Probability Density Function is the function of probability defined for various distributions of variables and is the less common topic in the study of probability throughout the academic journey of students. However, this function is very useful in many areas of real life such as predicting rainfall, financial modelling such as the stock market, income disparity in social sciences, etc.
This article explores the topic of the Probability Density Function in detail including its definition, condition for existence of this function, as well as various examples.
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Probability Density Function is used for calculating the probabilities for continuous random variables. When the cumulative distribution function (CDF) is differentiated we get the probability density function (PDF). Both functions are used to represent the probability distribution of a continuous random variable.
The probability density function is defined over a specific range. By differentiating CDF we get PDF and by integrating the probability density function we can get the cumulative density function.
Probability density function is the function that represents the density of probability for a continuous random variable over the specified ranges.
Probability Density Function is abbreviated as PDF and for a continuous random variable X, Probability Density Function is denoted by f(x).
PDF of the random variable is obtained by differentiating CDF (Cumulative Distribution Function) of X. The probability density function should be a positive for all possible values of the variable. The total area between the density curve and the x-axis should be equal to 1.
Let X be the continuous random variable with probability density function f(x). For a function to be valid probability function should satisfy below conditions.
So, the PDF should be the non-negative and piecewise continuous function whose total value evaluates to 1.
Let X be a continuous random variable and the probability density function pdf is given by f(x) = x – 1 , 0 < x ≤ 5. We have to find P (1 < x ≤ 2).
Let Y be a continuous random variable and F(y) be the cumulative distribution function (CDF) of Y. Then, the probability density function (PDF) f(y) of Y is obtained by differentiating the CDF of Y.
If we want to calculate the probability for X lying between the interval a and b, then we can use the following formula:
P (a ≤ X ≤ b) = F(b) – F(a) = [Tex]\bold<\int\limits^_f(x)dx> [/Tex]
A Probability Density Function (PDF) is a function that describes the likelihood of a continuous random variable taking on a particular value. Unlike discrete random variables, where probabilities are assigned to specific outcomes, continuous random variables can take on any value within a range. Probability Density Function (PDF) tells us
To find the probability from the probability density function we have to follow some steps.
Step 1: First check the PDF is valid or not using the necessary conditions.
Step 2: If the PDF is valid, use the formula and write the required probability and limits.
Step 3: Divide the integration according to the given PDF.
Step 4: Solve all integrations.
Step 5: The resultant value gives the required probability.
If X is continuous random variable and f(x) be the probability density function. The probability for the random variable is given by area under the pdf curve. The graph of PDF looks like bell curve, with the probability of X given by area below the curve. The following graph gives the probability for X lying between interval a and b.
Let f(x) be the probability density function for continuous random variable x. Following are some probability density function properties:
f(x) ≥ 0, ∀ x ∈ R
P(X = a) = P (a ≤ X ≤ a) = [Tex]\bold<\int\limits^_f(x)dx> [/Tex] = 0
Mean of the probability density function refers to the average value of the random variable. The mean is also called as expected value or expectation. It is denoted by μ or E[X] where, X is random variable.
Mean of the probability density function f(x) for the continuous random variable X is given by:
Median is the value which divides the probability density function graph into two equal halves. If x = M is the median then, area under curve from -∞ to M and area under curve from M to ∞ are equal which gives the median value = 1/2.
Median of the probability density function f(x) is given by:
Variance of probability density function refers to the squared deviation from the mean of a random variable. It is denoted by Var(X) where, X is random variable.
