In this exercise, we construct an example of a sequence of random variables \({Z_n}\) such that but \(\Pr \left( {\mathop {\lim }\limits_{n \to \infty } \;{Z_n} = 0} \right) = 0\).That is, \({Z_n}\)converges in probability to 0, but \({Z_n}\)does not converge to 0 with probability 1. Indeed, \({Z_n}\)converges to 0 with probability 0.
Let Xbe a random variable having the uniform distribution on the interval\(\left[ {0,1} \right]\). We will construct a sequence of functions \({h_n}\left( x \right)\;for\;n = 1,2,...\)and define\({Z_n} = {h_n}\left( X \right)\). Each function \({h_n}\) will take only two values, 0 and 1. The set of x where \({h_n}\left( x \right) = 1\) is determined by dividing the interval \(\left[ {0,1} \right]\)into k non-overlapping subintervals of length\(\frac{1}{k}\;for\;k = 1,2,...\)arranging these intervals in sequence, and letting \({h_n}\left( x \right) = 1\) on the nth interval in the sequence for \(n = 1,2,...\)For each k, there are k non overlapping subintervals, so the number of subintervals with lengths \(1,\frac{1}{2},\frac{1}{3},...,\frac{1}{k}\) is \(1 + 2 + 3 + ... + k = \frac{{k\left( {k + 1} \right)}}{2}\)
The remainder of the construction is based on this formula. The first interval in the sequence has length 1, the next two have length 陆, the next three have length 1/3, etc.
- For each\(n = 1,2,...\), proved that there is a unique positive integer \({k_n}\) such that \(\frac{{\left( {{k_n} - 1} \right){k_n}}}{2} < n \le \frac{{{k_n}\left( {{k_n} + 1} \right)}}{2}\)
- For each\(n = 1,2,..\;\), let\({j_n} = n - \frac{{\left( {{k_n} - 1} \right){k_n}}}{2}\). Show that \({j_n}\) takes the values \(1,...,{k_n}\;\)as n runs through\(1 + \frac{{\left( {{k_n} - 1} \right){k_n}}}{2},...,\frac{{{k_n}\left( {{k_n} + 1} \right)}}{2}\).
- Define \({h_n}\left( x \right) = \left\{ \begin{array}{l}1\;if\;\frac{{\left( {{j_n} - 1} \right)}}{{{k_n}}} \le x < \frac{{{j_n}}}{{{k_n}}},\\0\;if\;not\end{array} \right.\)
Show that, for every \(x \in \left[ {0,1} \right),\;{h_n}\left( x \right) = 1\) for one and only one n among \(1 + \frac{{\left( {{k_n} - 1} \right){k_n}}}{2},...,\frac{{{k_n}\left( {{k_n} + 1} \right)}}{2}\)
- Show that \({Z_n} = {h_n}\left( X \right)\)takes the value 1 infinitely often with probability 1 \(\)
- Show that(6.2.18) holds.
- Show that\(\Pr \left( {{Z_n} = 0} \right) = 1 - \frac{1}{{{K_n}}}\)and \(\mathop {\lim }\limits_{n \to \infty } \;{k_n} = \infty \;\).
- Show that
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