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# 数学代写|随机过程Stochastic Porcess代考|BASIC THEOREMS

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## 数学代写|随机过程代写Stochastic Porcess代考|BASIC THEOREMS

In Section 4.1 we observed that the set of possible values of a random variable might be finite, infinite but countable, or uncountable. For example, let $X, Y$, and $Z$ be three random variables representing the respective number of tails in flipping a coin twice, the number of flips until the first heads, and the amount of next year’s rainfall. Then the sets of possible values for $X, Y$, and $Z$ are the finite set ${0,1,2}$, the countable set ${1,2,3,4, \ldots}$, and the uncountable set ${x: x \geq 0}$, respectively. Whenever the set of possible values that a random variable $X$ can assume is at most countable, $X$ is called discrete. Therefore, $X$ is discrete if either the set of its possible values is finite or it is countably infinite. To each discrete random variable, a real-valued function $p: \mathbf{R} \rightarrow \mathbf{R}$, defined by $p(x)=P(X=x)$, is assigned and is called the probability mass function of $X$. (It is also called the probability function of $X$ or the discrete probability function of $X$.) Since the set of values of $X$ is countable, $p(x)$ is positive at most for a countable set. It is zero elsewhere; that is, if possible values of $X$ are $x_1, x_2, x_3, \ldots$, then $p\left(x_i\right) \geq 0(i=1,2,3, \ldots)$ and $p(x)=0$ if $x \notin\left{x_1, x_2, x_3, \ldots\right}$. Now, clearly, the occurrence of the event $\left{x_1, x_2, x_3, \ldots\right}$ is certain. Therefore, we have that $\sum_{i=1}^{\infty} P\left(X=x_i\right)=1$ or, equivalently, $\sum_{i=1}^{\infty} p\left(x_i\right)=1$.

Definition 4.3 The probability mass function $p$ of a random variable $X$ whose set of possible values is $\left{x_1, x_2, x_3, \ldots\right}$ is a function from $\mathbf{R}$ to $\mathbf{R}$ that satisfies the following properties.
(a) $p(x)=0$ if $x \notin\left{x_1, x_2, x_3, \ldots\right}$.
(b) $p\left(x_i\right)=P\left(X=x_i\right)$ and hence $p\left(x_i\right) \geq 0(i=1,2,3, \ldots)$.
(c) $\sum_{i=1}^{\infty} p\left(x_i\right)=1$.

## 数学代写|随机过程代写Stochastic Porcess代考|EXPECTATIONS OF DISCRETE RANDOM VARIABLES

Theorem 1.4 For any event $A, P\left(A^c\right)=1-P(A)$.
Proof: $\quad$ Since $A A^c=\emptyset, A$ and $A^c$ are mutually exclusive. Thus
$$P\left(A \cup A^c\right)=P(A)+P\left(A^c\right)$$
But $A \cup A^c=S$ and $P(S)=1$, so
$$1=P(S)=P\left(A \cup A^c\right)=P(A)+P\left(A^c\right)$$
Therefore, $P\left(A^c\right)=1-P(A)$.
This theorem states that the probability of nonoccurrence of the event $A$ is 1 minus the probability of its occurrence. For example, consider $S={(i, j): 1 \leq i \leq 6,1 \leq j \leq 6}$, the sample space of tossing two fair dice. If $A$ is the event of getting a sum of 4 , then $A={(1,3),(2,2),(3,1)}$ and $P(A)=3 / 36$. Theorem 1.4 states that the probability of $A^c$, the event of not getting a sum of 4 , which is harder to count, is $1-3 / 36=33 / 36$. As another example, consider the experiment of selecting a random number from the set ${1,2,3, \ldots, 1000}$. By Example 1.14, the probability that the number selected is divisible by 3 is $333 / 1000$. Thus by Theorem 1.4 , the probability that it is not divisible by 3 , a quantity harder to find directly, is $1-333 / 1000=667 / 1000$.
Theorem 1.5 If $A \subseteq B$, then
$$P(B-A)=P\left(B A^c\right)=P(B)-P(A)$$
Proof: $A \subseteq B$ implies that $B=(B-A) \cup A$ (see Figure 1.4). But $(B-A) A=\emptyset$. So the events $B-A$ and $A$ are mutually exclusive, and $P(B)=P((B-A) \cup A)=$ $P(B-A)+P(A)$. This gives $P(B-A)=P(B)-P(A)$.

## 数学代写|随机过程代写Stochastic Porcess代考|BASIC THEOREMS

4.3随机变量$X$的可能值集为$\left{x_1, x_2, x_3, \ldots\right}$，其概率质量函数$p$是一个从$\mathbf{R}$到$\mathbf{R}$的函数，满足以下性质。
(a) $p(x)=0$如果$x \notin\left{x_1, x_2, x_3, \ldots\right}$。
(b) $p\left(x_i\right)=P\left(X=x_i\right)$，因此$p\left(x_i\right) \geq 0(i=1,2,3, \ldots)$。
(c) $\sum_{i=1}^{\infty} p\left(x_i\right)=1$。

## 数学代写|随机过程代写Stochastic Porcess代考|EXPECTATIONS OF DISCRETE RANDOM VARIABLES

$$P\left(A \cup A^c\right)=P(A)+P\left(A^c\right)$$

$$1=P(S)=P\left(A \cup A^c\right)=P(A)+P\left(A^c\right)$$

$$P(B-A)=P\left(B A^c\right)=P(B)-P(A)$$

## MATLAB代写

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