Solution to Review Question by Qiang Gao, updated at Sep 16, 2017
Chapter 2 Large-Sample Theory Section 1 Review of Limit Theorems for Sequences of Random Variables ...
Review Question 2.1.1 (Usual convergence vs. convergence in probability) A sequence of real numbers is a trivial example of a sequence of random variables. Is it true that
$$
\lim_{n \to \infty} z_n = \alpha
\implies
\operatorname*{plim}_{n \to \infty} z_n = \alpha ?
$$
Hint : Look at the definition of $$ \operatorname{plim} $$. Since $$ \lim_{n \to \infty} z_n = \alpha $$, $$ | z_n - \alpha | < \varepsilon $$ for $$ n $$ sufficiently large.
By definition, $$ \lim_{n \to \infty} z_n = \alpha $$ means, for any $$ \varepsilon > 0 $$, for $$n$$ sufficiently large,
$$
| z_n - \alpha | < \varepsilon.
\tag{1}
$$
Considering $$ z_n $$ as a trivial random variable, (1) is equivalent to
$$
\begin{gather}
\operatorname{Prob} (| z_n - \alpha | < \varepsilon) = 1, \\
\operatorname{Prob} (| z_n - \alpha | > \varepsilon) = 0, \\
\lim_{n \to \infty} \operatorname{Prob} (| z_n - \alpha | > \varepsilon) = 0.
\tag{2}
\end{gather}
$$
Then $$ \operatorname*{plim}_{n \to \infty} z_n = \alpha $$ by definition.
A sequence of random scalars $$ {z_n} $$ converges in probability to a constant $$ \alpha $$ if, for any $$ \varepsilon > 0 $$,
$$
\lim_{n \to \infty} \operatorname{Prob} ( | z_n - \alpha | > \varepsilon ) = 0.
$$
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