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Let $Y_1,...,Y_T$ be iid. random variables with $E_P(Y_1)=0$ and $P(Y_1\neq 0)>0$. Consider the filtration generated by $Y$, i. e. $\mathcal{F}_0=\{\emptyset, \Omega\}$ and $\mathcal{F}_t=\sigma(Y_1,...,Y_t)$. Furthermore, define $S_t=Y_1+\cdots +Y_t$. So I know that $S$ is a martingale with respect to $(\mathcal{F}_t)_{t=0,...,T}$.

I have to show that $S$ is not a martingale with respect to the "enhanced" filtration $\mathcal{F}^*_t=\sigma(\mathcal{F}_t,S_T)$.

My attempt: So, if we know at time $t$ the value of $S_T$ consider $\mathbb{E}(S_T\mid \mathcal{F}^*_{T-1})$. If $S$ was a martingale, it would be $=S_{T-1}$ but since we know the end value it would intuitively be $S_T$. How can I make this more rigorous?

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  • $\begingroup$ Cancelling the word "intuitively" makes it imho fully rigorous. $\endgroup$
    – Kurt G.
    Commented Sep 3, 2023 at 18:03
  • $\begingroup$ Thanks, @KurtG. The argument is convincing, at least for me. It depends on the fact that the filtration models the knowledge. $\endgroup$
    – Analysis
    Commented Sep 3, 2023 at 18:21
  • $\begingroup$ So it does. With that enhanced filtration we look into the future (an enticing and horrifying possibility at the same time). Then the martingale condition leads to $S_T=S_{T-1}$ which contradicts that the $Y$'s are random and not necessarily zero. $\endgroup$
    – Kurt G.
    Commented Sep 3, 2023 at 18:25
  • $\begingroup$ @KurtG. In finance, it's a way to model insider information. $\endgroup$
    – Analysis
    Commented Sep 3, 2023 at 18:43
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    $\begingroup$ True. Horrifying as I said. $\endgroup$
    – Kurt G.
    Commented Sep 3, 2023 at 18:53

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