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Let $M=(M_t)_{1\le t\le 2}$ be a continuous (resp. right-continuous) martingale. Denote $x:=\mathbb E[M_1]\in\mathbb R$. Can we construct on some probability space a continuous (resp. right-continuous) martingale $N=(N_t)_{0\le t\le 2}$ such that

$$N_0=x \quad \mbox{and} \quad \mbox{Law}(M_t: 1\le t\le 2)=\mbox{Law}(N_t: 1\le t\le 2)?$$

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  • $\begingroup$ @user479223 Thanks for pointing out this typo. It has been corrected $\endgroup$
    – Fawen90
    Commented Feb 13 at 21:28
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    $\begingroup$ For right continuous, what is wrong with $N_t=x$ if $t<1$ and $N_t=M_t$ for $t\geq 1$? $\endgroup$
    – user479223
    Commented Feb 13 at 21:31
  • $\begingroup$ @user479223 I agree that it is straightforward for the right-continuous martingale $\endgroup$
    – Fawen90
    Commented Feb 13 at 21:48
  • $\begingroup$ For continuous, I believe Dubins-Schwarz works. There is a (possibly random) $A_t$ with $A_1=1$ a.s. so that $M_t=W_{A_t}$. So then let $N_t=W_t$ for $t<1$ and $W_{A_t}$ for $t\geq 1$. In this case we have a time changed Brownian motion on $[0,2]$ which is a martingale. $\endgroup$
    – user479223
    Commented Feb 13 at 22:24
  • $\begingroup$ You have $N_1=W_1$. I guess you wanted to say something else. $\endgroup$
    – Iosif Pinelis
    Commented Feb 13 at 22:40

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