Figure 1The definition of the probability density function is presented starting from a histogram , then a probability histogram . Examples with solutions related to probability density are also included.
We use 2560 data values generated using Google sheets to present the idea of moving from histogram of frequencies to a histogram of probabilities . The data represents the time (in hours) taken to finish a project and hence is assumed to be continuous.
The time here is considered as a continuous random variable.
The histogram obtained using google sheets is shown below. The classes on the horizontal axis have a width equal to 1 and the frequencies are on the vertical axis.
The sum of all frequencies is equal to the total number of data values which is equal to 2560.
Figure 1
Figure 2
Figure 3
Figure 4
It is clear from the above probability histograms and the trendline, that if we have a large number of data values which is the case of a continuous random variable, as the class width becomes smaller, the trendline is a function that may be used to calculate probabilities using the area between the x-axis and the curve of this function.
This trendline function is called the probability density function.
If function \( f_{X}(x) \) is the probability density function of a random variable \( X \), then the probability that \( X \) is greater than or equal to \( a \) and smaller than or equal to \( b \) written as \( P( a \le X \le b) \)is given by the area between the x-axis, the curve and the vertical lines \( x = a \) and \( x = b \).
\[ \displaystyle P(a \le X \le b) = \text{Area between the curve, the x-axis, and x = a to x = b} \]
Figure 5
The (PDF) probability density functions \( f_{X}(x) \) of a continuous random variable \( X \) has the following properties:
1 - \( f_{X}(x) \ge 0 \)
2 - \( \displaystyle P(-\infty \lt X \lt -\infty) = 1 \), total area between curve of \( f_{X}(x) \) and x-axis is equal to 1.
Using integrals in calculus, we can write
\[ \displaystyle P(a \le X \le b) = \int_a^b f_{X}(x) \; dx \]
which is the area between the curve of the PDF, the x-axis, and \( x = a \) to \( x = b \).
\[ \displaystyle \int_{-\infty}^{\infty} f_{X}(x) \; dx = 1\]
which is the total area between the x-axis and the curve of the PDF
Example 1
A random variable \( X \) has a uniform probability density functions \( f_{X}(x) \) given by
\[ \begin{equation}
f_{X}(x) = \left\{
\begin{array}{l l}
\dfrac{k}{b-a} & \quad a \le x \le b\\
0 & \quad x \lt a \text{ or } x \gt b
\end{array} \right.
\end{equation}\]
where \( k \) is a positive constant.
a - Plot the graph of \( f_{X}(x) \).
b - Use the plot above and the properties of the probability density functions to find \(k\).
c - Use integrals in calculus to show that the total area between the x-axis and the curve is equal to \( 1 \).
d - Let \( a = 3 \) and \( b = 8 \), find the probability \( P( 4 \le X \le 7) \).
Solution
a-
The graph of the probability density function \( f_{X}(x) \) is shown below.
Figure 6
Example 2
A random variable \( X \) has a uniform probability density functions \( f_{X}(x) \) given by
\[ \begin{equation}
f_{X}(x) = \left\{
\begin{array}{l l}
e^{-k x} & \quad x \ge 0\\
0 & \quad x \lt 0
\end{array} \right.
\end{equation}\]
where \( k \) is a positive constant.
a - Find \( k \).
b - Find \( a \) so that \( P( 0 \le X \le a) = 0.9\) .
Solution
a-
The property of the probability density function that \( f_{X}(x) \ge 0 \) is satisfied.
For the second property of the probability density function to be satisfied, the total area between the curve of \( f_{X}(x) \) and the x-axis must be equal to \( 1 \).
The area \( A \) between the x-axis and the curve of of \( f_{X}(x) \) is given by the integral
\(\displaystyle \qquad A = \int_{-\infty}^{\infty} f_{X}(x) \; dx \)
Rewrite the above as
\( \displaystyle\qquad A = \int_{-\infty}^{0} f_{X}(x) \; dx + \int_{0}^{\infty} f_{X}(x) \; dx \)
\( f_{X}(x) = 0 \) in the interval \( (-\infty , 0 ) \), hence the above simplifies to
\(\displaystyle \qquad A = \int_{0}^{\infty} f_{X}(x) \; dx \)
The above integral is an improper one and may be written as
\( \displaystyle \qquad A = \lim_{\; b\to\infty} \int_{0}^{b} f_{X}(x) \; dx \)
Substitute \( f_{X}(x) \) by \( e^{-k x} \) and evaluate the integral
\( \displaystyle \qquad A = \lim_{\; b\to\infty} \int_{0}^{b} e^{-k x} \; dx \)
\( \displaystyle \qquad A = \lim_{\; b\to\infty} \left[ -\frac{1}{k}e^{-kx} \right]_0^b \)
\( \displaystyle \qquad A = \lim_{\; b\to\infty} \left[ -\frac{1}{k}e^{-k b} + \frac{1}{k} e^{0} \right] \)
Simplify
\( \displaystyle \qquad A = - \lim_{\; b\to\infty} \frac{1}{k}e^{-k b} + \frac{1}{k} \)
Since \( k \) is positive, \( \displaystyle \lim_{\; b\to\infty} \frac{1}{k}e^{-k b} = 0 \) and hence
\( \displaystyle \qquad A = \dfrac{1}{k} \)
The total area \( A = 1 \), hence the equation
\( \qquad \dfrac{1}{k} = 1 \)
Solve for \( k \) to obtain
\( \qquad k = 1 \)
b-
\( \displaystyle \qquad P( 0 \le X \le a) = \int_0^a f_{X}(x) \; dx \)
Since \( k = 1 \), \( f_{X} (x) = e^{- x} \) for \( x \ge 0 \), hence
\( \displaystyle \qquad P( 0 \le X \le a) = \int_0^a e^{-x} \; dx \)
Evaluate the above integral.
\( \displaystyle \qquad P( 0 \le X \le a) = \left[-e^{-x} \right]_0^a = -e^{-a} + 1 \)
Since \( P( 0 \le X \le a) = 0.9 \) ; we write the equation
\( \qquad -e^{-a} + 1 = 0.9 \)
Solve for \( a \)
\( \qquad e^{-a} = 0.1 \)
\( \qquad a = - \ln 0.1 \approx 2.3 \)