# Martingale property for two stochastic processes?

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!

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 $$u.
Now, $$\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$$).