Is the following process a martingale?
Let $X$ denotes a process defined by the following: $$X_{t}=\sigma_{t} W_{t},$$ where $W$ is a Wiener process, $\sigma_{t}=\begin{cases} 1.1 & \text{if}\;X_{t}-X_{t-1}>0\\ 0.9 & \text{if}\;X_{t}-X_{t-1}\leq0\\ 1 & \text{if}\;t\in\left[0,1\right] \end{cases}$. I wanted to check if $X$ whether a martingale or not, by the definition of the martingale.
If $t\in\left[0,1\right]$, then it is clear $X=W$ is a martingale, I have bigger problems with the $1<s<t-1$ and $1<t-1\leq s\leq t$ cases.
I also think if $1<s\leq t-1$, then $X_{t}-X_{s}$ is independent from $\mathcal{F}_{s}$, but I am not certain in this statement. If this is the case, then $$\mathbb{E}\left[X_{t}\mid\mathcal{F}_{s}\right] =\mathbb{E}\left(\sigma_{t}W_{t}-\sigma_{s}W_{s}+\sigma_{s}W_{s}\mid\mathcal{F}_{s}\right)=\mathbb{E}\left(\sigma_{t}W_{t}-\sigma_{s}W_{s}\right)+\sigma_{s}W_{s}=$$
$$=\mathbb{E}\left(\sigma_{t}W_{t}\right)-\mathbb{E}\left(\sigma_{s}W_{s}\right)+\sigma_{s}W_{s}=\ldots,$$ where $$\mathbb{E}\left(\sigma_{t}W_{t}\right) =\mathbb{E}\left(\sigma_{t}W_{t}\cdot1\right)=\mathbb{E}\left(\sigma_{t}W_{t}\cdot\left(\chi_{\left\{ X_{t}-X_{t-1}>0\right\} }+\chi_{\left\{ X_{t}-X_{t-1}\leq0\right\} }\right)\right)=$$
$$=\mathbb{E}\left(\sigma_{t}W_{t}\chi_{\left\{ X_{t}-X_{t-1}>0\right\} }\right)+\mathbb{E}\left(\sigma_{t}W_{t}\chi_{\left\{ X_{t}-X_{t-1}\leq0\right\} }\right)=\mathbb{E}\left(1.1W_{t}\right)+\mathbb{E}\left(0.9W_{t}\right)=0.$$ So $$\mathbb{E}\left[X_{t}\mid\mathcal{F}_{s}\right]=\ldots=\mathbb{E}\left(\sigma_{t}W_{t}\right)-\mathbb{E}\left(\sigma_{s}W_{s}\right)+\sigma_{s}W_{s}=0-0+\sigma_{s}W_{s}=\sigma_{s}W_{s}.$$
I am not sure if it is true, and I still don't know the $1<t-1\leq s\leq t$ part.