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Let $(\Omega,\mathcal A,\operatorname P)$ be a probability space, $(\mathcal F_t)_{t\ge0}$ be a filtration on $(\Omega,\mathcal A,\operatorname P)$ and $(X_t)_{t\ge0}$ and $(Y_t)_{t\ge0}$ be real-valued $\mathcal F$-progressive processes on $(\Omega,\mathcal A,\operatorname P)$ with $$\operatorname E\left[|X_t|\int_0^t|Y_s|\:{\rm d}s\right]<\infty\tag1$$ and $$\operatorname E\left[\int_0^t|X_sY_s|\:{\rm d}s\right]<\infty\tag2$$ for all $t\ge0$. Assume that $X$ is an $\mathcal F$-martingale.

How can we show that $$M:=X\int_0^{\;\cdot\;}Y_s\:{\rm d}s-\int_0^{\;\cdot\;}X_sY_s\:{\rm d}s$$ is an $\mathcal F$-martingale.

Using that $X$ is an $\mathcal F$-martingale, I was able to write $$\operatorname E\left[M_t\mid\mathcal F_s\right]=M_s\color{blue}{+\operatorname E\left[X_t\int_s^tY_r\:{\rm d}r\mid\mathcal F_s\right]-\operatorname E\left[\int_s^tX_rY_r\:{\rm d}r\mid\mathcal F_s\right]}\tag3.$$

How can we show that the blue term is $0$?

0xbadf00d
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1 Answers1

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$$\mathbb E\left[X_t\int_s^t Y_r\,\mathrm d r\mid \mathcal F_s\right]-\mathbb E\left[\int_s^t X_rY_r\,\mathrm d r\mid \mathcal F_s\right]=\mathbb E\left[\int_s^t Y_r(X_t-X_r)\,\mathrm d r\mid \mathcal F_s\right]$$$$\underset{(*)}{=}\int_s^t\mathbb E\big[Y_r(X_t-X_r)\mid \mathcal F_s\big]\,\mathrm d r.$$ A reference for $(*)$ is given here. Using the fact that $X$ is a martingale and the tower property of the conditional expectation, it follows that $$\mathbb E[Y_r(X_t-X_r)\mid \mathcal F_s]=\mathbb E\Big[\mathbb E\big[Y_r(X_t-X_r)\mid\mathcal F_r\big]\mid \mathcal F_s\Big]=0.$$

Surb
  • 55,662