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Define $B_t=W_t - tW_1$. Let $\mathcal{F}_t$ and $\mathcal{G}_t$ be the filtrations generated by $\{W_t\}$ and $\{B_t\}$ respectively. Is $B_t$ a martingale wrt (i) $\mathcal{F}_t$ and (ii) $\mathcal{G}_t$?

For $\mathcal{F}_t$:

\begin{align} E[B_t\mid \mathcal{F}_s]=W_s -t E[W_1\mid \mathcal{F}_s]=W_s(1-t)\ne B_s. \end{align}

For $\mathcal{G}_t$:

\begin{align} E[B_t\mid \mathcal{G}_s]= E[W_s -sW_1 + W_t - W_s +(s-t)W_1\mid \mathcal{G}_s]= B_s+E[W_t-W_s\mid \mathcal{G}_s]+(s-t)E[W_1\mid \mathcal{G}_s]= B_s. \end{align}

Is the above reasoning correct?

Jose Avilez
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Bravo
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1 Answers1

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It doesn't seem right to me. It seems that you assume that $E[W_1 \, | \, \mathcal{G}_s] = E[W_1] = 0$, but this is not the case. Notice that $W_1$ must be $\mathcal{G}_s$-measurable (otherwise $B_t$ would not be adapted to its own filtration), so that in fact $E[W_1 \, | \, \mathcal{G}_s] = W_1$. As a result, your computation should yield: $$E[B_t \, | \, \mathcal{G}_s ] = B_s + (s-t)W_1$$

Jose Avilez
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