# Strict stationarity implies bounded in probability

My definition of bounded in probabilty is following:

A random sequence $\{ x_t : t \in \mathbb{Z} \}$ is said to be bounded in probability, if $$\lim_{c \rightarrow \infty} \sup_{t \in \mathbb{Z}} \Pr\left( |x_t|>c\right) =0.$$ In a paper of A. Klivecka regarding the random GARCH it is said right after this definition:

Clearly, any strictly stationary sequence is bounded in probability. Note that any sequence $\{ x_t : t \in \mathbb{Z} \}$ is bounded in probability if $\sup_{t \in \mathbb{Z}} \mathbb{E}\left|x_t\right|^p < \infty$ for some $p>0$ (by Chebychevs inequality).

Why are these staments true? Some more details would be really nice.

• If a sequence $\left(x_t\right)_{t\in\mathbb Z}$ is strictly stationary, then in particular, $x_t$ has the same distribution as $x_0$ for all $t\in\mathbb Z$. Consequently, $\left\lvert x_t\right\rvert$ has the same distribution as $\left\lvert x_0\right\rvert$ for all $t\in\mathbb Z$ and for all $c$, $$\sup_{t\in\mathbb Z}\Pr\left(\left\lvert x_t\right\rvert\gt c\right)=\Pr\left(\left\lvert x_0\right\rvert\gt c\right).$$ We conclude using the fact that for any real valued random variable $Y$, $\lim_{c\to +\infty}\Pr\left(\left\lvert Y\right\rvert\gt c\right)=0$.
• We use $\Pr\left(\left\lvert x_t\right\rvert\gt c\right)=\Pr\left(\left\lvert x_t\right\rvert^p\gt c^p\right)$ and inequality $\Pr\left(\left\lvert Y\right\rvert\gt x\right)\leqslant x^{-1}\mathbb E\left\lvert Y\right\rvert$ (which is a consequence of the integration of the pointwise inequality $x\mathbf 1\left\{\left\lvert Y\right\rvert\gt x\right\}\leqslant \left\lvert Y\right\rvert)$), with $x=c^p$ and $Y=x_t$.
• I thoguht about using Chebychev like this: $\lim_{c \rightarrow \infty} \sup_{t \in \mathbb{Z}} \mathbb{P} (|X_t| > c) \leq \lim_{c \rightarrow \infty} \frac{\sup_{t \in \mathbb{Z}} \mathbb{E} {|X_t|^p}}{c^p} =0$. Would that be also a possible solution? Jun 15, 2018 at 22:31