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

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

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

To clarify the concept of expectation, consider a casino game in which the probability of losing $\$ 1$per game is 0.6 , and the probabilities of winning$\$1, \$ 2$, and$\$3$ per game are $0.3,0.08$, and 0.02 , respectively. The gain or loss of a gambler who plays this game only a few times depends on his luck more than anything else. For example, in one play of the game, a lucky gambler might win $\$ 3$, but he has a$60 \%$chance of losing$\$1$. However, if a gambler decides to play the game a large number of times, his loss or gain depends more on the number of plays than on his luck. A calculating player argues that if he plays the game $n$ times, for a large $n$, then in approximately $(0.6) n$ games he will lose $\$ 1$per game, and in approximately$(0.3) n,(0.08) n$, and$(0.02) n$games he will win$\$1, \$ 2$, and$\$3$, respectively. Therefore, his total gain is
$$(0.6) n \cdot(-1)+(0.3) n \cdot 1+(0.08) n \cdot 2+(0.02) n \cdot 3=(-0.08) n .$$
This gives an average of $\$-0.08$, or about 8 cents of loss per game. The more the gambler plays, the less luck interferes and the closer his loss comes to$\$0.08$ per game. If $X$ is the random variable denoting the gain in one play, then the number -0.08 is called the expected value of $X$. We write $E(X)=-0.08 . E(X)$ is the average value of $X$. That is, if we play the game $n$ times and find the average of the values of $X$, then as $n \rightarrow \infty, E(X)$ is obtained. Since, for this game, $E(X)<0$, we have that, on the average, the more we play, the more we lose. If for some game $E(X)=0$, then in the long run the player neither loses nor wins. Such games are called fair. In this example, $X$ is a discrete random variable with the set of possible values ${-1,1,2,3}$. The probability mass function of $X, p(x)$, is given by
\begin{tabular}{c|cccc}
$i$ & -1 & 1 & 2 & 3 \
\hline$p(i)=P(X=i)$ & 0.6 & 0.3 & 0.08 & 0.02
\end{tabular}
and $p(x)=0$ if $x \notin{-1,1,2,3}$. Dividing both sides of (4.1) by $n$, we obtain
$$(0.6) \cdot(-1)+(0.3) \cdot 1+(0.08) \cdot 2+(0.02) \cdot 3=-0.08 \text {. }$$

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

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

$$(0.6) n \cdot(-1)+(0.3) n \cdot 1+(0.08) n \cdot 2+(0.02) n \cdot 3=(-0.08) n .$$

\begin{tabular}{c|cccc}
$i$ & -1 & 1 & 2 & 3 \hline$p(i)=P(X=i)$ & 0.6 & 0.3 & 0.08 & 0.02
\end{tabular}

$$(0.6) \cdot(-1)+(0.3) \cdot 1+(0.08) \cdot 2+(0.02) \cdot 3=-0.08 \text {. }$$

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