Is it true that if $X(t)$ is a martingale then $X^{2}(t)-E[X^{2}(t)]$ is also a martingale?

Question

I'm assuming that $X^{2}(t)-E[X^{2}(t)]$ is integrable.

Answer 1

No, not in general.

For a discrete time example, let $S_n$ be a simple random walk, and let $Z = 0$ or $1$ with probability $1/2$ and independent of $S_n$. Set $X_n = Z S_n$; this is a martingale. Intuitively, $X_n$ corresponds to the following gambling game: flip a coin at time 0. If it comes up tails, bet $1 on coin flips forever. If it comes up heads, never play at all.

If you set $Y_n = X_n^2 - E[X_n^2] = X_n^2 - \frac{1}{2}n$, and consider for example the event $A = \{Y_1 = -\frac{1}{2}\} = \{Z=0\}$ which is $\mathcal{F}_1$-measurable, you can easily observe $E[Y_2 - Y_1 \mid A] = -\frac{1}{2} \ne 0$. So $Y_n$ is not a martingale. (Intuitively, by observing $Y_1$ you can see whether the coin $Z$ came up heads or tails. Conditionally on the coin having been tails, $Y_n$ is going to strictly decrease forever, which a martingale cannot do.)