Question

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 ½, the next three have length 1/3, etc.

1. For each\(n = 1,2,...\), proved that there is a unique positive integer \({k_n}\) such that \(\frac{{\left( {{k_n} - 1} \right){k_n}}}{2} < n \le \frac{{{k_n}\left( {{k_n} + 1} \right)}}{2}\)
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\(1 + \frac{{\left( {{k_n} - 1} \right){k_n}}}{2},...,\frac{{{k_n}\left( {{k_n} + 1} \right)}}{2}\).
3. 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 \(1 + \frac{{\left( {{k_n} - 1} \right){k_n}}}{2},...,\frac{{{k_n}\left( {{k_n} + 1} \right)}}{2}\)

4. Show that \({Z_n} = {h_n}\left( X \right)\)takes the value 1 infinitely often with probability 1 \(\)

5. Show that(6.2.18) holds.

6. Show that\(\Pr \left( {{Z_n} = 0} \right) = 1 - \frac{1}{{{K_n}}}\)and \(\mathop {\lim }\limits_{n \to \infty } \;{k_n} = \infty \;\).
7. Show that

Short answer

1. It is proved that there is unique positive integer\({k_n}\)such that \(\frac{{\left( {{k_n} - 1} \right){k_n}}}{2} < n \le \frac{{{k_n}\left( {{k_n} + 1} \right)}}{2}\)
2. It is proved that\({j_n}\)takes the values\(1,2,...,{k_n}\)for the function\({j_n} = n - \frac{{\left( {{k_n} - 1} \right){k_n}}}{2}\)as n runs through\(1 + \frac{{\left( {{k_n} - 1} \right){k_n}}}{2},...,\frac{{{k_n}\left( {{k_n} + 1} \right)}}{2}\).

It is proved that, for every\(x \in \left[ {0,1} \right),\;{h_n}\left( x \right) = 1\)for one and only one n among\(1 + \frac{{\left( {{k_n} - 1} \right){k_n}}}{2},...,\frac{{{k_n}\left( {{k_n} + 1} \right)}}{2}\).

4. It is proved that \({Z_n}\) takes the value 1 infinitely often with probability 1.
5. It is proved that\(\Pr \left( {\mathop {\lim }\limits_{n \to \infty } \;{Z_n} = 0} \right) = 0\)
6. It is proved that\(\Pr \left( {{Z_n} = 0} \right) = 1 - \frac{1}{{{K_n}}}\)and\(\mathop {\lim }\limits_{n \to \infty } \;{k_n} = \infty \;\)
7. It is proved that.

Step-by-step solution

Given information

Let X be a random variable with moment generating function\(\psi \) .

Then, for every real t,

\(P\left( {X \ge t} \right) \le \mathop {\min }\limits_{s > 0} \exp \left( { - st} \right)\psi \left( s \right)\)

This theorem is most useful when X is the sum of the n i.i.d. random variables each with finite m.g.f. and when \(t = nu\) for a large value of n and some fixed u.

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 the probability to 0, but \({Z_n}\) does not converges to 0 with probability 1. Indeed \({Z_n}\) converges to 0 with probability 0.

Let X be the random variable having uniform distribution on the interval\(\left[ {0,1} \right]\).

Construct a sequence of function \({h_n}\left( x \right)\)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 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 overlapping subintervals, so the number of 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\(\frac{1}{2}\), the next three have the length\(\frac{1}{3}\)etc.

Proving that there is unique positive integer \({k_n}\) such that \(\frac{{\left( {{k_n} - 1} \right){k_n}}}{2} < n \le \frac{{{k_n}\left( {{k_n} + 1} \right)}}{2}\).

Convergence in Probability:

A sequence \({Z_1},{Z_2},...\)of random variables converges to b in probability if for every number\( \in > 0\).

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

This property is denoted by

And it is sometimes stated simply as \({Z_n}\) converges to b in probability.

Law of large numbers:

Suppose\({X_1},...,{X_n}\)form a random sample from a distribution for which the mean is \(\mu \) and for which the variance is finite. Let \({\overline X _n}\) denote the sample mean. Then .

Moment Generating function (m.g.f.):

Let X be a random variable. For each real number t. define

\(\psi \left( t \right) = E\left( {{e^{tx}}} \right)\)

The number \(\psi \left( t \right)\) is called moment generating function of X.

