If $(x_n)$ is a martingale difference then $(x_n\mathbf{1}_{\{ |x_n|\le a_n \}})$ is a martingale?

Question

Let $(x_n,\mathcal{F}_n, n\ge 1)$ be a martingale diference. Is $(x_n\mathbf{1}_{\{ |x_n|\le a_n \}},\mathcal{F}_n, n\ge 1)$ a martingale and why?? $a_n$ is a constant.

Answer 1

In general no. For example, take $(x_n)_{n\geqslant 1}$ an i.i.d. sequence of integrable random variables and $\mathcal F_n:=\sigma(x_1,\dots,x_n)$. In this case, $(x_n,\mathcal F_n)$ is a martingale differences sequence.

However, by independence, $$\mathbb E\left[x_n\mathbf 1\{|x_n|\leqslant a_n\}\mid\mathcal F_{n-1}\right]=\mathbb E\left[x_n\mathbf 1\{|x_n|\leqslant a_n\}\right]=\mathbb E\left[x_1\mathbf 1\{|x_1|\leqslant a_n\}\right]$$ which is not $0$ or $x_{n-1}\mathbf 1\{|x_{n-1}|\leqslant a_{n-1}\}$ in general.

Nevertheless, in the setting of the question, $(y_n,\mathcal F_n)$ is a martingale differences sequence, where $$y_n=x_n\mathbf 1\{|x_n|\leqslant a_n\}-\mathbb E\left[x_n\mathbf 1\{|x_n|\leqslant a_n\}\mid\mathcal F_{n-1}\right].$$