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# 数学代写|组合学代写Combinatorics代考|Math145 Law of Large Numbers

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## 数学代写|组合学代写Combinatorics代考|Law of Large Numbers

In this section we shall give some explanation of the approximate equalities (12.1.1) and (12.1.2). The quantities $\alpha_n(1)$ and $\beta_n(1)$ that appear there are in fact realizations of the binomial random variables $\mathcal{B}\left(n, \frac{1}{2}\right)$ and $\mathcal{B}\left(n, \frac{1}{6}\right)$. For some realizations of the experiments considered, the differences
$$\left|\alpha_n(1)-\frac{1}{2}\right|, \quad\left|\beta_n(1)-\frac{1}{6}\right|$$
may be significantly greater than 0 , for example greater than $\varepsilon>0$. The question is: what is the probability of such an event?

Theorem 12.5.1. Bernoulli’s law of large numbers. Let $S_n \in \mathcal{B}(n, p)$ be a binomial random variable. Then, for every $\varepsilon>0$ the following inequality holds:
$$P\left{\left|\frac{S_n}{n}-p\right| \geqslant \varepsilon\right} \leqslant \frac{p(1-p)}{n \varepsilon^2} .$$
Note that inequality (12.5.1) implies that $P\left{\left|S_n / n-p\right| \geqslant \varepsilon\right} \rightarrow 0$ as $n \rightarrow \infty$. This fact, known as the law of large numbers, was proved by Jacob Bernoulli (1654-1705), and published firstly in his work Ars Conjectandi in 1713. Bernoulli’s law of large numbers is the first limit theorem in Probability Theory and one of the basic results of this theory.

We shall prove here a more general result, known as Chebyshev’s law of large numbers. For this purpose we need the following lemma.

Lemma 12.5.2. Chebyshev’s inequality. For every random variable $X$ with finite variance $\sigma^2$ and every $\varepsilon>0$ the following inequality holds true:
$$P{|X-E X|} \geqslant \varepsilon} \leqslant \frac{\sigma^2}{\varepsilon^2} .$$
Proof. Let $X$ be a discrete random variable with the set of values $\mathcal{S}_X$. Let $\varepsilon>0$, and consider the following partition of the set $\mathcal{S}_X$ :
$$\mathcal{S}_X=\left{a_1, a_2, \ldots\right} \cup\left{b_1, b_2, \ldots\right},$$

such that $\left|a_i-E X\right|<\varepsilon$ for all $i$ ‘s, and $\left|b_j-E X\right| \geqslant \varepsilon$ for all $j$ ‘s. Then we have
\begin{aligned} \sigma^2 &=\sum_i\left(a_i-E X\right)^2 P\left{X=a_i\right}+\sum_j\left(b_j-E X\right)^2 P\left{X=b_j\right} \ & \geqslant \sum_i\left(b_j-E X\right)^2 P\left{X=b_j\right} \geqslant \varepsilon^2 \sum_j P\left{X=b_j\right} \ &=\varepsilon^2 P{|X-E X| \geqslant \varepsilon} . \end{aligned}

## 数学代写|组合学代写Combinatorics代考|Basic Combinatorial Configurations

13.1. Let $p_k(n)$ be the number of permutations of the set ${1,2, \ldots, n}$ that have exactly $k$ fixed points. Prove the following equalities:
(a) $k p_n(k)=n p_{n-1}(k-1)$, where $1 \leqslant k \leqslant n$;
(b) $\sum_{k=m}^n \frac{k !}{(k-m) !} p_n(k)=n$ !, where $0 \leqslant m \leqslant n$.
13.2. Is there a permutation of the set $\mathbb{N}_n={1,2, \ldots, n}$ such that for any two positive integers $i, j \in \mathbb{N}_n$ their arithmetic mean is not equal to a number that is placed between $i$ and $j$ in this permutation?
13.3. Let a finite $0-1$ sequence be given such that the following conditions are satisfied:
(a) Any two subsequences which consist of 5 consecutive terms of the given sequence are not equal to each other (the subsequences may overlap partially).
(b) If we add 0 or 1 at the end of the given sequence, then the new sequence does not have property (a) anymore.

Prove that the subsequences consisting of the first four and the last four terms of the starting sequence coincide.
13.4. Prove that there exist $2^{n+1}$ positive integers such that the following three conditions are satisfied:

(a) Each of these positive integers has only the digits 1 and 2 in its decimal representation.
(b) Each of these positive integers has $2^n$ digits;
(c) Every two of these $2^{n+1}$ positive integers have different digits at least in $2^{n-1}$ positions.

## 数学代写|组合学代写Combinatorics代考|大数定律

$$\left|\alpha_n(1)-\frac{1}{2}\right|, \quad\left|\beta_n(1)-\frac{1}{6}\right|$$

$$P\left{\left|\frac{S_n}{n}-p\right| \geqslant \varepsilon\right} \leqslant \frac{p(1-p)}{n \varepsilon^2} .$$

## MATLAB代写

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