I haven't any idea how to attack this problem:
Let $M$ a continuous martingale and a Gaussian process. Prove that there exists a (deterministic) continuous function $f$ on $\mathbb{R}_{+}$ such that $M^2_t-f(t)$ is a martingale.
Can someone help me?
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Hint to get started: What would $f$ have to be? (Maybe you can express it in terms of the covariance function of $M$.) – Nate Eldredge Oct 30 '16 at 22:05
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In my opinion $f(t)=E(M_t^2)$ but I haven't idea how to show it – foubw Oct 30 '16 at 22:17
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I also know that such a function exists then $f(t)-f(s)=E(M_t^2-M^2_s)=E((M_t-M_s)^2)$ – foubw Oct 30 '16 at 22:29
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Can you give me a sketch of solution for the existence? – foubw Oct 30 '16 at 23:00
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Yes, $f(t) = E[M_t^2]$ is right. If $M_t^2-f(t)$ is a martingale (which, say, is zero at $t=0$) then $E[M_t^2 - f(t)] = 0$ for all $t$. – Nate Eldredge Oct 31 '16 at 00:56
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But how can I show that f(t) exists? – foubw Oct 31 '16 at 06:42
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Why need to be 0 at zero? – foubw Oct 31 '16 at 06:43
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I am saying you should define $f(t) = E[M_t^2]$ and then try to show that this choice makes $M_t^2 - f(t)$ into a martingale. There is no loss of generality in assuming $M_0^2 - f(0) = 0$ because you can add any constant to $f(t)$ without breaking the desired property. – Nate Eldredge Oct 31 '16 at 13:41