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Something is weird from a proof that I am reading:

The well-known theorem of characterization of quadratic variation states that:

Suppose $X$ is a continuous local martingale and $A$ is a continuous adapted process of bounded variation. Then the process $X^2-A$ is a local martingale if and only if $A = \langle X \rangle + A_0$.

The ''if'' direction of the proof starts by claiming that

There exists a sequence of stopping times $(T_n)$ that reduces $X$ to a continuous martingale such that $\sup_{t \geq 0} \mathbb{E} ((X^{T_n}_t)^2) < +\infty$, so $(X^{T_n})^2 - \langle X^{T_n} \rangle$ is a martingale.

My concern is: Why is it true that $\sup_{t \geq 0} \mathbb{E} ((X^{T_n}_t)^2) < +\infty$ ?

There is a theorem that states that

If $X$ is a continuous local martingale such that $X_0$ is bounded, then the sequence of stopping times defined by $S_n = \inf \{ t \geq 0 : |X_t | >n \}$ reduces $X$ to a bounded martingale, for each $n$.

However, in this case, $X_0$ is not assumed to be bounded. Hence, how does the claim follow?

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  • $\begingroup$ If $X_0 \notin L^2$ is not bounded, then there is no chance to define such a sequence of stopping times (just consider the "constant" martingale $X_t := X_0$). $\endgroup$
    – saz
    Commented Feb 25, 2015 at 8:28
  • $\begingroup$ That constant martingale that you have mentioned is a martingale if $ X_0 \in L^1 $. But how does this relate to my problem? $\endgroup$
    – Richard
    Commented Feb 25, 2015 at 8:53

1 Answer 1

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If $X_0$ is not bounded, then there does not necessarily exist a sequence of stopping times $(T_n)_n$ such that

$$\sup_{t \geq 0} \mathbb{E}((X_t^{T_n})^2) < \infty. \tag{1}$$

Indeed: For $X_0 \in L^1 \backslash L^2$ the process $X_t := X_0$ defines a continuous martingale, but obviosuly there does not a exist a sequence of stopping times $(T_n)_n$ such that $(1)$ holds.

However, in the original statement (i.e. the one about the uniqueness of the quadratic variation) we may assume without loss of generality that $X_0 =0$. Indeed: If $X_0 \neq 0$, then can consider the martingale

$$Y_t := X_t-X_0$$

instead. (Note that $\langle X \rangle = \langle Y \rangle$.)

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  • $\begingroup$ So by the result for $X_0 =0$, for general $(X_t)$, we know that $((X_t - X_0)^2 -A_t)$ is a local martingale $\implies A= \langle X \rangle + A_0$. But we have to show that $(X^2_t -A_t)$ is a local martingale $\implies A= \langle X \rangle + A_0$. $\endgroup$
    – Richard
    Commented Feb 25, 2015 at 22:03
  • $\begingroup$ and the problem to this issue is that we don't know whether $(X_t X_0)$ is a local martingale. $\endgroup$
    – Richard
    Commented Feb 25, 2015 at 22:06
  • $\begingroup$ Do you mean $X_t-X_0$? Why do you think that is is not a local martingale? $\endgroup$
    – saz
    Commented Feb 26, 2015 at 6:36
  • $\begingroup$ No, we are given that $(X^2_t - A_t)$ is a local martingale and have to show that $((X^2_t -2X_t X_0 + X^2_0) -A_t) = ((X_t-X_0)^2-A_t)$ is also a local martingale, in order to conclude that $A= \langle X \rangle + A_0.$ $\endgroup$
    – Richard
    Commented Feb 26, 2015 at 12:23
  • $\begingroup$ @Richard What's your definition of a local martingale? (There are (slightly) different ones; usually it's that $(X_t^{T_n} 1_{\{T_n>0\}})_{t \geq 0}$ is a martingale for a sequence of stopping times $T_n \uparrow \infty$.) $\endgroup$
    – saz
    Commented Feb 26, 2015 at 16:34

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