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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.

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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.

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  • $\begingroup$ Thank you very much! However, when it comes to the distribution of $X_{t}$, it is usual to use the theorem: if $f(t)$ is a deterministic square integrable function, then the stochastic integral $\int ^{t}_{0}f(\tau)dW_{\tau}$ is a normal random variable of mean 0 and variance $\int ^{t} _{0} |f(\tau)|^{2}d\tau$. Therefore, $X_{t} \sim N(0, \int ^{t}_{0}(t-\tau)^{2}d\tau)$. That is to say, $t-\tau$ should be a deterministic square integrable function, which is also the condition for the martingality. How to explain the distribution of $X_{t}$? $\endgroup$ – cmd1991 Jan 21 '16 at 9:22
  • $\begingroup$ @cmd1991: The distribution of $\int_0^t(t-\tau)dW_{\tau}$ is normal, as the limit in $L^2$ or in probability of normal random variables is normal. Though the martingality condition fails, however, Ito's isometry formula still holds. That is, your distribution is fine. $\endgroup$ – Gordon Jan 21 '16 at 13:41
  • $\begingroup$ How does the martingality fail? It looks square integrable to me... $\endgroup$ – user9403 Jan 21 '16 at 22:56
  • $\begingroup$ @user9403: Martingality requires that the integrand is independent of the integral limit. $\endgroup$ – Gordon Jan 22 '16 at 0:14
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    $\begingroup$ @cmd1991: Any stochastic calculus book has a discussion of the martingale property for $\int_0^tX_{\tau}dW_{\tau}$. However, you will not find any martingale property for $\int_0^t X(\tau, t)dW_{\tau}$. See, for example, the book by Karatzas and Shreve: amazon.com/Brownian-Stochastic-Calculus-Graduate-Mathematics/dp/…. $\endgroup$ – Gordon Jan 22 '16 at 13:45
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The two processes are not pathwise equal. Here is a simulation (sample path) of the two processes $(t-\tau)dW_{\tau}$ and $W_\tau d\tau$: Martingale and BM Average

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

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  • $\begingroup$ What are the two processes? Do you mean the $(t-\tau)dW_{\tau}$ and $W_{\tau}d\tau$? $\endgroup$ – cmd1991 Jan 22 '16 at 3:25
  • $\begingroup$ Yes, I've edited my answer to clarify. $\endgroup$ – user9403 Jan 22 '16 at 10:36
  • $\begingroup$ @cmd1991: $(t-\tau)dW_{\tau}$ and $W_{\tau}d\tau$ are not processes. They are some incremental changes. $\endgroup$ – Gordon Jan 22 '16 at 13:55

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