4
$\begingroup$

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.

$\endgroup$

1 Answer 1

3
$\begingroup$
  • 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 $Y=x_t$.

$\endgroup$
1
  • $\begingroup$ 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? $\endgroup$ Commented Jun 15, 2018 at 22:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.