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Certain mathematical expectations are so important that they have special names. In this section we consider two of them: the mean and the variance.
Mean Value
If X is a random variable with p.d.f. $f\left(x\right)$ of the discrete type and space R = $\left({b}_{1},{b}_{2},{b}_{3}\mathrm{,...}\right)$ , then $E\left(X\right)={\displaystyle \sum _{R}xf\left(x\right)={b}_{1}f\left({b}_{1}\right)+{b}_{2}f\left({b}_{2}\right)+{b}_{3}}f\left({b}_{3}\right)+\mathrm{...}$ is the weighted average of the numbers belonging to R , where the weights are given by the p.d.f. $f\left(x\right)$ .
We call $E\left(X\right)$ the mean of X (or the mean of the distribution ) and denote it by $\mu $ . That is, $\mu =E\left(X\right)$ .
Let X have the p.d.f.
$$f\left(x\right)=\{\begin{array}{l}\frac{1}{8},x=\mathrm{0,3,}\\ \frac{3}{8},x=\mathrm{1,2.}\end{array}$$
The mean of X is
$$\mu =E[X=0\left(\frac{1}{8}\right)+1\left(\frac{3}{8}\right)+2\left(\frac{3}{8}\right)+3\left(\frac{1}{8}\right)=\frac{3}{2}.$$
The example below shows that if the outcomes of X are equally likely (i.e., each of the outcomes has the same probability), then the mean of X is the arithmetic average of these outcomes.
Roll a fair die and let X denote the outcome. Thus X has the p.d.f. $$f\left(x\right)=\frac{1}{6},x=1,2,3,4,5,6.$$ Then,
$$E\left(X\right)={\displaystyle \sum _{x=1}^{6}x\left(\frac{1}{6}\right)=\frac{1+2+3+4+5+6}{6}=\frac{7}{2},}$$
which is the arithmetic average of the first six positive integers.
Variance
It was denoted that the mean $\mu =E\left(X\right)$ is the centroid of a system of weights of measure of the central location of the probability distribution of X . A measure of the dispersion or spread of a distribution is defined as follows:
If $u\left(x\right)={\left(x-\mu \right)}^{2}$ and $E\left[{\left(X-\mu \right)}^{2}\right]$ exists, the variance , frequently denoted by ${\sigma}^{2}$ or $Var\left(X\right)$ , of a random variable X of the discrete type (or variance of the distribution) is defined by
The positive square root of the variance is called the standard deviation of X and is denoted by
Let the p.d.f. of X by defined by $$f\left(x\right)=\frac{x}{6},x=1,2,3.$$
The mean of X is
$$\mu =E\left(X\right)=1\left(\frac{1}{6}\right)+2\left(\frac{2}{6}\right)+3\left(\frac{3}{6}\right)=\frac{7}{3}.$$
To find the variance and standard deviation of X we first find
$$E\left({X}^{2}\right)={1}^{2}\left(\frac{1}{6}\right)+{2}^{2}\left(\frac{2}{6}\right)+{3}^{2}\left(\frac{3}{6}\right)=\frac{36}{6}=6.$$
Thus the variance of X is $${\sigma}^{2}=E\left({X}^{2}\right)-{\mu}^{2}=6-{\left(\frac{7}{3}\right)}^{2}=\frac{5}{9},$$
and the standard deviation of X is $$\sigma =\sqrt{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$9$}\right.}=\mathrm{0.745.}$$
Let X be a random variable with mean ${\mu}_{x}$ and variance ${\sigma}_{x}^{2}$ . Of course, $Y=aX+b$ , where a and b are constants, is a random variable, too. The mean of Y is
$${\mu}_{Y}=E\left(Y\right)=E\left(aX+b\right)=aE\left(X\right)+b=a{\mu}_{X}+b.\text{}$$
Moreover, the variance of Y is
$${\sigma}_{Y}^{2}=E\left[{\left(Y-{\mu}_{Y}\right)}^{2}\right]=E\left[{\left(aX+b-a{\mu}_{X}-b\right)}^{2}\right]=E\left[{a}^{2}{\left(X-{\mu}_{X}\right)}^{2}\right]={a}^{2}{\sigma}_{X}^{2}.$$
Moments of the distribution
Let r be a positive integer. If $$E\left({X}^{r}\right)={\displaystyle \sum _{R}{x}^{r}f\left(x\right)}$$ exists, it is called the r th moment of the distribution about the origin. The expression moment has its origin in the study of mechanics.
In addition, the expectation $$E\left[{\left(X-b\right)}^{r}\right]={\displaystyle \sum _{R}{x}^{r}f\left(x\right)}$$ is called the r th moment of the distribution about b . For a given positive integer r.
$$E\left[{\left(X\right)}_{r}\right]=E\left[X\left(X-1\right)\left(X-2\right)\cdot \cdot \cdot \left(X-r+1\right)\right]$$ is called the r th factorial moment .
There is another formula that can be used for computing the variance that uses the second factorial moment and sometimes simplifies the calculations.
First find the values of $E\left(X\right)$ and $E\left[X\left(X-1\right)\right]$ . Then $${\sigma}^{2}=E\left[X\left(X-1\right)\right]+E\left(X\right)-{\left[E\left(X\right)\right]}^{2},$$ since using the distributive property of E , this becomes $${\sigma}^{2}=E\left({X}^{2}\right)-E\left(X\right)+E\left(X\right)-{\left[E\left(X\right)\right]}^{2}=E\left({X}^{2}\right)-{\mu}^{2}.$$
Let continue with example 4 , it can be find that
$$E\left[X\left(X-1\right)\right]=1\left(0\right)\left(\frac{1}{6}\right)+2\left(1\right)\left(\frac{2}{6}\right)+3\left(2\right)\left(\frac{3}{6}\right)=\frac{22}{6}.$$
Thus $${\sigma}^{2}=E\left[X\left(X-1\right)\right]+E\left(X\right)-{\left[E\left(X\right)\right]}^{2}=\frac{22}{6}+\frac{7}{3}-{\left(\frac{7}{3}\right)}^{2}=\frac{5}{9}.$$
The symbol $\overline{x}$ represents the mean of the empirical distribution . It is seen that $\overline{x}$ is usually close in value to $\mu =E\left(X\right)$ ; thus, when $\mu $ is unknown, $\overline{x}$ will be used to estimate $\mu $ .
Similarly, the variance of the empirical distribution can be computed. Let v denote this variance so that it is equal to
$$v={{\displaystyle \sum _{i=1}^{n}\left({x}_{i}-\overline{x}\right)}}^{2}\frac{1}{n}={\displaystyle \sum _{i=1}^{n}{x}_{i}^{2}\frac{1}{n}-{\overline{x}}^{2}}=\frac{1}{n}{\displaystyle \sum _{i=1}^{n}{x}_{i}^{2}-{\overline{x}}^{2}}.$$
This last statement is true because, in general, $${\sigma}^{2}=E\left({X}^{2}\right)-{\mu}^{2}.$$
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