Statistical terminology  (Page 2/4)

Probability distribution function for a continuous random variable.

The cumulative distributon function.

Let $f\left(x\right)=x^2$ be the pdf for the random variable x defined between 0 and 1. The cumulative distribution function for any a is $$F\left(a\right)={\displaystyle \underset{0}{\overset{a}{\int }}{x}^{2}dx}=\frac{1}{3}{x^3|}_0^a=\frac{a^3}{3}.$$

Expected value calculation.

Let $f\left(x\right)=x^a$ be a pdf for $0\le x\le 1$ and $a>0.$ Let $g\left(x\right)=x^3.$ We can calculate $$E\left[g\left(x\right)\right]={\displaystyle \underset{0}{\overset{1}{\int }}\left(x^3\right)x^{a}dx}={\displaystyle \underset{0}{\overset{1}{\int }}{x}^{a+3}dx}=\frac{1}{a+4}{x^{a+4}|}_0^1=\frac{1}{a+4}.$$

Calculation of the population mean.

Assume we have the same pdf used in Example 7. The population mean for this distribution is $$\mu =E\left[x\right]={\displaystyle \underset{0}{\overset{1}{\int }}\left(x\right)x^{a}dx}={\displaystyle \underset{0}{\overset{1}{\int }}{x}^{a+1}dx}=\frac{1}{a+2}{x^{a+2}|}_0^1=\frac{1}{a+2}.$$

Calculation of the population variance using the expected value operator.

Define the variance operator, V , to be:

$$V\left(x\right)=E\left[{\left(x-\mu \right)}^2\right].$$

Then,

$$E\left[{\left(x-\mu \right)}^2\right]={\displaystyle \int {\left(x-\mu \right)}^{2}f\left(x\right)dx}.$$

Squaring the term in the integral gives: ${\displaystyle \int \left(x^2-2\mu x+{\mu }^2\right)f\left(x\right)dx}=E\left(x^2-2\mu x+{\mu }^2\right).$

Expand of the left-hand-side of this equality:

$${\displaystyle \int x^{2}f\left(x\right)dx}-{\displaystyle \int 2\mu xf\left(x\right)dx+}{\displaystyle \int {\mu }^{2}f\left(x\right)dx}=E\left(x^2\right)-E\left(2\mu x\right)+E\left({\mu }^2\right).$$

Thus, we have established that:

$$E\left[{\left(x-\mu \right)}^2\right]=E\left(x^2\right)-E\left(2\mu x\right)+E\left({\mu }^2\right).$$

Evaluating the last two terms gives

$$E\left(2\mu x\right)={\displaystyle \int 2\mu xf\left(x\right)dx}=2\mu {\displaystyle \int x}dx=2{\mu }^2$$

and

$$E\left({\mu }^2\right)={\displaystyle \int {\mu }^{2}f\left(x\right)dx}$$

or, since ${\displaystyle \int f\left(x\right)dx}=1,$ that $E\left({\mu }^2\right)={\mu }^2.$ Thus, $E\left[{\left(x-\mu \right)}^2\right]=E\left(x^2\right)-2{\mu }^2+{\mu }^2$ or

$$E\left[{\left(x-\mu \right)}^2\right]=E\left(x^2\right)-{\mu }^2.$$

For example, in Example 8 we found that $\mu =\frac{1}{a+2}.$ The expected value of $x^2$ is

$$E\left[x^2\right]={\displaystyle \underset{0}{\overset{1}{\int }}\left(x^2\right)x^{a}dx}={\displaystyle \underset{0}{\overset{1}{\int }}{x}^{a+2}dx}=\frac{1}{a+3}{x^{a+3}|}_0^1=\frac{1}{a+3}.$$

Thus, the variance of the distribution is

$$V\left(x\right)=\frac{1}{a+3}-{\left(\frac{1}{a+2}\right)}^2$$

or $$V\left(x\right)=\frac{{\left(a+2\right)}^2-\left(a+3\right)}{\left(a+3\right){\left(a+2\right)}^2}=\frac{a^2+3a+1}{\left(a+3\right){\left(a+2\right)}^2}.$$