Continuous distribution

Continuous distribution

Random variables of the continuous type

Random variables whose spaces are not composed of a countable number of points but are intervals or a union of intervals are said to be of the continuous type . Recall that the relative frequency histogram $h\left(x\right)$ associated with n observations of a random variable of that type is a nonnegative function defined so that the total area between its graph and the x axis equals one. In addition, $h\left(x\right)$ is constructed so that the integral

is an estimate of the probability $P\left(a<X<b\right)$ , where the interval $\left(a,b\right)$ is a subset of the space R of the random variable X .

Let now consider what happens to the function $h\left(x\right)$ in the limit, as n increases without bound and as the lengths of the class intervals decrease to zero. It is to be hoped that $h\left(x\right)$ will become closer and closer to some function, say $f\left(x\right)$ , that gives the true probabilities , such as $P\left(a<X<b\right)$ , through the integral

PROBABILITY DENSITY FUNCTION

We say that the probability density function (p.d.f.) of the random variable X of the continuous type, with space R that is an interval or union of intervals, is an integrable function $f\left(x\right)$ satisfying the following conditions:

• $f\left(x\right)>0$ , x belongs to R ,
• ${\displaystyle \underset{R}{\overset{}{\int }}f(x)dx}=\mathrm{1,}$
• The probability of the event A belongs to R is $P\left(X\right)\in A{\displaystyle \underset{A}{\overset{}{\int }}f(x)dx.}$

We can avoid repeated references to the space R of the random variable X , one shall adopt the same convention when describing probability density function of the continuous type as was in the discrete case.

Let extend the definition of the p.d.f. $f\left(x\right)$ to the entire set of real numbers by letting it equal zero when, x belongs to R . For example,

$$f\left(x\right)=\{\begin{array}{l}\frac{1}{40}{e}^{-x/40}\\ \mathrm{0,}elsewhere,\end{array}\mathrm{,0}\le x<\infty ,$$

has the properties of a p.d.f. of a continuous-type random variable x having support $\left(x:0\le x<\infty \right)$ . It will always be understood that $f(x)=0$ , when x belongs to R , even when this is not explicitly written out.

PROBABILITY DENSITY FUNCTION

For the fundamental theorem of calculus we have, for x values for which the derivative $F\text{'}\left(x\right)$ exists, that F ’( x )= f ( x ).

Also $F\text{'}\left(0\right)$ does not exist. Since there are no steps or jumps in a distribution function $F\left(x\right)$ , of the continuous type, it must be true that $$P\left(X=b\right)=0$$ for all real values of b . This agrees with the fact that the integral

$${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}$$ is taken to be zero in calculus. Thus we see that $$P\left(a\le X\le b\right)=P\left(a<X<b\right)=P\left(a\le X<b\right)=P\left(a<X\le b\right)=F\left(b\right)-F\left(a\right),$$ provided that X is a random variable of the continuous type. Moreover, we can change the definition of a p.d.f. of a random variable of the continuous type at a finite (actually countable) number of points without alerting the distribution of probability.

For illustration, $$f\left(x\right)=\{\begin{array}{l}\mathrm{0,}-\infty <x<\mathrm{0,}\\ \frac{1}{40}{e}^{-x/40}\mathrm{,0}\le x<\infty ,\end{array}$$ and $$f\left(x\right)=\{\begin{array}{l}\mathrm{0,}-\infty <x\le \mathrm{0,}\\ \frac{1}{40}{e}^{-x/40}\mathrm{,0}<x<\infty ,\end{array}$$

are equivalent in the computation of probabilities involving this random variable.

For illustration of computations of probabilities, consider

$$P\left(\frac{1}{2}<Y\le \frac{3}{4}\right)=G\left(\frac{3}{4}\right)-G\left(\frac{1}{2}\right)={\left(\frac{3}{4}\right)}^2-{\left(\frac{1}{2}\right)}^2=\frac{5}{16}$$

and $$P\left(\frac{1}{4}\le Y<2\right)=G\left(2\right)-G\left(\frac{1}{4}\right)=1-{\left(\frac{1}{4}\right)}^2=\frac{15}{16}.$$

For random variables of the continuous type, the p.d.f. does not have to be bounded. The restriction is that the area between the p.d.f. and the x axis must equal one. Furthermore, it should be noted that the p.d.f. of a random variable X of the continuous type does not need to be a continuous function.

For example,

$$f\left(x\right)=\{\begin{array}{l}\frac{1}{\mathrm{2,}}0<x<1or2<x<\mathrm{3,}\\ \mathrm{0,}elsewhere,\end{array}$$

enjoys the properties of a p.d.f. of a distribution of the continuous type, and yet $f\left(x\right)$ had discontinuities at $x=\mathrm{0,1,2,}$ and 3. However, the distribution function associates with a distribution of the continuous type is always a continuous function. For continuous type random variables, the definitions associated with mathematical expectation are the same as those in the discrete case except that integrals replace summations.

FOR ILLUSTRATION , let X be a random variable with a p.d.f. $f\left(x\right)$ . The expected value of X or mean of X is

$$\mu =E\left(X\right)={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}xf\left(x\right)dx.}$$ The variance of X is $${\sigma }^2=Var\left(X\right)={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\left(x-\mu \right)}^{2}f\left(x\right)dx.}$$

The standard deviation of X is $$\sigma =\sqrt{Var\left(X\right)}.$$