Explain and apply the concepts of random variables

Given a random experiment, with sample space, \(S\), we can define the possible values of \(S\) as a random variable. Random variables can be discrete or continuous.

A discrete random variable is a variable whose range of possible values is from a specified list.

A probability density function(sometimes called the probability mass function if the variable is discrete) is a function that describes the relationship between a random variable, \(X\), and the probability of each value of \(x\) occurring.

Example

The probability density function shown in the graph below can be written as:

$$ f\left( x \right) =\begin{cases} .2, & x=1,4 \\ .3, & x=2,3 \end{cases} $$

The probability density (or mass) function of a discrete random variable has the following properties:

  1. \(f \left(x \right) > 0,\quad x € S;\quad\) meaning all individual probabilities must be greater than zero
  2. \({ \Sigma }_{ x\in S }f\left( x \right)=1; \quad \)meaning the sum of all individual probabilities must equal one
  3. \(P\left( X\in A \right) ={ \Sigma }_{ x\in A }f\left( x \right) \quad \),where \(A\subset S;\quad \)meaning the probability of event \(A\) is the sum of the probabilities of the values of \(X\) in \(A\).

If we are interested in cumulative probabilities then we can use the cumulative distribution function, or cdf, which is defined by the function shown below:

$$ F\left( x \right) =P\left( X\le x \right) ,\quad \quad -\infty <x<\infty $$

Example

The cumulative distribution function for the probability density function shown above can be written as:

$$ F\left( x \right) =\begin{cases} 0, & x<1 \\ .2, & 1\le x<2 \\ .5, & 2\le x<3 \\ .8, & 3\le x<4 \\ 1, & x\ge 4 \end{cases} $$

A continuous random variable is a random variable that has an infinite number of possible values.

The probability density function, \(f(x)\), of a continuous random variable is a differential equation with the following properties:

  1. \(f\left( x \right)\ge 0\) meaning the function (all probabilities) is always greater than zero
  2. \(\int _{ -\infty }^{ \infty }{ f\left( x \right)dx=1 } \) meaning the sum of all the possible values (probabilities) is one

It is important to note that unlike the probability density function of a discrete random variable, the probability density function of a continuous random variable is not a probability. In order to calculate a probability using a probability density function, the following function must be used;

$$ P\left( X < a \right) =P\left( X\le a \right) =F\left( a \right) =\int _{ -\infty }^{ a }{ f\left( x \right) dx } $$

Also, it is important to note that the probability of any individual value of a probability density function is zero, as shown in the formula below:

$$ P\left( X=a \right) =\int _{ a }^{ a }{ f\left( x \right) dx=0 } $$

Example

Given the following probability density function of a continuous random variable:

$$ f\left( x \right) =\begin{cases} { x }^{ 2 }+C, & 0 < x < 1 \\ 0, & otherwise \end{cases} $$

  1. Calculate C.

    $$\begin{align*}& \int _{ 0 }^{ 1 }{ \left( { x }^{ 2 }+C \right) dx=1 } \\& { \left[ \frac { { x }^{ 3 } }{ 3 } +Cx \right] }_{ x=0 }^{ x=1 }=1 \\& {1}/{3} + C = 1 \\& C = {2}/{3} \\\end{align*}$$

  2. Calculate \(P (X > {\frac {1}{2})}\).

    $$ \begin{align*}& \int _{ { {1}/{2}} }^{ 1 }{ \left( { x }^{ 2 }+{ {2}/{3}} \right) dx } \\& { \left[ \frac { { x }^{ 3 } }{ 3 } +{ {2}/{3}}x \right] }_{ x={\frac {1}{2}} }^{ x=1 } \\& \left[ {1}/{3}+{2}/{3} \right] – \left[{1}/{24}+{1}/{3} \right] = {5}/{8} \\\end{align*}$$

Conditional probabilities of random variables can be calculated in a similar way to empirical conditional

probabilities as shown in the formula below:

$$ P\left( A|B \right) =\frac { P\left( A\cap B \right) }{ P\left( B \right) } $$

Let event \(A\) be \(X > 2\) and event \(B\) be \(X < 4\) then

$$ P\left( A|B \right) =\frac { P\left( 2<X<4 \right) }{ P\left( X<4 \right) } $$

Example

Given the following probability density function of a continuous random variable:

$$ f\left( x \right) =\begin{cases} { x }^{ 2 }+{2}/{3}, & 0 < x < 1 \\ 0, & otherwise \end{cases} $$

  1. Calculate \(P \left(X > {\frac {3}{4}} | X > {\frac {1}{2}} \right)\).

    $$\begin{align*}& P \left(X < {\frac {3}{4}} | X > {\frac {1}{2}} \right ) ={ P \left( {\frac {1}{2}} < X < {\frac {3}{4}} \right) } /{ P \left(X > {\frac {1}{2}} \right)} \\& P \left( {\frac {1}{2}} < X < {\frac {3}{4}} \right) = \\& \int _{ { 1 }/{ 2 } }^{ { 3 }/{ 4 } }{ \left( { x }^{ 2 }+{ 2 }/{ 3 } \right) dx } \\& { \left[ \frac { { x }^{ 3 } }{ 3 } +{ {2}/{3}}x \right] }_{ x={ {1}/{2}} }^{ x={3}/{4} } = .265625 \\& {.265625}/\left({5}/{8} \right) = .425 \\\end{align*}$$

 

Learning Outcome

Topic 2.a – b: Univariate Random Variables – Explain and apply the concepts of random variables, probability and probability density functions, cumulative distribution functions & Calculate conditional probabilities.


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