Variance of the probability density function f(x) for continuous random variable X is given by:
Var(X) = E [(X – μ) 2 ] = [Tex]\bold<\int\limits^<\infin>_<-\infin>(x-\mu)^2f(x)dx> [/Tex]
Standard Deviation is the square root of the variance. It is denoted by σ and is given by:
The key differences between Probability Density Function (PDF) and Cumulative Distribution Function (CDF) are listed in the following table:
| Aspect | Probability Density Function (PDF) | Cumulative Distribution Function (CDF) |
|---|---|---|
| Definition | The PDF gives the probability that a random variable takes on a specific value within a certain range. | The CDF gives the probability that a random variable is less than or equal to a specific value. |
| Range of Values | Defined for continuous random variables. | Defined for both continuous and discrete random variables. |
| Mathematical Expression | f(x), where f(x)≥0 and ∫ −∞ ∞ f(x)dx=1 | F(x), where 0≤F(x)≤1 for all x, and F(−∞)=0 and F(∞)=1 |
| Interpretation | Represents the likelihood of the random variable taking on a specific value. | Represents the probability that the random variable is less than or equal to a specific value. |
| Area Under the Curve | The area under the PDF curve over a certain interval gives the probability that the random variable falls within that interval. | The value of the CDF at a specific point gives the probability that the random variable is less than or equal to that point. |
| Relationship with CDF | The PDF can be obtained by differentiating the CDF with respect to the random variable. | The CDF can be obtained by integrating the PDF with respect to the random variable. |
| Probability Calculation | The probability of a random variable falling within a specific interval (a,b) is given by ∫abf(x)dx. | The probability of a random variable being less than or equal to a specific value x is given by F(x). |
| Properties | The PDF is always non-negative: f(x)≥0 for all x. The total area under the PDF curve is equal to 1. | The CDF is a monotonically increasing function: F(x 1 ) ≤ F(x 2 ) if x 1 ≤ x 2 . 0≤F(x)≤1 for all x. |
| Examples | Normal Distribution PDF: |
There are different types of probability density functions given below:
The uniform distribution is the distribution whose probability for equally likely events lies between a specified range. It is also called as rectangular distribution. The distribution is written as U(a, b) where, a is the minimum value and b is the maximum value.
If x is the variable which lies between a and b, then formula of PDF of uniform distribution is given by:
The binomial distribution is the distribution which has two parameters: n and p where, n is the total number of trials and p is the probability of success.
Let x be the variable, n is the total number of outcomes, p is the probability of success and q be the probability of failure, then probability density function for binomial distribution is given by:
P(x) = n Cx p x q n-x
The normal distribution is distribution that is symmetric about its mean. It is also called as Gaussian distribution. It is denoted as N ( [Tex]\bar[/Tex] , σ 2 ) where, [Tex]\bar[/Tex] is the mean and σ 2 is the variance. The graph of the normal distribution is bell like graph.
If x be the variable, [Tex]\bar[/Tex] is the mean, σ 2 is the variance and σ be the standard deviation, then formula for the PDF of Gaussian or normal distribution is given by:
In standard normal distribution mean = 0 and standard deviation = 1. So, the formula for the probability density function of the standard normal form is given by:
Chi-Squared distribution is the distribution defined as the sum of squares of k independent standard normal form. IT is denoted as X 2 (k).
The probability density function for Chi-squared distribution formula is given by:
The joint probability density function is the density function that is defined for the probability distribution for two or more random variables. It is denoted as f(x, y) = Probability [(X = x) and (Y = y)] where x and y are the possible values of random variable X and Y. We can get joint PDF by differentiating joint CDF. The joint PDF must be positive and integrate to 1 over the domain.
The PDF is the function defined for single variable whereas joint PDF is the function defined for two or more than two variables, and other key differences between these both concepts are listed in the following table:
PDF (Probability Density Function)
Probability Density Function is the probability function defined for single variable.
Joint Probability Density Function is the probability function defined for more than one variable.
It is denoted as f(x).
It is denoted as f (x, y, …).
Probability Density Function is obtained by differentiating the CDF.
Joint Probability Density Function is obtained by differentiating the joint CDF
It can be calculated by single integral.
It can be calculated using multiple integrals as there are multiple variables.
