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Let $(\Omega,(\mathcal{F}_t)_{t≥0}, \mathcal{P})$ be a filtered probability space. Let $(B_1(t), B_2(t))_{t≥0}$ be a two-dimensional Brownian motion. Let $ θ\in \mathbb{R}$ and let $$ X_t^θ = B_1(t) \cos θ − B_2(t) \sin θ $$

Why $X_t^θ$ is a martingale ?

Joel
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    If $M_1(t)$ and $M_2(t)$ are two independent martingales with respect to the same filtration, $a_1 M_1(t) + a_2 M_2(t)$ is a martingale with respect to that filtration for any $a_i \in \mathbb{R}$. – snar Jan 21 '20 at 22:15
  • https://math.stackexchange.com/a/22376/16192 – gt6989b Jan 21 '20 at 22:15

1 Answers1

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For any $t\geqslant 0$ we have \begin{align} \mathbb E[|X_t^\theta|] &= \mathbb E[|B_1(t)\cos\theta - B_2(t)\sin\theta|]\\ &\leqslant |\cos\theta|\mathbb E[|B_1(t)|] + |\sin\theta|\mathbb E[|B_2(t)|]\\ &\leqslant 2\mathbb E[|B_1(t)|]\\ &= 2\sqrt{\frac{2t}\pi}<\infty, \end{align} so that $\{X_t^\theta\}$ is integrable. As for the martingale condition, if $0\leqslant s<t$ then \begin{align} \mathbb E[X_t^\theta\mid \mathcal F_s] &= \mathbb E[(X_t^\theta-X_s^\theta+X_s^\theta)\mid\mathcal F_s]\\ &=\mathbb E[X_t^\theta-X_s^\theta\mid\mathcal F_s] + \mathbb E[X_s^\theta\mid\mathcal F_s]\\ &= \mathbb E[X_t^\theta-X_s^\theta] + X_s^\theta\\ &= \mathbb E[B_1(t)\cos\theta-B_2(t)\sin(\theta)] + X_s^\theta\\ &= \cos\theta\mathbb E[B_1(t)] + \sin\theta\mathbb E[B_2(t)] + X_s^\theta\\ &= X_s^\theta, \end{align} since $\mathbb E[B_1(t)] = \mathbb E[B_2(t)] = 0$.

Math1000
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