Posted on Categories:Probability theory, 数学代写, 概率论

# 数学代写|概率论代考Probability Theory代写|Various inequalities

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## 数学代写|概率论代考Probability Theory代写|Various inequalities

We begin with two very important inequalities about random variables.
Proposition 5.1.1. (Markov’s inequality.) If $X$ is a non-negative random variable, then for all $\alpha>0$,
$$\mathbf{P}(X \geq \alpha) \leq \mathbf{E}(X) / \alpha$$
In words, the probability that $X$ exceeds $\alpha$ is bounded above by its mean divided by $\alpha$.
Proof. Define a new random variable $Z$ by
$$Z(\omega)= \begin{cases}\alpha, & X(\omega) \geq \alpha \ 0, & X(\omega)<\alpha .\end{cases}$$
Then clearly $Z \leq X$, so that $\mathbf{E}(Z) \leq \mathbf{E}(X)$ by the order-preserving property. On the other hand, we compute that $\mathbf{E}(Z)=\alpha \mathbf{P}(X \geq \alpha)$. Hence, $\alpha \mathbf{P}(X \geq \alpha) \leq \mathbf{E}(X)$.

Markov’s inequality applies only to non-negative random variables, but it immediately implies another inequality which holds more generally:

Proposition 5.1.2. (Chebychev’s inequality.) Let $Y$ be an arbitrary random variable, with finite mean $\mu_Y$. Then for all $\alpha>0$,
$$\mathbf{P}\left(\left|Y-\mu_Y\right| \geq \alpha\right) \leq \operatorname{Var}(Y) / \alpha^2$$
In words, the probability that $Y$ differs from its mean by more than $\alpha$ is bounded above by its variance divided by $\alpha^2$.

## 数学代写|概率论代考Probability Theory代写|Convergence of random variables

If $Z, Z_1, Z_2, \ldots$ are random variables defined on some $(\Omega, \mathcal{F}, \mathbf{P})$, what does it mean to say that $\left{Z_n\right}$ converges to $Z$ as $n \rightarrow \infty$ ?

One notion we have already seen (cf. Theorem 4.2.2) is pointwise convergence, i.e. $\lim {n \rightarrow \infty} Z_n(\omega)=Z(\omega)$. A slightly weaker notion which often arises is convergence almost surely (or, a.s. or with probability 1 or w.p. 1 or almost everywhere), meaning that $\mathbf{P}\left(\lim {n \rightarrow \infty} Z_n=Z\right)=1$, i.e. that $\mathbf{P}\left{\omega \in \Omega: \lim _{n \rightarrow \infty} Z_n(\omega)=Z(\omega)\right}=1$. As an aid to establishing such convergence, we have the following:

Lemma 5.2.1. Let $Z, Z_1, Z_2, \ldots$ be random variables. Suppose for each $\epsilon>0$, we have $\mathbf{P}\left(\left|Z_n-Z\right| \geq \epsilon\right.$ i.o. $)=0$. Then $\mathbf{P}\left(Z_n \rightarrow Z\right)=1$, i.e. $\left{Z_n\right}$ converges to $Z$ almost surely.
Proof. It follows from Proposition A.3.3 that
$$\mathbf{P}\left(Z_n \rightarrow Z\right)=\mathbf{P}\left(\forall \epsilon>0,\left|Z_n-Z\right|<\epsilon \text { a.a. }\right)=1-\mathbf{P}\left(\exists \epsilon>0,\left|Z_n-Z\right| \geq \epsilon \text { i.o. }\right)$$

By countable subadditivity, we have that
$$\mathbf{P}\left(\exists \epsilon>0, \epsilon \text { rational, }\left|Z_n-Z\right| \geq \epsilon \text { i.o. }\right) \leq \sum_{\substack{\epsilon>0 \ \epsilon \text { rational }}} \mathbf{P}\left(\left|Z_n-Z\right| \geq \epsilon \text { i.o. }\right)=0 .$$
But given any $\epsilon>0$, there exists a rational $\epsilon^{\prime}>0$ with $\epsilon^{\prime}<\epsilon$. For this $\epsilon^{\prime}$, we have that $\left{\left|Z_n-Z\right| \geq \epsilon\right.$ i.o. $} \subseteq\left{\left|Z_n-Z\right| \geq \epsilon^{\prime}\right.$ i.o. $}$. It follows that $\mathbf{P}\left(\exists \epsilon>0,\left|Z_n-Z\right| \geq \epsilon\right.$ i.o. $) \leq \mathbf{P}\left(\exists \epsilon^{\prime}>0, \epsilon^{\prime}\right.$ rational, $\left|Z_n-Z\right| \geq \epsilon^{\prime}$ i.o. $)=0$, thus giving the result.

# 概率论代写

## 数学代写|概率论代考Probability Theory代写|Various inequalities

$$\mathbf{P}(X \geq \alpha) \leq \mathbf{E}(X) / \alpha$$

$$Z(\omega)= \begin{cases}\alpha, & X(\omega) \geq \alpha \ 0, & X(\omega)<\alpha .\end{cases}$$

$$\mathbf{P}\left(\left|Y-\mu_Y\right| \geq \alpha\right) \leq \operatorname{Var}(Y) / \alpha^2$$

## 数学代写|概率论代考Probability Theory代写|Convergence of random variables

$$\mathbf{P}\left(Z_n \rightarrow Z\right)=\mathbf{P}\left(\forall \epsilon>0,\left|Z_n-Z\right|<\epsilon \text { a.a. }\right)=1-\mathbf{P}\left(\exists \epsilon>0,\left|Z_n-Z\right| \geq \epsilon \text { i.o. }\right)$$

$$\mathbf{P}\left(\exists \epsilon>0, \epsilon \text { rational, }\left|Z_n-Z\right| \geq \epsilon \text { i.o. }\right) \leq \sum_{\substack{\epsilon>0 \ \epsilon \text { rational }}} \mathbf{P}\left(\left|Z_n-Z\right| \geq \epsilon \text { i.o. }\right)=0 .$$

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## MATLAB代写

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