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Last updated on October 6, 2025
A continuous random variable is a random variable that can have different values within a specific range. It is used to represent measurements like weight, height, and time. In this article, we will explore the concepts and properties of continuous random variables.
A continuous random variable has an infinite number of possible values within a certain range. In statistics and probability theory, a continuous random variable plays a crucial role where the results are measured instead of being counted. A random variable is considered continuous if it can take any value within a specific interval. It is characterized by its possible values, which are infinite or uncountable. Also, using a function known as the probability density function, we can determine if a probability will fall inside a specific interval.
To understand the probability distributions and conduct statistical analysis, we need to comprehend the various properties of a continuous random variable. Several important properties distinguish the continuous random variable from the discrete random variable. They include:
A probability density function f(x) defines a continuous random variable x. The relative probability for x to be a certain value x is represented by the PDF f(x). The PDF needs to meet two conditions. The first one is f(x) ≥ 0 for all x. It means the probability density function (PDF) must always be non-negative but can be zero for some values. The next condition is that the total probability over all possible values of x must be 1, meaning the sum of all probability values must be 100% or 1. This is written as:
\(\int_{-\infty}^{\infty} f(x)\, dx = 1 \)
The cumulative distribution function (CDF), represented as F(x), shows the likelihood that a continuous random variable X will take a value that is less than or equal to x:
\(F(x) = P(X \leq x) = \int_{-\infty}^{x} f(t)\,dt \)
The CDF is non-decreasing and continuous. The CDF approaches 0 as x moves closer to negative infinity.
lim F(x) = 0
x→∞
Likewise, the CDF approaches 1 as x moves closer to positive infinity.
lim F(x) = 1
x→∞
The moment generating function (MGF) of a continuous random variable X is represented as Mx (t). It is defined as:
\(M_X(t) = E\!\left(e^{tX}\right) = \int_{-\infty}^{\infty} e^{tx} f(x)\,dx \)
The MGF can be used to find all the moments of X, including its mean and variance if it exists.
The characteristic function of a continuous random variable X is expressed as ϕX(t). It is the Fourier transform of the probability density function (PDF).
\(\varphi_X(t) = E\!\left(e^{itX}\right) = \int_{-\infty}^{\infty} e^{itx} f(x)\,dx\)
It determines the distribution of X uniquely.
A continuous random variable X with PDF f(x) has the following expectation or mean:
\(E(X) = \int_{-\infty}^{\infty} x f(x)\,dx \)
The formula represents the expected value of X.
The variance of X measures the average of the squared deviations of the random variable from the mean. It is defined as:
\(\operatorname{Var}(X) = E\!\left[(X - E(X))^{2}\right] = \int_{-\infty}^{\infty} (x - \mu)^{2} f(x)\,dx \)
Here, μ = E (X) is the mean.
Where X's second moment is represented by E(X2) and it is denoted as:
\(E(X^2) = \int_{-\infty}^{\infty} x^2 f(x)\,dx \)
The probabilities associated with a continuous random variable are described by the probability density function (PDF) and cumulative probability function (CDF). Here are some key formulas related to continuous random variables.
PDF of a continuous random variable is a function that shows the likelihood of the variable taking on different values. It does not provide probabilities for specific values but rather for intervals. X is the continuous random variable, and the formula for the PDF, f(x), is:
f(x) = dF(x) / dx = F'(x)
Here, F(x) is the cumulative distribution function.
It describes the likelihood that a specific value, x, will be equal to or less than the random variable, X. CDF of a continuous random variable can be found by integrating the PDF. The formula is calculated between two points, a and b. The formula for CDF of continuous random variable is:
The mean of the continuous random variable, X, is its expected value. It is calculated as a weighted average of all possible values. Also, each value is weighted according to the probability density function (PDF). The formula is:
It measures how much the values of X deviate from the mean. It is the expected value of the squared differences between the variable and its mean. The formula is given as follows:
A random experiment’s numerical outcome is called a random variable. The two types of random variables are discrete random variable and continuous random variable. Let us look at the differences between them:
Continuous Random Variable
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Discrete Random Variable
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Can have any value within a specified range
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Can only have particular and separate values
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The possible values are infinite within a certain range. Example: All real values between 1 and 2
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The possible values are finite or countably infinite. Example: 1, 2, 3, and so on.
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A probability density function (PDF) describes a continuous random variable
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A probability mass function describes a discrete random variable
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The probability of a single value is zero (P (X = x) = 0
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The probability of a single value is non-zero (P (X = x) > 0
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It is represented by a smooth curve
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It is represented by bar graphs
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Examples include time, distance, temperature, height, and weight
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Examples include the number of children, the number of flowers, and the results of a die roll
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We use continuous random variables to model situations that involve measurements. For example, it is used when we want to know the possible amount of rainfall over a year or temperature on any given day. Following are the significant continuous random variables linked to certain probability distributions:
Uniform Random Variable
A uniform random variable represents a uniform distribution, which describes events with equal chances of happening. A uniform random variable’s PDF is as follows:
\(f(x) = \begin{cases} \dfrac{1}{b - a}, & a \le x \le b, \\ 0, & \text{otherwise.} \end{cases}\)
Here, a and b are the lower and upper bounds of the distribution.
