by Qiang Gao, updated at June 8, 2018
...
(a) Verify that the definition in the text of “$$ \mathbf{z}n \to{m.s.} \mathbf{z} $$” is equivalent to
Hint: $$ \operatorname{E} [ ( \mathbf{z}n - \mathbf{z} )'( \mathbf{z}n - \mathbf{z} ) ] =
\operatorname{E} [ ( z{n1} - z_1 )^2 ] + \cdots +
\operatorname{E} [ ( z{nK} - z_K )^2 ] $$, where
(b) Similarly, verify that the definition in the text of “$$ \mathbf{z}_n \to_p \boldsymbol{\alpha} $$” is equivalent to
for any $$ \varepsilon > 0 $$.
(a) If $$ \mathbf{z}n \to{m.s.} \mathbf{z} $$, by definition in the text,
then we can show
$$ \begin{align} & \color{white}{=} \lim_{n \to \infty} \operatorname{E} [ ( \mathbf{z}n - \mathbf{z} )'( \mathbf{z}n - \mathbf{z} ) ] \ & = \lim{n \to \infty} \left( \operatorname{E} [ ( z{n1} - z_1 )^2 ] + \cdots + \operatorname{E} [ ( z_{nK} - z_K )^2 ] \right) \ & = \lim_{n \to \infty} \operatorname{E} [ ( z_{n1} - z_1 )^2 ] + \cdots + \lim_{n \to \infty} \operatorname{E} [ ( z_{nK} - z_K )^2 ] \ & = 0. \end{align} $$
If $$ \lim_{n \to \infty} \operatorname{E} [ ( \mathbf{z}_n - \mathbf{z} )'( \mathbf{z}_n - \mathbf{z} ) ] = 0 $$, then
Because each terms in (1) are non-negative, (1) implies
which means $$ z_{nk} \to_{m.s.} z_k $$ for $$ k = 1, \ldots, K
(b) Similarly, if $$ \mathbf{z}_n \to_p \boldsymbol{\alpha}
then we can show
$$
\begin{gather}
\operatorname{Prob}
\left(
( z_{nk} - \alpha_k )^2 > \varepsilon^2
\right) \to 0,
\text{ for
\operatorname{Prob}
\left(
( z_{nk} - \alpha_k )^2 > \varepsilon
\right) \to 0,
\text{ for
\operatorname{Prob} \left( ( z_{n1} - \alpha_1 )^2 > \varepsilon \right) + \cdots + \operatorname{Prob} \left( ( z_{nK} - \alpha_K )^2 > \varepsilon \right) \to 0, && \text{(limit sum)} \
\left( \not\Rightarrow \operatorname{Prob} \left( (z_{n1} - \alpha_1)^2 + \cdots + (z_{nK} - \alpha_K)^2 > K \varepsilon \right) \to 0 \right) && \text{(bigger event)} \
\operatorname{Prob} \left( \bigcup_{k=1}^K \left( ( z_{nk} - \alpha_k )^2 > \varepsilon \right) \right) \to 0, && \text{(smaller event)} \
\operatorname{Prob} \left( \bigcap_{k=1}^K \left( ( z_{nk} - \alpha_k )^2 \leq \varepsilon \right) \right) \to 1, && \text{(complement event)} \
\operatorname{Prob} \left( ( z_{n1} - \alpha_1 )^2 + \cdots + ( z_{nK} - \alpha_K )^2 \leq K \varepsilon \right) \to 1, && \text{(bigger event)} \
\operatorname{Prob}
\left(
( z_{n1} - \alpha_1 )^2 + \cdots +
( z_{nK} - \alpha_K )^2 \leq \varepsilon
\right) \to 1,
&& \text{(any
\operatorname{Prob} \left( ( \mathbf{z}_n - \boldsymbol{\alpha} )' ( \mathbf{z}_n - \boldsymbol{\alpha} ) \leq \varepsilon \right) \to 1, && \text{(matrix notation)} \
\operatorname{Prob} \left( ( \mathbf{z}_n - \boldsymbol{\alpha} )' ( \mathbf{z}_n - \boldsymbol{\alpha} )
\varepsilon \right) \to 0. && \text{(complement event)} \end{gather} $$
If $$ \operatorname{Prob} \left( ( \mathbf{z}_n - \boldsymbol{\alpha} )' ( \mathbf{z}_n - \boldsymbol{\alpha} ) > \varepsilon \right) \to 0 $$, then we can show
$$ \begin{gather} \operatorname{Prob} \left( ( z_{n1} - \alpha_1 )^2 + \cdots + ( z_{nK} - \alpha_K )^2 > \varepsilon \right) \to 0, && \text{(scalar notation)} \
\operatorname{Prob} \left( \bigcup_{k=1}^K \left( ( z_{nk} - \alpha_k )^2 > \varepsilon \right) \right) \to 0, && \text{(smaller event)} \
\operatorname{Prob}
\left(
( z_{nk} - \alpha_k )^2 > \varepsilon
\right) \to 0
\text{ for
which means $$ z_{nk} \to_p \alpha_k $$ for $$ k = 1, \ldots, K
Copyright ©2018 by Qiang Gao