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A quick question : I want to know if the product of two continuous martingales under a brownian filtration could be a martingale ? I have particularly the product of an exponential martingale (come from a Girsanov changement of probability) and a general martingale (both under a common probability P).

Obviously, I have read carrefully the Did's answer here and the reference he gaves (Alexander Cherny : Some Particular Problems of Martingale Theory).

Again, since we are under a brownian filtration is it possible to get the product of two martingales (a priori non-independant) to be a martingale ? or it is not as stronger as stating the independance of the 2 processes ?

Maybe we can make use the martingale representation theorem and find another condition less stronger ?

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Al Bundy
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Consider the martingales $X_t:=\int_0^t 1_{\{B_s>0\}}\,dB_s$ and $Y_t:=\int_0^t 1_{\{B_s<0\}}\,dB_s$, where $(B_s)$ is a Brownian motion. The covariation of these two martingales is $\int_0^t1_{\{B_s>0\}}1_{\{B_s<0\}}\,ds=0$, so the product $X_tY_t$ is a martingale.

John Dawkins
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  • Great example, that means that independance is not required everytime. But I need (to know if it exists) something general with two martingales under a brownian filtration. One of them is an exponential martingal. – Al Bundy Mar 23 '17 at 07:08
  • or maybe it must be studied on a case-by-case basis. – Al Bundy Mar 23 '17 at 09:06
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    In general, the product of two Brownian martingales $X_t=\int_0^t H_s,dB_s$ and $Y_s=\int_0^t K_s,dB_s$ will be a (local) martingale if and only if $\int_0^t H_sK_s,ds=0$ for all $t>0$. – John Dawkins Mar 23 '17 at 12:55
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    @the-owner Let $X_t=\exp[\int_0^t1_{{B_s>0}}dB_s-\frac12\int_0^t1_{{B_s>0}}ds]$ and $Y_t=\exp[\int_0^t1_{{B_s<0}}dB_s-\frac12\int_0^t1_{{B_s<0}}ds]$, then $X_tY_t=\exp[B_t-\frac{t}2]$ and all $X,Y,XY$ are positive continuous martingales. – JGWang Mar 25 '17 at 08:11