Statistics Tutorials: Probability Distributions - All You Need to Know About Densities and Probability Distributions

EXAMPLE 1

f\left( x \right)=\left\{ \begin{align} & \frac{1}{2}\text{ for }x=1, \\ & \frac{1}{4}\text{ for }x=2, \\ & \frac{1}{8}\text{ for }x=3, \\ & \frac{1}{8}\text{ for }x=4 \\ \end{align} \right.

Let us see, we need to see if conditions (1) and (2) are met. First of all, notice that we have $f\left( x \right)\ge 0$for all values {1, 2, 3, 4}, which is the set of all possible values that X can take, since $f\left( 1 \right)=\frac{1}{2}>0$, $f\left( 2 \right)=\frac{1}{4}>0$, $f\left( 3 \right)=\frac{1}{8}>0$and $f\left( 4 \right)=\frac{1}{8}>0$. Therefore, condition (1) is met.

$\sum\limits_{i=1}^{4}{f\left( {{x}_{i}} \right)}=f\left( 1 \right)+f\left( 2 \right)+f\left( 3 \right)+f\left( 4 \right)=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{8}=1$

EXAMPLE 2

$\int\limits_{-\infty }^{\infty }{f\left( x \right)dx}=\int\limits_{0}^{2}{{{x}^{2}}dx}=\left. \frac{{{x}^{3}}}{3} \right|_{0}^{2}=\frac{{{2}^{3}}}{3}-\frac{{{0}^{3}}}{3}=\frac{8}{3}>1$

Finally, How to Compute Probabilities with Densities and Probability Functions?

$\Pr \left( X\in D \right)=\int\limits_{D}^{{}}{f\left( x \right)dx}$ $\Pr \left( X\in \left[ 1,5 \right] \right)=\Pr \left( 1\le X\le 5 \right)=\int\limits_{1}^{5}{f\left( x \right)dx}$

$\Pr \left( X\in D \right)=\Pr \left( X\in \left\{ {{b}_{1}},{{b}_{2}},...,{{b}_{k}} \right\} \right)=\sum\limits_{j=1}^{k}{f\left( {{b}_{j}} \right)}$

$\Pr \left( X\in \left\{ 1,2 \right\} \right)=f\left( 1 \right)+f\left( 2 \right)$