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Let $(\Omega, \mathcal{F},(\mathcal{F}_t)_{t≥0}, \mathbb{P})$ be a filtered probability space and let $(B_t)_{t≥0}$ be a Brownian motion with $B_0 = 0$.

Moreover assume that $\mathcal{F}_t := σ(B_s : 0 \leq s \leq t)$. Consider the two processes: $Y_t := \int_0^t B_u \, du$ , $ t\geq 0$ and $Zt := Y_t − tB_t \;$ , $ t\geq 0$.

How can I check the last property of martingale to show that the process $(Z_t)_{t≥0}$ is an $(\mathcal{F}_t)_{t\geq 0}$-martingale ? Thanks in advance for any help!

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Note first that $\mathbb{E}[B_u|\mathcal{F}_s]=B_s$ for all $u\geq s$ and $\mathbb{E}[B_u|\mathcal{F}_s]=B_u$ for all $u<s$.

Now, $$ \mathbb{E}[Z_t|\mathcal{F}_s] = \mathbb{E}\left[ \int_0^tB_udu-tB_t \Big| \mathcal{F}_s \right] \\ = \mathbb{E}\left[ \int_0^s B_udu-sB_t + \int_s^t B_udu-(t-s)B_t \Big| \mathcal{F}_s \right] \\ = \mathbb{E}\left[ \int_0^s B_udu-sB_t \Big| \mathcal{F}_s \right] + \mathbb{E}\left[ \int_s^t B_udu-(t-s)B_t \Big| \mathcal{F}_s\right] \quad(*)$$

The first expectations above simplifies to $$ \mathbb{E}\left[ \int_0^s B_udu-sB_t \Big| \mathcal{F}_s \right] = \int_0^s \mathbb{E}[B_u|\mathcal{F}_s ]du - \mathbb{E}[sB_t|\mathcal{F}_s] = \int_0^s B_u du -sB_s = Z_s $$

So it remains to show that the second expectations in $(*)$ is zero. I will leave that to you (simply use the fact that $\mathbb{E}[B_u|\mathcal{F}_s]=B_s$ for all $u\geq s$).

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