# All You Need to Know About Densities and Probability Distributions

### EXAMPLE 1

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

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 $$f\left( 1 \right)=\frac{1}{2}>$$, $$f\left( 2 \right)=\frac{1}{4}>0$$, $$f\left( 3 \right)=\frac{1}{8}>0$$ 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?

$\Pr \left( X\in D \right)=\int\limits_{D}^{{}}{f\left( x \right)dx}$ $\Pr \left( X\in \left[ 1,5 \right] \right)=\Pr \left( 1\le X\le 5 \right)=\int\limits_{1}^{5}{f\left( x \right)dx}$ $\Pr \left( X\in D \right)=\Pr \left( X\in \left\{ {{b}_{1}},{{b}_{2}},...,{{b}_{k}} \right\} \right)=\sum\limits_{j=1}^{k}{f\left( {{b}_{j}} \right)}$

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

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