Why a Time-Dependent Stochastic Integral Need Not Be a Martingale

Summary

The document asks whether the time integral of Brownian motion is a martingale. Although integration by parts rewrites it using a stochastic integral, the resulting integrand depends on the terminal time. The accepted answer explains that the familiar martingale result applies when the integrand is a suitable square-integrable process indexed by the integration variable, not when it changes with the endpoint in this way. Thus the rewrite does not establish that the original process is a martingale.

A second response emphasizes that the two expressions under discussion are not equal as processes along each path, even though they agree at the final time. The exchange gives a conceptual warning about applying stochastic-integral results without checking their conditions. It does not provide a full formal proof or a detailed filtration and integrability analysis, so readers should treat it as an explanation of the key distinction rather than a complete theorem statement.

Key ideas

• A stochastic integral’s martingale property depends on the integrand meeting the relevant conditions.
• An integrand that depends on the terminal time cannot be treated as a fixed admissible integrand in the same way.
• Two processes may agree at a final time without being pathwise equal throughout the interval.
• Integration by parts alone does not prove that the original time integral is a martingale.

Tags

# Question about the martingale property of stochastic integral


# Question about the martingale property of stochastic integral












Let $W_{t}$ be a Wiener process, and let $$X_{t} = \int^{t}_{0}W_{\tau}d\tau$$ Is $X_{t}$ a martingale? We can rewrite in differential form as $$dX_{t} = W_{t}dt$$ ,which means $X_{t}$ is a diffusion process with only drift part $W_{t}$ and therefore $X_{t}$ is not a martingale. However, by integration by parts, we have \begin{align} X_{t} &= t W_{t} - \int^{t}_{0}\tau dW_{\tau}\\ &=t \int^{t}_{0}dW_{\tau} - \int^{t}_{0}\tau dW_{\tau}\\ &= \int^{t}_{0}(t-\tau)dW_{\tau} \end{align} $t-\tau$ is a deterministic, square integrable function, according to the martingale property of stochastic integral, $X_{t}$ is a martingale. Now my question is which analysis above is the right one? Thanks in advanced.

## Answer by Gordon (score 4, accepted)

https://quant.stackexchange.com/a/22842

If $X_t$ is square integrable, then the integral \begin{align*} \int_0^t X_{\tau} dW_{\tau} \end{align*} is a martingale. Here, the integrand $X_{\tau}$ does not depend on the integral limit $t$. However, in your case, the integrand, $t-\tau$, depends on $t$, then the condition for the martingality of the integral fails.

## Answer by user9403 (score 1)

https://quant.stackexchange.com/a/22858

The two processes are not pathwise equal. Here is a simulation (sample path) of the two processes $(t-\tau)dW_{\tau}$ and $W_\tau d\tau$:

Note that both processes have the same value at the final time.

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