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Are the following statements equivalent?

$$a) X(t)/t\xrightarrow{a.s} c $$ $$b) X(t)\xrightarrow{a.s} t c $$

where $c$ is a constant and $X(t)$ is a sequence of random variable.

By definition, above statements are equivalent to the followings $$a) P(\lim_{t \to \infty}X(t)/t=c)=1$$ $$b) P(\lim_{t \to \infty}X(t)=tc)=1$$

which are equivalent to: $$a) P(\omega \in \Omega: \forall \epsilon>0, \,\exists T \, \mid \forall t>T,|X(t,\omega)/t-c|<\epsilon)=1$$ $$b) P(\omega \in \Omega: \forall \epsilon>0, \,\exists T \, \mid \forall t>T,|X(t,\omega)-tc|<\epsilon)=1$$

Now, they are equivalent. Right?

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  • $\begingroup$ @Cm7F7Bb what do you mean by how can? What is wrong with that? $\endgroup$ – Susan_Math123 Jul 8 '16 at 19:32
  • $\begingroup$ When $X(t)$ is a random variable , why not the limit can be dependent to $t$? (mathematically) $\endgroup$ – Susan_Math123 Jul 8 '16 at 19:34
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    $\begingroup$ Consider a simple example. Set $x_n=cn+1$, where $x_n$ is a real sequence and $c\in\mathbb R$, then $x_n/n\to c$ as $n\to\infty$. But the notation $x_n\to cn$ as $n\to\infty$ does not make sense, does it? $\endgroup$ – Cm7F7Bb Jul 8 '16 at 19:52
  • $\begingroup$ Related? math.stackexchange.com/questions/1522359/… $\endgroup$ – BCLC Jul 9 '16 at 17:01
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    $\begingroup$ @BCLC Please stop editing questions when you have a poor grasp of their content. To have the idea of replacing "$X(t)\to tc$ almost surely" by "$P(\lim[X(t)=tc])=1$" only shows one does not know what the notations mean. $\endgroup$ – Did Jul 10 '16 at 10:07
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Let's assume without any loss of generality that $t$ is a discrete index, that is, $X_t$ is a random variable on some probability space $(\Omega, \mathcal F, \mathbb P)$ for each positive integer $t$. The statement $$\mathbb P\left(\lim_{t\to\infty}\frac{X_t}t = c \right)=1 $$ means that there exists an event $\Omega_0\in\mathcal F$ with $\mathbb P(\Omega_0)=1$ such that $$\lim_{t\to\infty}\frac{X_t(\omega)}t = c $$ for all $\omega\in\Omega_0$.

In order to make sense of the ill-defined statement $X_t \stackrel{\mathrm{a.s.}}\longrightarrow tc$, we would first have to make sense of $\lim_{t\to\infty} X_t(\omega)=tc$ for a fixed $\omega\in\Omega$ - which itself is not a well-defined statement.

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  • $\begingroup$ Thanks. Can you please tell me when the statement $b$ is equivalent to $a$? $\endgroup$ – Susan_Math123 Jul 8 '16 at 22:52
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    $\begingroup$ @Su20200 My answer explains why $b$ is not a well-defined mathematical statement... $\endgroup$ – Math1000 Jul 8 '16 at 22:55
  • $\begingroup$ Thanks a lot. But, the main part of the answer which includes the last two sentences are not clear to me. Why in order to make sense of the ill-defined statement $X_t \stackrel{\mathrm{a.s.}}\longrightarrow tc$, we would first have to make sense of $\lim_{t\to\infty} X_t(\omega)=tc$ for a fixed $\omega\in\Omega$ ? Why it is not a well-defined statement. $\endgroup$ – Susan_Math123 Jul 11 '16 at 14:56

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