Normal Random Variable
A normal random variable is a continuous random variable that is used to model a normal distribution. If a normal distribution’s parameters are expressed as X ~ N (μ, σ2), then the following is the formula for the PDF:
\(f(x) = \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{1}{2}\left(\frac{x - \mu}{\sigma}\right)^2}\)
Here, μ is the mean
σ is the standard deviation
σ2 is the variance
Exponential Random Variable
An exponential distribution is a continuous probability distribution used to model processes in which a specific number of events occur continuously and independently at a constant average rate λ, where λ ≥ 0. The exponential random variable follows an exponential distribution. The following is the PDF of an exponential random variable:
\(f(x) = \begin{cases} \lambda e^{-\lambda x}, & x \ge 0, \\ 0, & x < 0. \end{cases} \)
Where,
\(\lambda \) > 0 is the rate parameter
x \(\ge\ \)0 because the exponential distribution models non-negative values
Within a specified range, a continuous random variable can have an infinite number of values. This concept is widely used in situations where measurements are involved. Here are some real-world applications of continuous random variables.
A continuous random variable is a random variable that can take any value within a specific range. Knowing the key properties and concepts of this variable helps us to solve various mathematical problems, and it improves our problem-solving skills. Here are some of the common mistakes and their helpful solutions to avoid these errors.
A random variable X has the probability density function (PDF): f (x) = k(5−x), 0 ≤ x ≤5. Find the value of k.
0.08
A function must adhere to the following basic rule in order to be considered a valid probability density function (PDF):
\(\int_{-\infty}^{\infty} f(x)\,dx = 1 \)
As we know, the given PDF is defined for 0 ≤ x ≤5, then the integral is taken over this range:
\(\int_{0}^{5} f(x)\,dx = 1 \)
Next, we can substitute f(x) = k (5 − x):
So, now we have to factor out the constant k:
\(k \int_{0}^{5} (5 - x)\,dx = 1 \)
Then, we have to integrate (5 − x) term by term:
∫ (5 − x) dx = ∫ 5 dx − ∫ x dx
Here, we have to use the basic integration rules:
∫ 5 dx = 5x
∫ x dx = x2 / 2
So the antiderivative is:
\(\int (5 - x)\,dx = 5x - \frac{x^2}{2} \)
Next, we evaluate the definite integral from 0 to 5:
\(\begin{align*} \int_{0}^{5} (5 - x)\,dx &= \left[ 5x - \frac{x^2}{2} \right]_0^5 \\ &= \left( 5(5) - \frac{5^2}{2} \right) - \left( 5(0) - \frac{0^2}{2} \right) \\ &= (25 - 12.5) - 0 \\ &= 12.5 \end{align*} \)
Substitute back to solve for k:
So, k (12.5) = 1
k = 1 / 12.5
k = 0.08
Hence, the value of k is 0.08.
For a continuous random variable with PDF: f (x) = 4x^2, 0 ≤ x ≤ 1. Find the value of E (X).
1
Given: f (x) = 4x2, 0 \(\le \) x \(\le \) 1
We have to find E (X) of X. Here, we can use the formula:
\(E(X) = \int_{a}^{b} x f(x)\,dx\)
Where a = 0 and b = 1
Now, we have to substitute f (x)
\(E(X) = \int_{0}^{1} x (4x^2)\,dx = \int_{0}^{1} 4x^3\,dx\)
Integrating,
\(\int 4x^3\,dx = x^4 + C\)
Evaluating the definite integral from 0 to 1,
\(E(X) = \left[ x^4 \right]_0^1 = 1^4 - 0^4 = 1 \)= 1
Find the mean E(X) for the uniform distribution U (2,10).
6
The formula for the mean of a uniform distribution U (a, b) is:
E (X) = (a + b) / 2
From U (2, 10), we have a = 2 and b = 10, so,
E (X) = (2 + 10) / 2
= 12 / 2 = 6
Hence, the mean is 6.
Find the median m such that P (X ≤ m) = 0.6 for the uniform distribution U (0, 5).
m = 3
Here, to find the value of the median, we have to use the formula:
F(m) = (m − a) / (b − a)
From U (0,5), this becomes:
F(m) = (m − 0) / 5 = m / 5
Setting F(m) = 0.6
m / 5 = 0.6
Next, we can solve for m:
m = 0.6 × 5
m = 3
The median is 3.
Find the median m such that P (X ≤ m) = 0.5 for the uniform distribution U (2, 8).
m = 5
F(m) = (m − a) / (b − a)
From U (2,8), this becomes:
F(m) = (m − 2) / (8 − 2) = (m − 2) / 6
Setting F(m) = 0.5:
(m − 2) / 6 = 0.5
Next, we can solve for m:
m − 2 = 0.5 × 6
m − 2 = 3
m = 5
So, the median is 5.
Jaipreet Kour Wazir is a data wizard with over 5 years of expertise in simplifying complex data concepts. From crunching numbers to crafting insightful visualizations, she turns raw data into compelling stories. Her journey from analytics to education ref
: She compares datasets to puzzle games—the more you play with them, the clearer the picture becomes!