Some of the applications of Probability Density function are:
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Example 1: If the probability density function is given as: [Tex]\bold x / 2 & 0\leq x < 4\\ 0 & x\geq4 \end> [/Tex] . Find P (1 ≤ X ≤ 2).
Solution:
Apply the formula and integrate the PDF.
P (1 ≤ X ≤ 2) = [Tex]\int\limits^_f(x)dx [/Tex]
f(x) = x / 2 for 0 ≤ x ≤ 4
⇒ P (1 ≤ X ≤ 2) = [Tex]\int\limits^_(x/2)dx [/Tex]
⇒ P (1 ≤ X ≤ 2) = [Tex]\frac\times\big [\frac \big ]^2_1 [/Tex]
⇒ P (1 ≤ X ≤ 2) = 3 / 4
Example 2: If the probability density function is given as: [Tex]\bold c(x – 1) & 0 < x < 5\\ 0 & x\geq5 \end> [/Tex] . Find c.
Solution:
For PDF:
[Tex]\int\limits^<\infin>_<-\infin>f(x)dx = 1\\ \Rightarrow\int\limits^_<-\infin>f(x)dx\hspace+\hspace \int\limits^_f(x)dx \hspace+\hspace \int\limits^<\infin>_f(x)dx = 1\\ \Rightarrow\int\limits^_<-\infin>0dx\hspace+\hspace \int\limits^_c(x-1)dx \hspace+\hspace \int\limits^<\infin>_0dx1\\ \Rightarrow 0 \hspace+\hspace c \big [\frac- x\big]^5_1 +0 =1\\ \Rightarrow c\big [\frac- x\big]^5_1\\ \Rightarrow 8c = 1\\ \Rightarrow c = \frac [/Tex]
Example 3: If the probability density function is given as: [Tex]\bold \fracx^2 & 0\leq x < 2\\ 0 & otherwise \end> [/Tex] . Find the mean.
Solution:
Example 4: If the probability density function is given as: [Tex]\bold <>x & 0\leq x < 1\\ 0 & otherwise \end> [/Tex] . verify if this is a valid probability density function.
To verify that f(x) is a valid PDF, it must satisfy two conditions:
f(x)≥0 for all x.
The integral of f(x) over its entire range must equal 1.
Checking f(x)≥0:
f(x)=2x is clearly non-negative for 0≤x≤10.
Integrating f(x) over its range:∫−∞∞f(x) dx=∫012x dx=[x2]01=12−02=1. [Tex]\int_<-\infty>^ <\infty>f(x) \, dx = \int_^ 2x \, dx = \left[ x^2 \right]_^ [/Tex]
= 1 2 – 0 2 = 1.
Since both conditions are satisfied, f(x) is a valid PDF.
Example 5: Given the probability density function f(x)= [Tex]\begin 3x^2 & \text 0 \leq x \leq 1 \\ 0 & \text \end[/Tex] , find the mean (expected value) of the distribution.
The mean of a continuous random variable X with PDF f(x) is given by:
E(X)= [Tex]\int_<-\infty>^ <\infty>x f(x) \, dx[/Tex] .
For the given PDF:
E(X)= = [Tex]\int_^ x \cdot 3x^2 \, dx[/Tex]
= [Tex]\int_^ 3x^3 \, dx [/Tex]
= [Tex]3 \left[ \frac \right]_^ [/Tex]
= [Tex]3 \cdot \frac [/Tex]
= [Tex] \frac [/Tex]
Example 6: Using the same PDF [Tex] f(x) = \begin 3x^2 & \text 0 \leq x \leq 1 \\ 0 & \text \end[/Tex] , find the variance of the distribution.
The variance of a continuous random variable X is given by:
[Tex]\text(X) = E(X^2) – [E(X)]^2.[/Tex]
We already have [Tex]E(X) = \frac[/Tex] .
