Section 3.5: What Is a Random Sequence?

Exercise 9. [HM20] Prove Lemma E. [Hint: Consider $\sum_{j=1}^{n}(y_{jn} - a)^2$.]

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Solution

Lemma E states that if

$$ \lim_{n\to\infty}\frac1n\sum_{j=1}^{n}y_{jn}=a, \qquad \lim_{n\to\infty}\frac1n\sum_{j=1}^{n}y_{jn}^{,2}=a^2, $$

then for every $\varepsilon>0$,

$$ \lim_{n\to\infty} \frac1n #{j:1\le j\le n,\ |y_{jn}-a|\ge\varepsilon} =0. $$

Consider

$$ \sum_{j=1}^{n}(y_{jn}-a)^2. $$

Expanding the square gives

$$ \frac1n\sum_{j=1}^{n}(y_{jn}-a)^2 = \frac1n\sum_{j=1}^{n}y_{jn}^{,2} -\frac{2a}{n}\sum_{j=1}^{n}y_{jn} +a^2. $$

By the hypotheses,

$$ \lim_{n\to\infty} \frac1n\sum_{j=1}^{n}(y_{jn}-a)^2 = a^2-2a^2+a^2 = 0. $$

Let

$$ N_n(\varepsilon) = #{j:1
$$ \le j\le n,\ |y_{jn}-a|\ge\varepsilon}. $$ For every index counted by $N_n(\varepsilon)$, $$ (y_{jn}-a)^2\ge \varepsilon^2. $$ Hence $$ \sum_{j=1}^{n}(y_{jn}-a)^2 \ge \varepsilon^2 N_n(\varepsilon). $$ Dividing by $n$ yields $$ \frac{N_n(\varepsilon)}{n} \le \frac1{\varepsilon^2} \frac1n \sum_{j=1}^{n}(y_{jn}-a)^2. $$ Taking limits and using the result already established, $$ 0 \le \limsup_{n\to\infty}\frac{N_n(\varepsilon)}{n} \le \frac1{\varepsilon^2} \lim_{n\to\infty} \frac1n \sum_{j=1}^{n}(y_{jn}-a)^2

$$ Therefore lim_{n\to\infty}\frac1n\sum_{j= $$ \lim_{n\to\infty}\frac{N_n(\varepsilon)}{n}=0. $$ This is precisely the conclusion of Lemma E. This completes the proof. ∎1}^n y_{jn}=a, \qquad \lim_{n\to\infty}\frac1n\sum_{j=1}^n y_{jn}^2=a^2, $$

then for every $\epsilon>0$ the proportion of indices $j$ for which

$$ |y_{jn}-a|\ge\epsilon $$

tends to $0$.

Consider the quantity suggested in the hint:

$$ \sum_{j=1}^{n}(y_{jn}-a)^2. $$

Expanding,

$$ \frac1n\sum_{j=1}^{n}(y_{jn}-a)^2 = \frac1n\sum_{j=1}^{n}y_{jn}^2 -\frac{2a}{n}\sum_{j=1}^{n}y_{jn} +a^2. $$

Taking limits and using the hypotheses,

$$ \lim_{n\to\infty} \frac1n\sum_{j=1}^{n}(y_{jn}-a)^2 = a^2-2a^2+a^2 =0. $$

Let

$$ N_n(\epsilon) = #{,j:1\le j\le n,\ |y_{jn}-a|\ge\epsilon,}. $$

For every index counted by $N_n(\epsilon)$ we have

$$ (y_{jn}-a)^2\ge\epsilon^2. $$

Hence

$$ \sum_{j=1}^{n}(y_{jn}-a)^2 \ge N_n(\epsilon),\epsilon^2. $$

Dividing by $n$,

$$ \frac{N_n(\epsilon)}{n} \le \frac1{\epsilon^2} \cdot \frac1n \sum_{j=1}^{n}(y_{jn}-a)^2. $$

The right-hand side tends to $0$, therefore

$$ \lim_{n\to\infty}\frac{N_n(\epsilon)}{n}=0. $$

Thus, for every $\epsilon>0$, the set of indices $j$ for which $|y_{jn}-a|\ge\epsilon$ has asymptotic density $0$. Equivalently, the values $y_{jn}$ approach $a$ with probability $1$ in the sense of Definition A.

This completes the proof.

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