统计代写|随机过程代写stochastic process代考|EXPECTATIONS OF DISCRETE RANDOM VARIABLES

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统计代写|随机过程代写stochastic process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
$$(0.6) \cdot(-1)+(0.3) \cdot 1+(0.08) \cdot 2+(0.02) \cdot 3=-0.08$$
Hence
$$-1 \cdot p(-1)+1 \cdot p(1)+2 \cdot p(2)+3 \cdot p(3)=-0.08$$
a relation showing that the expected value of $X$ can be calculated directly by summing up the product of possible values of $X$ by their probabilities. This and similar examples motivate the following general definition, which was first used casually by Pascal but introduced formally by Huygens in the late seventeenth century.

统计代写|随机过程代写stochastic process代考|VARIANCES AND MOMENTS OF DISCRETE RANDOM VARIABLES

Thus far, through many examples, we have explained the importance of mathematical expectation in detail. For instance, in Example 4.16, we have shown how expectation is applied in decision making. Also, in Example 4.17, concerning the lottery, we showed that the expected value of the winning amount per game gives an excellent estimation for the total amount a player will win if he or she plays a large number of times. In these and many other situations, mathematical expectation is the only quantity one needs to calculate. However, very frequently we face situations in which the expected value by itself does not say much. In such cases more information should be extracted from the probability mass function. As an example, suppose that we are interested in measuring a certain quantity. Let $X$ be the true value ${ }^{\dagger}$ of the quantity minus the value obtained by measurement. Then $X$ is the error of measurement. It is a random variable with expected value zero, the reason being that in measuring a quantity a very large number of times, positive and negative errors of the same magnitudes occur with equal probabilities. Now consider an experiment in which a quantity is measured several times, and the average of the errors is obtained to be a number close to zero. Can we conclude that the measurements are very close to the true value and thus are accurate? The answer is no because they might differ from the true value by relatively large quantities but be scattered both in positive and negative directions, resulting in zero expectation. Thus in this and similar cases, expectation by itself does not give adequate information, so additional measures for decision making are needed. One such quantity is the variance of a random variable.

Variance measures the average magnitude of the fluctuations of a random variable from its expected value. This is particularly important because random variables fluctuate from their expected values. To mathematically define the variance of a random variable $X$, the first temptation is to consider the expectation of the difference of $X$ from its expected value, that is, $E[X-E(X)]$. But the difficulty with this quantity is that the positive and negative deviations of $X$ from $E(X)$ cancel each other, and we always get 0 . This can be seen mathematically from the corollary of Theorem 4.2: Let $E(X)=\mu$; then
$$E[X-E(X)]=E(X-\mu)=E(X)-\mu=E(X)-E(X)=0 .$$

随机过程代考

统计代写|随机过程代写stochastic process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}

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

$$-1 \cdot p(-1)+1 \cdot p(1)+2 \cdot p(2)+3 \cdot p(3)=-0.08$$

统计代写|随机过程代写stochastic process代考|VARIANCES AND MOMENTS OF DISCRETE RANDOM VARIABLES

$$E[X-E(X)]=E(X-\mu)=E(X)-\mu=E(X)-E(X)=0 .$$

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