All You Need to Know About Densities and Probability Distributions


Notation

Distinction Between Discrete and Continuous Random Variables

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

EXAMPLE 1

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

ANSWER:

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

ANSWER:

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

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

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

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

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

This tutorial is brought to you courtesy of MyGeekyTutor.com



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