All You Need to Know About Densities and Probability Distributions


Distinction Between Discrete and Continuous Random Variables

Properties that Need to be Met by ALL Probability and Density Functions


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


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.

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



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

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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