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# 数学代写|随机过程Stochastic Porcess代考|CONTINUITY OF PROBABILITY FUNCTIONS

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## 数学代写|随机过程代写Stochastic Porcess代考|CONTINUITY OF PROBABILITY FUNCTIONS

Let $\mathbf{R}$ denote (here and everywhere else throughout the book) the set of all real numbers. We know from calculus that a function $f: \mathbf{R} \rightarrow \mathbf{R}$ is called continuous at a point $c \in \mathbf{R}$ if $\lim {x \rightarrow c} f(x)=f(c)$. It is called continuous on $\mathbf{R}$ if it is continuous at all points $c \in \mathbf{R}$. We also know that this definition is equivalent to the sequential criterion $f: \mathbf{R} \rightarrow \mathbf{R}$ is continuous on $\mathbf{R}$ if and only if, for every convergent sequence $\left{x_n\right}{n=1}^{\infty}$ in $\mathbf{R}$,
$$\lim {n \rightarrow \infty} f\left(x_n\right)=f\left(\lim {n \rightarrow \infty} x_n\right)$$
This property, in some sense, is shared by the probability function. To explain this, we need to introduce some definitions. But first recall that probability is a set function from $\mathcal{P}(S)$, the set of all possible events of the sample space $S$, to $[0,1]$.
A sequence $\left{E_n, n \geq 1\right}$ of events of a sample space is called increasing if
$$E_1 \subseteq E_2 \subseteq E_3 \subseteq \cdots \subseteq E_n \subseteq E_{n+1} \cdots ;$$

it is called decreasing if
$$E_1 \supseteq E_2 \supseteq E_3 \supseteq \cdots \supseteq E_n \supseteq E_{n+1} \supseteq \cdots$$
For an increasing sequence of events $\left{E_n, n \geq 1\right}$, by $\lim {n \rightarrow \infty} E_n$ we mean the event that at least one $E_i, 1 \leq i<\infty$ occurs. Therefore, $$\lim {n \rightarrow \infty} E_n=\bigcup_{i=1}^{\infty} E_i .$$
Similarly, for a decreasing sequence of events $\left{E_n, n \geq 1\right}$, by $\lim {n \rightarrow \infty} E_n$ we mean the event that every $E_i$ occurs. Thus in this case $$\lim {n \rightarrow \infty} E_n=\bigcap_{i=1}^{\infty} E_i .$$
The following theorem expresses the property of probability function that is analogous to (1.4).

## 数学代写|随机过程代写Stochastic Porcess代考|PROBABILITIES 0 AND 1

Events with probabilities 1 and 0 should not be misinterpreted. If $E$ and $F$ are events with probabilities 1 and 0 , respectively, it is not correct to say that $E$ is the sample space $S$ and $F$ is the empty set $\emptyset$. In fact, there are experiments in which there exist infinitely many events each with probability 1 , and infinitely many events each with probability 0 . An example follows.
Suppose that an experiment consists of selecting a random point from the interval $(0,1)$. Since every point in $(0,1)$ has a decimal representation such as
$0.529387043219721 \cdots$,
the experiment is equivalent to picking an endless decimal from $(0,1)$ at random (note that if a decimal terminates, all of its digits from some point on are 0 ). In such an experiment we want to compute the probability of selecting the point $1 / 3$. In other words, we want to compute the probability of choosing $0.333333 \cdots$ in a random selection of an endless decimal. Let $A_n$ be the event that the selected decimal has 3 as its first $n$ digits; then
$$A_1 \supset A_2 \supset A_3 \supset A_4 \supset \cdots \supset A_n \supset A_{n+1} \supset \cdots$$

since the occurrence of $A_{n+1}$ guarantees the occurrence of $A_n$. Now $P\left(A_1\right)=1 / 10$ because there are 10 choices $0,1,2, \ldots, 9$ for the first digit, and we want only one of them, namely 3 , to occur. $P\left(A_2\right)=1 / 100$ since there are 100 choices $00,01, \ldots, 09,10,11, \ldots, 19,20$, $\ldots, 99$ for the first two digits, and we want only one of them, 33, to occur. $P\left(A_3\right)=1 / 1000$ because there are 1000 choices $000,001, \ldots, 999$ for the first three digits, and we want only one of them, 333, to occur. Continuing this argument, we have $P\left(A_n\right)=(1 / 10)^n$. Since $\bigcap_{n=1}^{\infty} A_n={1 / 3}$, by Theorem 1.8 ,
$$P\left(\frac{1}{3} \text { is selected }\right)=P\left(\bigcap_{n=1}^{\infty} A_n\right)=\lim {n \rightarrow \infty} P\left(A_n\right)=\lim {n \rightarrow \infty}\left(\frac{1}{10}\right)^n=0 .$$
Note that there is nothing special about the point $1 / 3$. For any other point $0 . \alpha_1 \alpha_2 \alpha_3 \alpha_4 \cdots$ from $(0,1)$, the same argument could be used to show that the probability of its occurrence is 0 (define $A_n$ to be the event that the first $n$ digits of the selected decimal are $\alpha_1, \alpha_2, \ldots, \alpha_n$, respectively, and repeat the same argument). We have shown that in random selection of points from $(0,1)$, the probability of the occurrence of any particular point is 0 . Now for $t \in(0,1)$, let $B_t=(0,1)-{t}$. Then $P({t})=0$ implies that
$$P\left(B_t\right)=P\left({t}^c\right)=1-P({t})=1$$

## 数学代写|随机过程代写Stochastic Porcess代考|CONTINUITY OF PROBABILITY FUNCTIONS

$$\lim {n \rightarrow \infty} f\left(x_n\right)=f\left(\lim {n \rightarrow \infty} x_n\right)$$

$$E_1 \subseteq E_2 \subseteq E_3 \subseteq \cdots \subseteq E_n \subseteq E_{n+1} \cdots ;$$

$$E_1 \supseteq E_2 \supseteq E_3 \supseteq \cdots \supseteq E_n \supseteq E_{n+1} \supseteq \cdots$$

## 数学代写|随机过程代写Stochastic Porcess代考|PROBABILITIES 0 AND 1

$0.529387043219721 \cdots$，

$$A_1 \supset A_2 \supset A_3 \supset A_4 \supset \cdots \supset A_n \supset A_{n+1} \supset \cdots$$

$$P\left(\frac{1}{3} \text { is selected }\right)=P\left(\bigcap_{n=1}^{\infty} A_n\right)=\lim {n \rightarrow \infty} P\left(A_n\right)=\lim {n \rightarrow \infty}\left(\frac{1}{10}\right)^n=0 .$$

$$P\left(B_t\right)=P\left({t}^c\right)=1-P({t})=1$$

## MATLAB代写

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