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Consider $M= \left(M\right )_{t \geq0}, \ N=\left(N\right) _{t \geq0} \in \mathcal M_{c,loc} $ starting both from zero, such that, a.e.$ \langle M \rangle_t =\langle N \rangle_t, \ \forall t\geq 0$.

Have $M$ and $N$ the same law? If not give a contre exemple.

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No. $M$ and $N$ need not have the same distribution. For example, take $$ Z_t = \exp\left\{B_t - \frac{t}{2}\right\},\qquad M_t = Z_t - 1,\qquad N_t = 1 -Z_t. $$

The continuous martingales $M$ and $N$ have the same quadratic variation $\langle Z\rangle_t$, but since $Z_t$ tends to $0$ as $t \to +\infty$ a.s. they cannot have the same distribution.

Proof that $Z_t \xrightarrow[t\to\infty]{} 0$ a.s.

We know that $\dfrac{B_t}{t} \xrightarrow[t\to\infty]{} 0$ as a consequence of the strong law of large numbers (or the law of the iterated logarithm). From this we deduce that $$Z_t = \exp\left\{t \left(\frac{B_t}{t}-\frac{1}{2}\right) \right\}\xrightarrow[t\to\infty]{} 0\quad\text{a.s.} $$

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How to show that $Z_t \longrightarrow 0$ as $t \rightarrow + \infty$ ? –  Paul Jan 18 '13 at 23:40
    
+1. Neat. $ $ $ $ –  Did Jan 19 '13 at 11:52
    
How to show that $t Y_t \rightarrow \infty $ where $Y_t =\frac{B_t}{t} \rightarrow \infty$? I'm not confortable with the limt of the product of t tending to infinit with something wich tends to zero. I'm not convinced –  Paul Jan 19 '13 at 15:29
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Hmm. You might want to read again what I wrote. –  Siméon Jan 19 '13 at 18:30
    
@Ju'x : Sorry I make a mistake when writing. I'd like to say: How to assurate that $tY_t \rightarrow 0 $ a.s. if $Y_t \rightarrow 0 $ a.s, particularly when $Y_t =\frac{B_t}{t} $ ? –  Paul Feb 15 '13 at 18:30
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