The numbers\(\frac{{\left( {k - 1} \right)k}}{2}\)for\(k = 1,2,...\)form a strictly increasing sequence starting at 0.

Hence, every integer n falls between a unique pair of these numbers.

So,\({k_n}\)is the value of k such that n is larger than \(\frac{{\left( {k - 1} \right)k}}{2}\) but no larger than\(\frac{{k\left( {k + 1} \right)}}{2}\)

Proving that \({j_n}\)takes the values\(1,2,...,{k_n}\) for the function \({j_n} = n - \frac{{\left( {{k_n} - 1} \right){k_n}}}{2}\)as n runs through \(1 + \frac{{\left( {{k_n} - 1} \right){k_n}}}{2},...,\frac{{{k_n}\left( {{k_n} + 1} \right)}}{2}\)

Let,

\({j_n} = n - \frac{{\left( {{k_n} - 1} \right){k_n}}}{2}\)

Clearly \({j_n}\)is the excess of n over the lower bound in part (a).

Hence,

\({j_n}\) runs from 1 up to the difference between the bounds, which is easily seen to be\({k_n}\).

Proving that, for every \(x \in \left[ {0,1} \right),\;{h_n}\left( x \right) = 1\) for one and only one n among \(1 + \frac{{\left( {{k_n} - 1} \right){k_n}}}{2},...,\frac{{{k_n}\left( {{k_n} + 1} \right)}}{2}\)

Let,

\({h_n}\left( x \right) = \left\{ \begin{array}{l}1{\rm{ }}if{\rm{ }}\frac{{\left( {{j_n} - 1} \right)}}{{{k_n}}} \le x < \frac{{{j_n}}}{{{k_n}}}\\0{\rm{ }}if{\rm{ }}not\end{array} \right.\)

The intervals where \({h_n}\)equals 1 are defined to be disjoint for\({j_n} = 1,...,k\), and they cover the whole interval\(\left[ {0,1} \right)\).

Hence, for each x,

\({h_n}\left( x \right) = 1\)for one and only one of these intervals, which correspond to n between the bounds in part a.

Proving that \(\Pr \left( {\mathop {\lim }\limits_{n \to \infty } \;{Z_n} = 0} \right) = 0\)

For every

\(x \in \left[ {0,1} \right)\),

\({h_n}\left( x \right) = 1\)for one n between the bounds in part a.

Since, there are infinitely many values of\({k_n}\),\({h_n}\left( x \right) = 1\) infinitely often for every\(x \in \left[ {0,1} \right)\), and \(P\left( {x \in \left[ {0,1} \right)} \right) = 1\).

Proving that \(\Pr \left( {X \ge t} \right) \le \mathop {\min }\limits_{s > 0} \;\exp \left( { - st} \right)\psi \left( s \right)\)

For every \( \in > 0\;\left| {{Z_n} - 0} \right| > \in \) whenever\({Z_n} = 1\).

Since,

\(\Pr \left( {{Z_n} = 1\;\inf initely\;often} \right) = 1\), the probability is 1 that \({Z_n}\) fails to converge to 0.

Hence, the probability is 0 that \({Z_n}\) does converge to 0.

Proving that \(\Pr \left( {{Z_n} = 0} \right) = 1 - \frac{1}{{{K_n}}}\)and \(\mathop {\lim }\limits_{n \to \infty } \;{k_n} = \infty \;\)

Let,

\({h_n}\left( x \right) = 1\)on an interval of length \(\frac{1}{{{k_n}}}\).

Hence, for each n,

\(\Pr \left( {\left| {{Z_n} - 0} \right| > \in } \right) = \frac{1}{{{k_n}}}\)

Therefore,

\(P\left( {{Z_n} = 0} \right) = 1 - \frac{1}{{{k_n}}}\)

And

\(\mathop {\lim }\limits_{n \to \infty } {k_n} = \infty \)

Proving that

Let,

\({h_n}\left( x \right) = 1\)on an interval of length \(\frac{1}{{{k_n}}}\) .

Hence, for each n,

\(\Pr \left( {\left| {{Z_n} - 0} \right| > \in } \right) = \frac{1}{{{k_n}}}\)

It tends to 0,

Thus,