditional expectation at time $s$:
$$\mathbb{E}[I_t \mid \mathcal{F}_s] = \mathbb{E}\left[ \color{blue}{\int_0^s H_u \, dW_u} + \color{red}{\int_s^t H_u \, dW_u} \Bigg\vert{} \mathcal{F}_s \right]$$
Split by linearity:$$\mathbb{E}[I_t \mid \mathcal{F}_s] = \mathbb{E}\left[ \color{blue}{\int_0^s H_u \, dW_u} \Bigg\vert{} \mathcal{F}_s \right] + \mathbb{E}\left[ \color{red}{\int_s^t H_u \, dW_u} \Bigg\vert{} \mathcal{F}_s \right]$$
Evaluate both parts:
- the past: since $\int_0^s H_u \, dW_u$ ($I_s$) has already happened by time $s$, its conditional expectation given current information is just itself:$$\mathbb{E}\left[ \color{blue}{\int_0^s H_u \, dW_u} \Bigg\vert{} \mathcal{F}_s \right] = \color{blue}{I_s}$$
- the future: by the fundamental property of Itô integrals over future intervals ($s$ to $t$), the expected value of future Brownian increments given current information is $0$ (the definition of a fair game with no drift):$$\mathbb{E}\left[ \color{red}{\int_s^t H_u \, dW_u} \Bigg\vert{} \mathcal{F}_s \right] = \color{red}{0}$$
Combine both of them:
$$\mathbb{E}[I_t \mid \mathcal{F}_s] = \color{blue}{I_s} + \color{red}{0} = I_s$$ Because the conditional expectation of the future Itô integral equals its current value ($\mathbb{E}[I_t \mid \mathcal{F}_s] = I_s$), the Itô integral is a martingale. This mathematically guarantees that we cannot systematically make or lose expected profit over time using a non-anticipating trading strategy in a frictionless market. $\blacksquare$
### My Question
If $\mathbb{E}[I_t \mid \mathcal{F}_s] = I_s$ means that the expected future value equals our current value, not zero. Therefore,
- if our current Itô integral value is $I_s = 0$, the expected future value is $0$?
- if our current value is positive (e.g., $I_s = \\\$500$ in profit) or negative (e.g., $I_s = -\\\$200$ in loss), our expected future value is whatever our current value is ($500$ or $-200$)?
- it's a "fair game" because our expected change from right now into the future is zero ($\mathbb{E}[I_t - I_s \mid \mathcal{F}_s] = 0$). Therefore, we don't have an automatic upward drift or downward pull, as we are expected to stay right where we currently are on average?
Did I get this correct?Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.