Now, we need to find E(X 2 ):
[Tex]E(X^2) = \int_<-\infty>^ <\infty>x^2 f(x) \, dx[/Tex]
[Tex] = \int_^ x^2 \cdot 3x^2 \, dx [/Tex]
[Tex]= \int_^ 3x^4 \, dx [/Tex]
[Tex]= 3 \left[ \frac \right]_^[/Tex]
[Tex]= 3 \cdot \frac [/Tex]
[Tex]= \frac.[/Tex]
Q 1: Let f(x) be a probability density function given by:
Verify that f(x) is a valid probability density function.
Q 2: Let f(x) be a probability density function given by:
Calculate the probability that X ≤ 1.
Q 3: Let f(x) be a probability density function given by:
Find the cumulative distribution function (CDF) F(x) for x ≥ 0.
Q 4: Given the probability density function f(x) of a continuous random variable X:
Q 5: Find the cumulative distribution function (CDF) for the PDF [Tex]f(x) = \begin 2x & \text 0 \leq x \leq 1 \\ 0 & \text \end.[/Tex]
Q 6: Given the function [Tex]f(x) = \begin k(1-x^2) & \text -1 \leq x \leq 1 \\ 0 & \text \end[/Tex] , find the value of k that makes f(x) a valid PDF.
Q 7: Using the same PDF [Tex]f(x) = \begin \frac e^ & \text x \geq 0 \\ 0 & \text \endotherwise[/Tex] , find the variance Var(X).
Q 8: Given the PDF [Tex]f(x) = \begin 3(1-x)^2 & \text 0 \leq x \leq 1 \\ 0 & \text \end[/Tex] , find the cumulative distribution function (CDF) F(x).
Q 9: For the PDF [Tex]f(x) = \begin \frac(x – x^2) & \text 0 \leq x \leq 1 \\ 0 & \text \end[/Tex] , calculate the probability that X is between 0.2 and 0.8, i.e., [Tex]P(0.2 \leq X \leq 0.8)[/Tex] .
Q 10: For the PDF [Tex]f(x) = \begin \frac e^ & \text x \geq 0 \\ 0 & \text \end[/Tex] , calculate the expected value E(X).
The probability density function is the function that defines the density of the probabilities of a continuous random variable over given range. It is denoted by f(x) where, x is the continuous random variable.
A PDF is used for continuous random variables, where the probability of any single, exact value is zero. A Probability Mass Function (PMF) is used for discrete random variables, where the probability of specific values can be non-zero.
The probability density function formula for the continuous random variable X in interval (a, b):
P (a ≤ X ≤ b) = [Tex]\int\limits^_f(x)dx [/Tex]
The mean of the probability density function can be calculated by following formula:
E[X] = μ = [Tex]\int\limits^<\infin>_<-\infin>xf(x)dx [/Tex]
No, a PDF cannot have negative values. The value of a PDF for a given point in its domain represents the probability density at that point and must be non-negative.
Area under a PDF curve within a certain interval represents the probability that the random variable falls within that interval.
Mean (or expected value) of a distribution using a PDF is found by integrating the product of the variable and its PDF over the entire range of the variable.
Cumulative Distribution Function (CDF) is the integral of the PDF. It represents the probability that the variable takes a value less than or equal to a certain value.
Yes, a PDF can be greater than 1 for a narrow interval because it represents probability density, not probability. The total area under the PDF curve over all possible values must equal 1.
A PDF is normalized by ensuring that the integral of the PDF over all possible values of the random variable equals 1. This normalization condition ensures that the PDF correctly represents a probability distribution.
The variance of a distribution from a PDF is calculated by integrating the square of the difference between the variable and its mean, multiplied by the PDF, over the entire range of the variable.
The CDF, F(x)F, is the integral of the PDF from [Tex]-\infty[/Tex] to x: [Tex]F(x) = \int_<-\infty>^ f(t) \, dt[/Tex]
Conversely, the PDF is the derivative of the CDF: [Tex]f(x)= \fracF(x)[/Tex] .
