Question

Prove Theorem 6.2.5.

Short answer

It is proved that If and if \(g\left( z \right)\)is a function that is continuous at z=b then

Step-by-step solution

Given information

It is given that and \(g\left( z \right)\) is a function that is continuous at z=b .

Proving that

Given that \(g\left( z \right)\) is continuous at z=b then there existing \(\delta \) such that \(\left| {z - b} \right| < \delta \) implies that \(\left| {g\left( z \right) - g\left( b \right)} \right| < \in \) for some constant \( \in > 0\).

Let’s consider the definition of convergence in probability as if a sequence of\(\begin{array}{l}{Z_1},{Z_2},...\\\end{array}\)of random variables converges to b in probability if for every number\( \in > 0\), then\(\mathop {\lim }\limits_{n \to \infty } \Pr \left( {\left| {{Z_n} - b} \right| < \in } \right) = 1\)

This property is denoted by

Therefore,

As, , hence \(\mathop {\lim }\limits_{n \to \infty } \Pr \left( {\left| {{Z_n} - b} \right| < \in } \right) = 1\)

But \(\left\{ {\left| {{Z_n} - b} \right| < \delta } \right\} \subset \left\{ {\left| {g\left( {{Z_n}} \right) - g\left( b \right)} \right| < \in } \right\}\)

Therefore,\(\Pr \left( {\left| {g\left( {{Z_n}} \right) - g\left( b \right)} \right| < \in } \right) \ge \Pr \left( {\left| {{Z_n} - b} \right| < \delta } \right)\)

Now,

\(\Pr \left( {\left| {{Z_n} - b} \right| < \delta } \right) \to 1\;as\;n \to \infty \)

Therefore,

\(\Pr \left( {\left| {g\left( {{Z_n}} \right) - g\left( b \right)} \right| < \in } \right) \to 1\;as\;n \to \infty \)

Hence, it is proved that if and if \(g\left( z \right)\)is a function that is continuous at z=b then