Question

Let \({Z_1},{Z_2},...\) be a sequence of random variables, and suppose that, for

\(n = 1,2,...\), the distribution of\({Z_n}\)is as follows:

\(\Pr \left( {{Z_n} = {n^2}} \right) = \frac{1}{n}\)and \(\Pr \left( {{Z_n} = 0} \right) = 1 - \frac{1}{n}\). Show that \(\mathop {\lim }\limits_{n \to \infty } E\left( {{Z_n}} \right) = \infty \,but\;{Z_n}\xrightarrow{P}0\).

Short answer

It is prove that.

Step-by-step solution

Given information

\({Z_1},{Z_2},...\)is a sequence of random variables and the distribution of\({Z_n}\)is given as follows

\(\Pr \left( {{Z_n} = {n^2}} \right) = \frac{1}{n}\)and\(\Pr \left( {{Z_n} = 0} \right) = 1 - \frac{1}{n}\).

Showing that

Let,

\(\begin{array}{c}E\left( {{Z_n}} \right) = {n^2}.\Pr \left( {{Z_n} = {n^2}} \right) + 0.\Pr \left( {{Z_n} = 0} \right)\\ = {n^2}.\frac{1}{n} + 0\left( {1 - \frac{1}{n}} \right)\\ = n.\end{array}\)

Therefore,

\(\mathop {\lim }\limits_{n \to \infty } E\left( {{Z_n}} \right) = \infty \)

Also, for any given\( \in > 0\),

\(\begin{array}{c}\Pr \left( {\left| {{z_n}} \right| < \in } \right) = \Pr \left( {{Z_n} = 0} \right)\\ = 1 - \frac{1}{n}\end{array}\)

Hence,

\(\mathop {\lim }\limits_{n \to \infty } \Pr \left( {\left| {{z_n}} \right| < \in } \right) = 1\) which means that