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To me this is a bit of a curve-ball as we usually only deal with the form $E[\cdot \mid Y_n]$.

The usual way to go about would be to prove $E[X_{n+1} \mid Y_n^2] = X_n$. I do not quite know where to start with regards to this due to the mentioned curve-ball.

$\textbf{My Attempt...of sorts}$

If $\{Y_n\}$ is some random variable along with the fact that $X_n$ is a martingale with respect to $\{Y_n\}$, we can assume $Y_n = X_n$.

Our problem then becomes to prove $E[X_{n+1} \mid X_n^2] = X_n$.

From this previously asked question, we see that $$E(X | X^{2}) = g(|X|) = |X| \frac{f(|X|) - f(-|X|)}{f(|X|) + f(-|X|)}. $$

From this we would conclude that $E[X_{n+1} \mid X_n^2] \neq X_n$ and thus $X_n$ is not a martningale with respect to $\{Y_n^2\}$.

Would you please guide me in answering this and perhaps trying to understand it in a simple manner. Understanding is key as I am going to try and attempt to answer similar questions such as whether $X_n^2$ is a martingale with respect to $\{Y_n^2\}$. Thank you very much.

Yes
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Let $\left(Y_i\right)_{i\geqslant 1}$ be an i.i.d. sequence of random variables, where $Y_1$ takes the values $-1$ and $1$ with probability $1/2$. The sequence $\left(X_n\right)_{n\geqslant 1}$ defined by $X_n=\sum_{i=1}^nY_i$ is a martingale for the natural filtration generated by the $Y_i$. But the one generated by $Y_i^2$ is the trivial filtration.

Davide Giraudo
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  • Thank you so very much. Just to check that I understand: $Y_i^2$ is the trivial filtration because knowing the square means we still know $Y_i$. Does getting $Y_i$ from $Y_i^2$ not result in loss of sign though with regards to the original $Y_i$ as one will always get $+1$ and never $-1$? – Yes Sep 03 '18 at 21:06
  • Davide Giraudo : I am not sure how you can say that in your example $E[X_{n+1} |Y_n]=E[\sum_{î=1}^{n+1} |Y_n]=\sum_{î=1}^{n}Y_i=X_n$ it doesn't appear to be true to me am I mising something ? Regards – TheBridge Sep 04 '18 at 06:23